Sainavle

Subtraction Of Two Vectors

The general rule for subtraction in algebra can be represented in two ways: 5-3 = 2 and 5 + (-3) = 2. This implies that the result of subtraction of a positive quantity from a number is the same as the result of adding a negative quantity of the same value to the number. This rule is followed in vector subtraction too.

If the difference between two vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{d}\), then \(\vec{a}\) – \(\vec{b}\) = \(\vec{d}\) and also \(\vec{a}\) + (-\(\vec{b}\)) = \(\vec{d}\), where –\(\vec{b}\) is the opposite vector of \(\vec{b}\)

Line segments OA and OB represent the magnitudes and directions of the vectors \(\vec{a}\) and \(\vec{b}\) respectively. BO is extended up to D, so that OD = BO.

This makes \(\overrightarrow{O D}\) and \(\overrightarrow{O B}\) two opposite vectors i.e., \(\overrightarrow{O D}\)  = –\(\overrightarrow{O B}\) = -b .

WBBSE Class 11 Vector Subtraction Notes

In the parallelogram OAED, the diagonal \(\overrightarrow{O E}\) gives the resultant of \(\vec{a}\) and –\(\vec{b}\) which is the difference of \(\vec{a}\) and \(\vec{b}\).

Hence, \(\overrightarrow{O A}+\overrightarrow{O D}=\overrightarrow{O E}\)

or, \(\vec{a}+(-\vec{b})=\vec{d}\)

or, \(\vec{a}-\vec{b}=\vec{d}\)

Read and Learn More: Class 11 Physics Notes

Relation Between The Resultant And The Difference Of Two Vectors: The diagonal \(\overrightarrow{O F}\) of the parallelogram OAFB represents the resultant of two vectors \(\vec{a}\) and \(\vec{a}\).

Now parallelograms OAFB and OAED are congruent. Therefore, it is evident that \(\overrightarrow{O E}\) and \(\overrightarrow{O F}\) can be represented by the diagonals (\(\overrightarrow{B A}\) and \(\overrightarrow{O F}\)) of the same parallelogram.

Hence, we may conclude that, if a diagonal of a parallelogram represents the resultant of two vectors, the other diagonal would represent the difference between them, with proper assignment of directions.

 

Relation Between The Resultant And The Difference Of Two Vectors Special Cases:

1. When the angle between the two vectors \(\vec{a}\) and \(\vec{b}\) is 90° i.e., they are a pair of orthogonal vectors, the parallelograms change into rectangles, for which the diagonals are equal and make equal angles with the base vector \(\vec{a}\).

Hence, the resultant and the difference of these two vectors are equal in magnitude and they are inclined at the same angle as the base vector but in opposite directions.

Analytically, \(\left.\begin{array}{l}
c^2=a^2+b^2 \text { or, } c=\sqrt{a^2+b^2} \\
d^2=a^2+b^2 \text { or, } d=\sqrt{a^2+b^2}
\end{array}\right\}\)….(1)

⇒ \(\left.\begin{array}{r}
\tan \theta=\frac{b}{a} \\
\tan \theta^{\prime}=-\frac{b}{a}
\end{array}\right\}\)…..(2)

∴ θ = θ’

2. For two orthogonal vectors of equal value, i.e., when a = b, the parallelogram changes into a square for which the diagonals are of equal length and are mutu¬ally perpendicular. In this case, the resultant and the difference of the two vectors are of equal magnitude; they are mutually orthogonal as well.

Putting a = b in equation (2), tanθ =1 or, θ = 45° and tanθ’ = -1 or, θ = -45°

Hence, the angle between the resultant and the difference of the two vectors =45° + 45° = 90°.

Applications of Vector Subtraction in Physics

Zero Of Null Vector: A vector having a magnitude zero (0) and no fixed direction is a null vector. If the initial and the terminal points of a line segment representing a vector coincide, then the vector is called a zero or null vector.

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Indispensability Of The Use Of A Null Vector: if a particle starting from a point returns to the same point after some time, its displacement, which is a vector quantity, is a zero vector. Also, two equal and opposite forces acting on a body have a zero resultant, which is a zero vector \(\vec{0}\), i.e., \(\vec{F}-\vec{F}=\overrightarrow{0}\). Hence the difference of two equal vectors is a zero vector.

Again, the acceleration of a particle moving with uniform velocity is zero. This is also another example of a zero or null vector, as acceleration is a vector quantity. Hence, the use of zero or null vectors is indispensable in vector algebra.

The following two operations give rise to a zero vector.

1. When a vector is multiplied by zero, the result is a zero vector. Thus, \(0(\vec{A})=\overrightarrow{0}.\)
2. When the negative of a vector is added to the vector itself, the result is a zero vector, thus \(\vec{A}+(-\vec{A})=\overrightarrow{0}\).

Properties Of Zero Or Null Vector:

1. The addition or subtraction of a zero vector from a vector results in the same vector. Thus \(\vec{A} \pm \overrightarrow{0}=\vec{A}.\)
2. The multiplication of a non-zero real number with a zero vector is again a zero vector. If n is a non-zero real number, then \(n(\overrightarrow{0})=\overrightarrow{0}.\)
3. If n₁ and n₂ are two different non-zero real numbers, then the relation \(n_1 \vec{A}=n_2 \vec{B}\) can hold only if both \(\vec{A}\) and \(\vec{B}\) are zero vector.

Position And Displacement Vectors: In the chapter One-dimensional Motion, we defined the position and the displacement vectors as two important properties related to the motion of a particle. Here, we shall discuss some important points about them vectorically.

Position vector Definition: A vector used to specify the position of a point with reference to the origin of the coordinate system is called a position vector.

Consider a point P having coordinates (x, y, z). If O is the origin then \(\overrightarrow{O P}\) is called the position vector, \(\vec{r}\). The distance between the origin and the point gives the magnitude of the position vector. Here, \(\overrightarrow{O P}=\vec{r}\) (position vector).

Thus, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)]

Hence, magnitude of \(\vec{r}\) is given by \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

Displacement Definition: The vector connecting the initial and final positions of a particle is called the displacement vector.

Suppose a particle is moved from point A(x₁, y₁, z₁) to another point B(x₂, y₂, z₂).

The shortest distance between the points A and B is the displacement. So \(\overrightarrow{A B}\) represents the displacement vector, \(\vec{r}\) as shown.

Graphical Method for Vector Subtraction

Applying triangle law of vectors to the ΔOAB we get, \(\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B} \quad \text { or, } \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{r}_2-\vec{r}_1=\vec{s}\)

The length of the straight line AB gives the magnitude of the displacement vector. The direction of the vector is along the direction of motion of the particle.

With respect to the origin O, \(\vec{r}_1=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\) = initial position vector

and \(\vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) = final position vector

∴ \(\overrightarrow{A B}=\vec{s}=\left(\vec{r}_2-\vec{r}_1\right)=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the displacement vector is independent of the choice of origin of the cartesian coordinate system.

∴ \(|\overrightarrow{A B}|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The points to be noted here are:

1. The displacement of a particle is the vector difference between its final and initial position vectors.
2. The direction of \(\overrightarrow{A B}\) is different from that of \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\). So, in general, the displacement vector is directed neither along the initial nor along the final position vector of a particle. In other words, generally ‘position’ and ‘change of position’ are vectors that are not in the same direction.
3. Velocity is defined as the rate of displacement with time, i. e., the rate of change of position with time. There is nothing like a ‘change of displacement’; hence, the velocity vector always lies along the displacement vector. The average velocity of the particle during the time spent while going from A to B, is directed along \(\overrightarrow{A B}\).

Position And Displacement Vectors Numerical Examples

Mathematical Formulation of Vector Subtraction

Example 1. A car is traveling towards the east at 10 m · s^-1. It takes 10 seconds to change its direction of motion to the north and continues with the same magnitude of velocity. Find the magnitude and direction of the average acceleration of the car.
Solution:

Change in the velocity of the car

= final velocity – initial velocity

= (10 m • s^-1 towards north) – (10 m • s^-1 towards east)

= \(\overrightarrow{A B}-\overrightarrow{O A}=\overrightarrow{A B}+\overrightarrow{A O}=\overrightarrow{A C}\) [from parallelogram law of vector addition]

∴ AC² =AB² + BC² =AB² + AO²

= 10² + 10² = 100 + 100 = 200

∴ AC = 10√2 m · s^-1

Average acceleration = \(\frac{\text { change in velocity }}{\text { time }}\)

= \(\frac{10 \sqrt{2}}{10}=\sqrt{2} \mathrm{~m} \cdot \mathrm{s}^{-2}\)

It is directed along \(\overrightarrow{A C}\) that makes angle 6 with \(\overrightarrow{A B}\), and \(\tan \theta=\frac{B C}{A B}=\frac{A O}{A B}=\frac{10}{10}=1=\tan 45^{\circ}\)

∴ \(\theta=45^{\circ}\)

Hence, average acceleration is directed towards the north-west.

Real-Life Examples of Vector Subtraction

Example 2. A boy runs 210 m along the corridor of his school and turns; right at the end of the corridor runs 180 m to the end of the building and then turns right and runs 30 m.

1. Construct a vector diagram that represents this motion. Indicate your choice of unit vectors,
2. What is the direction and magnitude of the straight line between start and finish?

Solution:

1. We choose, \(\hat{i}\) = unit vector along x-axis, \(\hat{j}\) = unit vector along y-axis,

The consecutive displacements are, \(\overrightarrow{A B}=210 \hat{i} \mathrm{~m}, \overrightarrow{B C}=-180 \hat{j} \mathrm{~m}, \overrightarrow{C D}=-30 \hat{i} \mathrm{~m}\)

2. Resultant, \(\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}=(180 \hat{i}-180 \hat{j}) \mathrm{m}\)

∴ \(\tan \theta=\frac{E D}{A E}=\frac{B C}{A E}=\frac{-180}{180}=-1\)

or, θ = 45°

So \(\overrightarrow{A D}\) is inclined at 45° with the x-axis in the negative direction.

∴ \(|\overrightarrow{A D}|=\sqrt{(180)^2+(-180)^2}=180 \sqrt{2} \mathrm{~m}\)

Short Answer Questions on Vector Subtraction

Example 3. A particle is moving in a circular path with a uniform speed v. Show that, when the particle traverses through an angle of 120°, the change in its velocity is √3v.
Solution:

The initial velocity \(\overrightarrow{v_1}\) and the final velocity \(\overrightarrow{v_2}\) are shown.

They are drawn from the same initial point.

Here, \(\left|\overrightarrow{v_1}\right|=\left|\vec{v}_2\right|=v\)

Change in velocity = \(\vec{v}_2-\vec{v}_1\)

∴ \(\left|\vec{v}_2-\vec{v}_1\right|=\sqrt{v_1^2+v_2^2-2 v_1 v_2 \cos 120^{\circ}}\)

= \(\sqrt{v^2+v^2-2 v \cdot v \cdot\left(-\frac{1}{2}\right)}=\sqrt{3 v^2}=\sqrt{3} \nu\)

Refraction Of Light At Spherical Surface Lens Very Short Questions And Answers

Question 1. Air bubbles are formed in water. Will it behave like a converging or a diverging lens?
Answer: Like advertising lens

Question 2. Under what conditions will the first and the second principal focal lengths of a lens be equal?
Answer: If the same medium exists on both sides of the lens

Question 3. For which type of lens is the optical center situated outside the lens?
Answer: For convexo-concave and concavo-convex lens]

Question 4. Where is -the optical center of a lens situated if only one surface of it is spherical?
Answer: At the intersecting point of the principal axis and the spherical surface

Question 5. The optical center of a lens may lie outside the lens Is the statement true or false?
Answer: True

Question 6. A portion of a lens is broken. Will we get a complete image of an object with such a lens?
Answer: Yes; but the intensity of the image will be lesser

WBBSE Class 12 Refraction of Light VSAQ

Question 7. Can a lens be used in the medium in which it is made?
Answer: No

Question 8. If the refractive index of the material of a lens is equal to that of its surrounding medium what will happen?
Answer: The lens, will behave as a plane glass plate]

Question 9. A convex lens is placed on a plane mirror. 1fa point source of light is placed at the focus of the lens where Will the image of the source of light be formed?
Answer: At the same focus

Question 10. Where does the optical center of a convexo-concave lens
Answer: Outside the lens

Question 11. Which principal focus is effective for the formation of images?
Answer: Secondary

Question 12. When will the image of an object formed by a convex lens be virtual?
Answer: If the object distance is less than

Question 13. If an object is placed at the focus of a convex lens where will its image be formed? .
Answer: At infinity

Question 14. Where should an object be placed concerning a convex lens to get an image of equal size?
Answer: At a distance of 2f from the lens

Question 15. Half of a biconvex lens is covered with black paper. Will a complete image of any object be formed?
Answer: Yes

Question 16. What is the nature of the image formed by a concave lens? Which mirror forms images of the same nature?
Answer: Virtual, Convex

Question 17. A concave lens has a focal length/. Where will the image be formed if an object is kept in the focus?
Answer: Towards the object at a distance

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Question 18. If the distances of a real object and its real image in the case of a convex lens are u and v respectively, what will be the nature of the v~u graph?
Answer: Rectangular hyperbola

Question 19. If the distances of a real object and its real image in the case of a convex lens are u and v respectively, what will be the nature of the graph?
Answer: Straight Line

Question 20. Magnification of a concave lens is always less than 1 is this statement true or false?
Answer: True

Question 21. What should be the minimum distance between a screen and an object such that a convex lens placed in between them can form a real image?
Answer: 4 times the focal length of the lens

Question 22. What type of image does a convex lens form if the object is virtual?
Answer: Real

Question 23. What is the minimum distance between an object and its real image for a convex lens of focal length 15 cm?
Answer: 60 cm

Very Short Answer Questions on Spherical Lenses

Question 24. What should be the position of an object relative to the biconvex lens so that this lens behaves like a magnifying glass?
Answer: Between the optical center and focus of the lens

Question 25. If a biconvex lens of focal length f is cut into two halves along its principal axis, what will be the focal length of each half?
Answer: f

Question 26. A convex lens is placed in contact with a concave lens of greater focal length. How this lens combination will act?
Answer: It will act as a convex lens

Question 27. Two thin convex lenses of equal focal length (f) are kept in contact with each other co-axially. What will be their equivalent focal length?
Answer: \(\frac{f}{2}\)

Question 28. A thin convex lens and a thin concave lens each of focal length f are kept in contact with each other. What will be the equivalent focal length of the combination?
Answer: Infinity

Question 29. What is the relation of the radii of curvature of a lens with its power?
Answer: P = \((\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Question 30. What is the S1 unit of power of a lens?
Answer: Dioptre or m^-1

Question 31. Two lenses of powers 0.5 D and 0.2 D are kept in contact with each other. What is the equivalent power of this combination of lenses?
Answer: 0.7D

Question 32. What is the focal length of a lens of power 2.5 D?
Answer: 40 cm

Common VSAQs on Laws of Refraction

Question 33. A glass lens is immersed in water. How is the power of the lens
Answer: The focal length will increase and the power will decrease

Question 34. If two convex lenses of powers and D are kept in contact with each other, what will be the equivalent power of the lens combination? affected?
Answer: D1+D2

Question 35. When the radii of curvature of the biconvex or biconcave lens increases, how is the power affected?
Answer: Power decreases

Question 36. A uniform biconvex lens is cut into two halves along its principal axis. If the original lens power is 2D, what will be the power of each of the two haves?
Answer: 2D

Question 37. What is the power of the lens of a sunglass?
Answer: Zero

Question 38. Will the focal length and power of a lens increase or decrease when it is immersed in water from air?
Answer: Focal length increases but power decreases

Question 39. A biconvex lens is cut into two halves perpendicular to its principal axis. If the power of the original lens is 4 D, what will be the power of each half?
Answer: 2D

Question 40. If the radii of curvature of the two surfaces of a concavoconvex lens are equal, what will be the focal length and the
Answer: Infinite, zero

Question 41. What will be the focal length and power of a plane glass
Answer: Infinite, zero

Practice VSAQs on Image Formation by Lenses

Question 42. What is the focal length and power of a parallel glass slab?
Answer: Infinity and zero respectively

Question 43. What type of lens is used in the camera?
Answer: Single or a combination of convex lenses

Question 44. What is the nature of the image formed in the camera?
Answer: Real and inverted

Question 45. If the diaphragm of a camera increases by f, how is the image affected?
Answer: Image becomes distinct.

Question 46. Under what condition will a concave lens behave as a convex
Answer: The lens has to be kept in a medium that is optically denser than the material of the lens.

Question 47. What type of lens is an air bubble in water?
Answer: The shape of an air bubble in water is very near to that of a thick biconvex lens. But as the refractive index of water is greater than that of air, the bubble acts as a concave lens.

Conceptual VSAQs on Focal Length and Magnification

Question 48. If half a portion of a biconvex lens is covered by black paper, then state whether the formed image is complete or not.
Answer: A complete image is formed for refraction by the remaining half portion of the lens. But in this case, the brightness of the image is decreased

Question 49. 1n which case the focal length first and second principal focus of a lens will be equal?
Answer: If the surrounding medium of a lens is the same then the focal length of the first and second principal focus of a lens becomes equal

Question 50. Explain what happens when the aperture of a camera lens is reduced by Wring
Answer:

• If the aperture of a lens is reduced the marginal rays are obstructed. Only the axial rays after refraction through the lens form the image.
• So, the image is free from spherical aberration and is distinct.

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Refraction Of Light At Spherical Surface Lens Short Questions And Answers

Question 1. An unsymmetrical thin lens forms the Image of a point object on its own axis. If the lens is inverted, will the position of the image be changed?
Answer:

The position of an image formed by a lens depends on the object’s distance and the focal length of the lens. The first and the second principal focal lengths of the lens are equal. So, if the object distance from the lens is unchanged and if the lens is inverted the position of the image will not be changed?

Question 2. The radii of curvature of the two surfaces of a concavoconvex lens are equal Determine the focal length and power of the lens. Where is this type of lens used
Answer:

If the focal length of a lens is f, we have from lens maker’s formula

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Now, according to the question, r₁ = r₂

Again, power of the lens P = \(\frac{1}{f}=\frac{1}{\infty}\) = 0

This type of lens made of coloured glass is used in single

Question 3. A convex lens is placed in contact with of focal length greater than Its focal length. What will be the nature of the lens combination?
Answer:

Suppose, the focal lengths of the convex lens and the concave lens are f₁ and f₂ respectively.If the focal length of the combination of the lenses is F

∴ \(\frac{1}{F}=+\frac{1}{f_1}+\frac{1}{-f_2}=\frac{1}{f_1}-\frac{1}{f_2}\)

Now, according to the question, f₁ < f₂ and so, \(\frac{1}{f_1}>\frac{1}{f_2}\)

∴ F>0

So, the lens combination will act as a convex lens i.e., a converging lens.

Question 4. A parallel beam of rays is focused at a point by a convex lens. A concave lens of focal length equal to the focal length of the convex lens is placed in contact with it. Find the final position of the image.
Answer:

Let the focal lengths of the convex and concave lens be f and the focal length of the lens combination be F

Therefore \(\frac{1}{F}=\frac{1}{f}+\frac{1}{-f}\) Or, F = ∞

So, the focal length of the lens combination is infinity, i.e., the image will form at infinite distance

WBBSE Class 12 Refraction of Light Short Q&A

Question 5. A convex lens is placed on a plain mirror. If a point light source Is placed on Its focus then where will the Image be formed? Draw a suitable ray diagram.
Answer:

Since the source is at the focus of the lens so, rays emitted from a source and after refraction by the lens become parallel, These parallel rays are incident on the plain mirror normally and retrace the same path, f After that those rays refracted by the lens meet at the focal point. Therefore the image is formed at the point where the source’ is present. The image is a real image

Question 6. Under what circumstances can a concave lens act as a converging lens?
Answer:

If the surrounding medium of the concave lens is denser than that of the material of the lens, the refractive index of the material of the lens relative to the medium becomes less than In this condition the concave lens acts as a converging lens. Under similar conditions, a convex lens acts as a diverging lens

Question 7. If a convex lens of refractive index 1. 5 is immersed in a liquid of refractive index 1.65, it acts as a concave lens. Explain the reason. 
Answer:

We know, that the refractive index for the material of a convex lens is less than the refractive index of the surrounding medium the convex lens acts as a concave lens. Since in this case the refractive index for the material of the lens is less than the refractive index of the surrounding liquid medium, the convex lens acts as a concave lens.

Question 8. The focal length of a uniform convex lens. If the lens is cut into two pieces along its principal axis, what will be the focal length of each half?
Answer:

If the lens is cut into two pieces along its principal axis, the radius of curvature of the surfaces of each half of the lens remains the same. Thus focal length of each half of the lens does not change i.e., their focal length will be f.

Question 9. What is die significance off-number camera lens? 
Answer:

f-number of camera lenses indicates the ratio of the focal length of the lens and the diameter of its aperture. If the f-number of a lens of definite focal length decreases, the diameter of the aperture of the lens increases. So, a large amount of light enters through the lens, and the image becomes indistinct

Short Answer Questions on Lens Properties

Question 10. A camera lens is covered by a black striped glass and a white-coloured donkey is photographed using it. Does the photograph look like a zebra?
Answer:

The photograph does not look like a zebra. If the camera lens Is covered by a black striped glass then its aperture is decreased. As a result relatively small number of light rays take part in the formation of the image. Therefore the brightness of the image is decreased but the character of the image is never changed

Question 11. A spherical mirror has only one focus but the lens has two foci. Explain
Answer:

In the case of a spherical mirror, only one point is found on its axis where if an object is placed then the image is formed at infinite and vice versa. This point is the focus of the spherical mirror. But for lerns two such points are found on its axis on two sides of the lens having one such point. So the spherical mirror has one focus but the lens has two focuses.

Question 12. Double-convex lenses are to be manufactured from a glass of refractive index 1. 55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:

In the case of a double-convex lens, the lens maker’s formula is given by

⇒ \(\frac{1}{f}=(\mu-1) \cdot \frac{2}{r}\)

r = 2f (μ-1 ) 2 × 20 (1.55 -1 )= 22 cm

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Question 13. What is the focal length of the combination of a convex lens of a focal length of 30 cm in contact with a concave lens of a focal length of 20 cm? is the system a converging or a diverging lens? ignore the thickness of the lenses.
Answer:

For the combination of two thin lenses, the focal length is given b

⇒ \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\)

Here, f₁ = 30 cm and f₂ = -20 cm

⇒ \(\frac{1}{f}=(\mu-1) \cdot \frac{2}{r}\)

The focal length of the combination lens is 60 cm.

Since the value off is negative, the combination will act as a concave lens

Question 14. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away employing a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Answer:

For a real image, the least distance between the object and image (D) should be four times the focal length (f) i.e., 4f = D

⇒ \(\frac{D}{4}=\frac{3}{4}\)m

= 0.75 m

Common Short Questions on Refraction Laws

Question 15. A screen is placed 90 cm from an object. The image of the object is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:

Focal length f = \(\frac{D^2-x^2}{4 D}=\frac{90^2-20^2}{4 \times 90}\)

= 21.44 cm

Question 16. A plano-concave lens is made of glass of refractive index1.5 and the radius of curvature of its curved surface is 50cm. What is the power of the lens?
Answer:

The radius of curvature of the plane surface, r→∞, radius of curvature of the concave surface, r₂ = 50cm = 0.5m

Power of the lens

P = \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

= \((1.5-1)\left(0-\frac{1}{0.5}\right)\)

= \(0.5 \times\left(\frac{1}{0.5}\right)\)

= – 1 dpt

Question 17. You are given three lenses L₁, L₂, and L₃ each of focal length 20 cm, An object Is kept at 40 cm in front of L₁ as shown. The final real Image Is formed at the focus I of L₁. Find the separations between L₁, L₂, and L₃.

We consider the magnitudes only of the distance in lens L₁,

u= PB = 40 cm = 2 × 20 cm = 2f

Then, v = 2f = 2 × 20 cm = 40 cm = PB₁

Given, the final image B2 is at the focus of I3; so,

RB₂ = 20 cm.

Then, the rays between L₂ and L₃ must be parallel.

This shows that is the focus of L₂; so, B₁Q = 20 cm.

Therefore, the separation between L₁ and L₂

= PQ = PB₁ + B₁Q = 40 + 20 = 60 cm

Again, the parallel rays between L₂ and L₃ indicate that these two lenses may be kept at any separation between them.

Practice Short Questions on Image Formation by Lenses

Question 18. A convex lens of focal length f₁ is kept in control with a concave lens of focal length f₂. Find the focal length of the. The focal length of a convex lens Is positive, but that of a concave lens is negative.

If F is the focal length of the combination, the

⇒ \(\frac{1}{F}=\frac{1}{4 f_1}+\frac{1}{-f_2}=\frac{1}{f_1}-\frac{1}{f_2}=\frac{f_2-f_1}{f_1 f_2}\)

∴ \(\frac{f_1 f_2}{f_2-f_1}\)

Question 19. A biconvex lens made of a transparent material with a refractive index of 1.5 is immersed in water with a refractive index of 1.33. Will the lens behave as a converging or diverging lens? Give
Answer:

The refractive indices of the material of the biconvex lens and air are 1.5 and 1.0 respectively. As 1.5 > 1, the lens acts as a converging lens in air. When it is immersed in water with a refractive index1.33, we still have 1.5 > 1.33. So the lens still behaves as a converging lens.

Question 20. A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam oflight coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
Answer:

The beam of light coming parallel to the principal axis of a convex lens converges at the focus Fl of the lens. So P is the position of the object P relative to the concave mirror.

Given, OO’ = 50 cm , OP = OF₁ = 20 cm

O’P = 50-20 = 30 cm

Also given, O’ F₂ = 10 cm = focal length of the mirror

Appropriate signs for the minor:

f = O’ F₂ = – 10 cm , u = O’ P = – 30 cm

Then the mirror equation is

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

or, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{10}+\frac{1}{30}\)

⇒ \(\frac{-3+1}{30}=-\frac{1}{15}\)

Or, v = -15 cm

So, the final image is formal at Q, where O’Q = -15 cm

Conceptual Short Questions on Focal Length and Magnification

Question 21. A biconvex lens of a glass of refractive index 1.5 having a focal length of 20 cm is placed in a medium of refractive index 1.65. Find its focal length. What should be the value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass?
Answer:

Focal length of the lens in air /Focal length of the lens in the other medium

⇒  \(\frac{\text { refractive index of glass with respect to medium }-1}{\text { refractive index of glass with respect to air }-1}\)

The required focal length = \(\frac{1.5-1}{\frac{1.5}{1.65}-1} \times 20\)

= – 110 cm

The value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass should be 1.65.

Question 22. A beam oflight converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is

1. A convex lens of focal length 20 cm,
2. A concave lens of focal length 16 cm?

Here, the object is virtual

1. u = + 12 cm , f = + 20 cm

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Or, \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{12}+\frac{1}{20}\)

∴ Or, \(\frac{1}{v}=\frac{8}{60}\)

∴ v = 7.5 cm

Hence, the image formed is real and at a distance of 7.5 cm on the right of the lens

2. Here, u = +12cm, f= -16cm

∴ \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{12}-\frac{1}{16}\)

Or, \(\frac{1}{v}=\frac{1}{48}\)

∴ v = 48 cm

Hence, the image formed is real, and at 48 cm on the right of the lens,

Real-Life Scenarios Involving Refraction Questions

Question 23.

1.  A screen is placed at a distance of 100 cm from a The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used
2. A covering lens is kept coaxially in contact with a diverging lens, both the lenses being of equal focal length. What is the focal length of the combination?

Answer:

The distance between the object and the screen (image) is 100 cm. The first position of the lens is L₁ and the second position is L₂.

Now, for the first position of the lens,

Object distance = u cm

Image distance, v = (100- u) cm

∴ \(\frac{1}{f}=\frac{1}{100-u}-\frac{1}{-u}=\frac{1}{100-u}+\frac{1}{u}\)…………… (1)

Now, for the second position of the lens,

Object distance = (u + 20) cm

Image distance, v = (100-u-20)cm

⇒ \(\frac{1}{f}=\frac{1}{80-u}-\frac{1}{-(u+20)}=\frac{1}{80-u}+\frac{1}{u+20}\)…………… (2)

Therefore, from equations (1) and (2), we get

⇒ \(\frac{100-u}+\frac{1}{u}=\frac{1}{80-u}+\frac{1}{u+20}\)

Or, 200 u = 8000

Or, u = 40 cm

v = 100 -u = 100 – 40 = 60 cm

Therefore, from the lens formula

⇒ \(\frac{1}{100-u}+\frac{1}{u}=\frac{1}{80-u}+\frac{1}{u+20}\)

f = 24 cm

The focal length of the converging lens is f and that of the diverging lens is -f.

If f_e is the equivalent focal length of the combination then

⇒ \(\frac{1}{f_e}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}-\frac{1}{f}\) = 0

∴ f_e = ∞

Examples of Applications of Spherical Lenses

Question 24. A double convex lens is made of a glass of a refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20cm
Answer:

The focal length of a double convex lens with both faces of equal radius of curvature is given by

⇒ \(\frac{1}{f}=(\mu-1) \cdot \frac{2}{r}\)

Given: f = 20 cm, u = .1.55 , r = ?

∴ \(\frac{1}{20}=(1.55-1) \cdot \frac{2}{r}\)

Or, r = 2 × 0.55 × 20 = 22 cm

∴ The required radius of curvature of the convex lens is 22 cm.

Question 25. Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer a satisfactory explanation for this. At home, Mrs. Sing raised the same question to her daughter Anuja who explained why plastic lenses were thicker

1. Write two qualities displayed each by Anuja and her mother
2. How do you explain this fact using lens maker’s formula?

Answer:

1. Anuja has explained it precisely because she has proper knowledge about it and solves her mother’s query. Mrs. Rashmi Singh is surprised and is hunting for suitable information

2. From lens maker’s formula.

⇒ \(\frac{1}{f}=(\mu-1) \cdot\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Where μ is the refractive index for the material of the lens.

Since, the refractive index of glass (ug) > refractive index of plastic (up)

⇒ \(\mu_g-1>\mu_p-1\)

Taking \(\frac{1}{R}=\frac{1}{r_1}-\frac{1}{r_2}\), we get \(\frac{1}{f}=(\mu-1) \cdot \frac{1}{R}\)

Since the focal lengths in the two cases are the same,

⇒ \(\left(\mu_g-1\right) \cdot \frac{1}{R_g}=\left(\mu_p-1\right) \cdot \frac{1}{R_p}\)

Or, \(\frac{R_g}{R_p}=\frac{\mu_g-1}{\mu_p-1}>1\)

Or, R_g> R_p

∴ The thickness of the plastic lens > the thickness of the glass lens.

WBCHSE Class 12 Physics MCQs

Refraction Of Light At Spherical Surface Lens Multiple Choice Questions

Question 1. What sort of lens will an air bubble in water behave like?

1. Biconvex
2. Concavo-convex
3. Biconcave
4. Convexo-concave

Answer: 4. Convexo-concave

Question 2. Observe the behavior of the light rays as shown in the relation of n₁ and n₂ is

1. n₂>n₁
2. n₁ >>n₂
3. n > n
4. n = n

Answer:

Question 3. An object behaves like a convex lens in air and a concave lens in water. The refractive index of the material of the object is

1. Less than air
2. More than both water and air
3. More than air but less than water
4. Almost equal to water

Answer: 3. More than air but less than water

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. The optical centre of a lens is a fixed point whose position is

1. Within the lens
2. Outside the lens
3. On the principal axis of the lens
4. At the focus of the lens

Answer: 3. On the principal axis of the lens

Question 5. A convex lens of focal length 20 cm is placed on a plane mirror. A point object is placed at a distance of 20 cm above the lens along its axis. What will be die final image distance from the lens?

1. 10 cm
2. 1
3. Infinity
4. 20 cm
5. 0

Answer: 3. 20 cm

WBBSE Class 12 Refraction at Spherical Surfaces MCQs

Question 6. If an object is placed at the focus of a concave lens, the image will be formed

1. At infinity
2. At the mid-point of the optical centre and the focus
3. At the optical centre
4. At the focus

Answer: 2. At the mid-point of the optical centre and the focus

Question 7. The focal length of a convex lens is if an object is placed at a distance u from the lens, the condition of formation of an inverted image of equal size as the object is

1. u = 2f
2. u>2f
3. f<u<2f
4. 0 <u<f

Answer: 1. u = 2f

Question 8. The focal length of a convex lens is If an object is placed at a distance u from the lens, the condition of formation of a diminished inverted image is

1. u = 2f
2. u>2f
3. f< u < 2f
4. 0 < u <f

Answer: 2. u>2f

Question 9. The focal length of a convex lens. If an object be placed at a distance u from the lens, the condition of formation of a virtual image is

1. u = f
2. u > 2f
3. f< u < 2f
4. 0 <u<f

Answer: 4. 0 <u<f

Question 10. The focal length of a convex lens.If an object is placed at a distance u from the lens, the condition of formation of an image at infinity is

1. u = f
2. u>2f
3. f< u < 2f
4. 0 <u<f

Answer: 1.  u = f

WBCHSE class 12 physics MCQs 

Question 11. The focal length of a convex lens. If an object be placed at a distance u from the lens, the condition of formation of an inverted magnified image is

1. u=f
2. u > 2f
3. f< u < 2f
4. 0 < u<f

Answer: 3. f< u < 2f

Question 12. The focal length of a convex lens. If an object is placed at a distance u from the lens, the condition of formation of a magnified virtual image is

1. u = f
2. u>2fF
3. f< u< 2f
4. 0<u<f

Answer: 4. 0<u<f

Short Questions on Spherical Lenses

Question 13. The focal length of a concave lens is f. If an object is placed at a distance u from the lens, the condition of formation of a diminished image is

1. u = 0
2. 0 < u < oo
3. u < 0 and |u| <|f|
4. A diminished image will not be formed under any condition

Answer: 2. 0 < u < ∞

Question 14. The focal length of a concave lens is f.  If an object is placed at a distance u from the lens, the condition of formation of a real image is

1. u = 0
2. 0 < u < oo
3. u < 0 and| u| <|f|
4. Real image will not be formed under any condition

Answer: 3. u < 0 and| u| <|f|

Question 15. A point object is placed at the centre of a glass sphere. If the radius of the sphere is 6 cm and the refractive index of the material is 1.5, then the distance of the virtual image from the surface of the sphere will be

1. 2cm
2. 4 cm
3. 6 cm
4. 12 cm

Answer: 3. 6 m

Question 16. An object is placed at a distance of 20 cm from a convex lens of focal length 0 cm. The image distance is

1. 20 cm
2. 6.67 cm
3. 10 cm
4. 30 cm

Answer: 1. 20 cm

Question 17. The size of the image of an object which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance at 26 cm from the convex lens, the real size of the image would be

1. 1.25 cm
2. 2.5 cm
3. 1.05 cm
4. 2 cm

Answer: 2. 2.5 cm

Question 18. A convex lens image of focal of a length object.30 cm produces 5 times the magnified real image of an object, What is the object distance?

1. 36 cm
2. 25cm
3. 30cm
4. 150 cm

Answer: 1. 36 cm

Question 19. If the object distance from the first focus of a convex lens is x and the image distance from the second focus of the lens is x’, the graph between x and x’ will be a

1. Hyperbola
2. Parabola
3. Circle
4. Ellipse

Answer: 1. Hyperbola

Practice MCQs on Types of Lenses

Question 20. The focal length of a lens made of glass in air is 10 cm. What will be the focal length of the lens in water? The refractive index of glass =1.51 and the refractive index of water = 1.33.
Answer:

1. 18.84 cm
2. 36 cm
3. 18 cm
4. 37.7 cm

Answer: 4. 37.7 cm

Question 21. If a lens is surrounded by a medium denser than air, the focal length of the lens

1. Decreases
2. 1ncreases
3. Remains same
4. Cannot be determined

Answer: 2. Increases

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 22. If the focal length of a symmetrical convex lens is equal to the radius of curvature of the lens, the refractive index of the material of the lens is

1. 0.5
2. 1.1
3. -1.5
4. 1.5

Answer: 4. 1.5

Refraction at spherical surfaces class 12 MCQs 

Question 23. The refractive index of the material of a double equi-convex lens is 1.5. If R be its radius of curvature then its focal length is

1. 0
2. 2R
3. R1
4. R

Answer: 4. R

Question 24. What type of lens is used in sunglasses?

1. A concavo-convex lens whose radii of curvature of the two surfaces are equal
2. A biconcave lens whose radii of curvature of the two surfaces are equal
3. A biconcave lens whose radii of curvature of the two surfaces are unequal
4. Plano-concave lens

Answer: 1. A concavo-convex lens whose radii of curvature of the two surfaces are equal

Question 25. An equal-convex lens is divided into two halves along) XOX’ and (ii) YOY’ as shown in the.  Suppose the focal lengths of the complete lens, of each half portion of

Case  1- And of each half portion of the
Case 2  – Respectively in this case the correct statement is

1. f = 2f; f”= f
2. f = 2f; f”= f
3. f = 2f; f”= 2f
4. f = f; f”= f

Answer: 4. f = f; f”= f

Question 26. A convex lens is immersed in a liquid whose refractive index is equal to that of the material of the lens. Under this condition the focal length of the lens
Answer:

1. Will decrease but will not be zero
2. Will remain unchanged
3. Will be zero
4. Will be of infinite value

Answer: 4. Will be of infinite value

Question 27. A convex lens made of glass has a focal length of 0.15 m in the air. if the refractive indices of glass and water are respectively \(\frac{3}{2}\) and \(\frac{4}{3}\), then the focal length of the lens immersed in water will be

1. 0.45 m
2. 0.15 m
3. 0.30 m
4. 0.6 m

Answer: 4. 0.6 m

Question 28. Which of the following is true for rays coming from infinity incident on the lens

1. Two images are formed.
2. A continuous image is formed between the focal points of the upper and lower lens
3. One image is formed
4. None of the above

Answer: 1.  Two images are formed.

Important Definitions in Refraction Q&A

Question 29. A beam of parallel rays after refraction in a convex lens converges at a point.If a concave lens of the same focal length is placed lit contact with the convex lens where will the image be shifted?

1. At infinity
2. At 2f
3. Between 0 and f
4. Between f and 2f

Answer: 1.  At infinity

Question 30. The ratio of powers of a thin convex and a thin concave lens is \(\frac{3}{2}\). When they are to contact, the equivalent focal length is 30 cm, Individual focal lengths are

1. 75 cm – 50 cm
2. 10 cm -15 cm
3. 15 cm -10 cm
4. 50 cm – 75 cm

Answer: 2. 10 cm -15 cm

Question 31. Two thin equal-convex lenses each of focal length 0,2 m are placed coaxially with their optical centres 0.5 in apart. Then the focal length of the combination

1. – 0.4 m
2. 0.4 m
3. -0.1 m
4. 0.1 m

Answer:

Refraction at spherical surfaces class 12 MCQs 

Question 32. Mercury (Hg) is coated on the plane part of a plano-convex lens. The refractive index of the lenses μ and the radius of curvature is R. The system behaves as u concave mirror whose radius of curvature is

1. μR
2. \(\frac{R}{2(\mu-1)}\)
3. \(\frac{R^2}{\mu}\)
4. \(\frac{(\mu+1)}{(\mu-1)} n\)

Answer: 2. \(\frac{R^2}{\mu}\)

Question 33. If the distance between the object and the screen is less than four times the focal length of the lens then,

1. A real image of the object will be formed on the screen for only one particular position of the lens
2. Two virtual images of the object will be formed on the screen for two different positions of the lens
3. No image will be formed on the screen for any position of the lens
4. Two real images of the object will be formed on the screen for two different positions of the lens

Answer: 3

Question 34. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?

1. Virtual, upright, height = 1 cm
2. Virtual, upright, height = 0.5 cm
3. Real, inverted, height = 4 cm
4. Real, inverted, height =1 cm

Answer: 3. Real, inverted, height = 4 cm

Question 35. A convex lens forms a real image of an object for its two different positions on the screen. If the height of the images in two cases are 8 cm and 2 cm, then the height of the object will be.

1. 2cm
2. 4cm
3. 8cm
4. 16cm

Answer: 2. 4cm

Question 36. If x₁ is the size of the magnified image and x₂ is the size of the diminished image in the lens displacement method, then the size of the object is

1. \(\sqrt{x_1 x_2}\)
2. x₁x₂
3. x²₁x₂
4. x₁x²₂

Answer: 1. \(\sqrt{x_1 x_2}\)

Examples of Applications of Lenses in Optics

Question 37. To determine the focal length of a thin convex lens, if red light is used instead of blue light the focal length of the lens

1. Increases
2. Remains same
3. Decreases
4. Cannot be determined

Answer: 1.  Increases

Class 12 physics refraction questions 

Question 38. Two thin lenses of focal lengths f₁, and f₂ are kept in contact co-axially. The power of the combination is given by

1. \(\sqrt{\frac{f_1}{f_2}}\)
2. \(\sqrt{\frac{f_2}{f_1}}\)
3. \(\frac{f_1+f_2}{2}\)
4. \(\frac{f_1+f_2}{f_1 f_2}\)

Answer: 4. \(\frac{f_1+f_2}{f_1 f_2}\)

Question 39. An asymmetric double convex lens is cut in two equal parts by . a plane perpendicular to the principal axis. If the power of the original lens is 4 D, the power of the halved portion will be

1. 2D
2. 3D
3. 4D
4. 5D

Answer: 2D

Question 40. A thin glass (refractive index, (μ =1.5) lens has an optical power of -5 D in air. Its optical power in a liquid medium with a refractive index of 1.6 will be

1. 1D
2. -1D
3. 25D
4. -25D

Answer: 1D

Question 41. A plano-concave lens of glass (μ = 1.5) has a radius of curvature of its curved face of 50 cm. The power of the lens is

1. -1.0 D
2. – 0.2 D
3. 1.0 D
4. 0.2 D

Answer: 1 . -1.0 D

Question 42. A plano-concave lens is made of glass of a refractive index of 1.5 and the radius of curvature of its curved surface is

1. +0.5D
2. – 0.5D
3. – 2D
4. +2D

Answer: 2. – 0.5D

Question 43. The f-number of a lens is f-15. By this statement we mean

1. Focal length of lens = \(\frac{1}{15} \times\) aperture of the lens
2. Aperture of the lens = \(\frac{1}{15} \times\) Focal length of lens
3. Aperture of the lens = \(\frac{1}{15^2}\) Focal length of lens
4. Focal length of lens = \(\frac{1}{15^2}\) aperture of the lens

Answer: 2. Aperture of the lens = \(\frac{1}{15} \times\) Focal length of lens

Question 44. Two thin lenses of focal lengths 20 cm and 25 cm are placed in contact The effective power of the combination is

1. 9D
2. 2D
3. 3D
4. 7D

Answer: 1. 9D

Class 12 physics refraction questions 

Question 45. The radius of curvature of the curried surface of a planoconvex lens is 20 cm. If the tire refractive index of the material of the lens is 1. 5, it will

1. Act as a convex lens only for the objects that lie on its curved side
2. Act as a concave lens for objects that lie on its curved side
3. Act as a convex lens irrespective of the side on which the object lies
4. Act as a concave lens irrespective of the side on which the object lies

Answer: 3. Act as a convex lens irrespective of the side on which the object lies

Question 46. An object approaches a convex lens from the left of the lens with a uniform speed of 5m. s^-1 and stops at the focus. The image

1. Moves away from the lens with a uniform speed 5m. s^-1
2. Moves away from the lens with uniform acceleration
3. Moves away from the lens with a non-uniform acceleration
4. Moves towards the lens with a non-uniform

Answer: 3. Moves away from the lens with a non-uniform acceleration

Question 47. A convex lens Is used to form an image on a screen When the upper half of the Jens is converted by an opaque screen, then

1. Half of the Image will disappear
2. A complete Image will be formed
3. The intensity of the Image will decrease
4. Intensity will remain the same

Answer: 2 And 3

Question 48. A convex lens of the glass of refractive index μ₁ is immersed in a liquid of refractive index μ₂. It will behave as

1. Converging lens if μ₁ > μ₂
2. Converging lens if μ₁< μ₂
3. Diverging lens if μ₁> μ₂
4. Diverging Jens if μ₁< μ₂

Answer: 1 And 4

Question 49. An object and a screen are fixed at a distance d apart. When a lens of focal length f is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are M₁ and M₂

1. d>2f
2. d>4f
3. M₁M₂
4. |M₁|- |M₂| = 1

Answer: 2 And 3

Spherical lens multiple choice questions 

Question 50. A thin, symmetric double-convex lens of power P is cut into thin parts A, B, and C as, the power of

1. A is P
2. A is 2P
3. B is \(\frac{P}{2}\)
4. B is \(\frac{P}{4}\)

Answer: 1 And 3

Question 51. Consider three converging lenses L₁ , L₂ and L₃ having identical geometrical construction. The index of refraction of L₁ and L₂ are μ₁ and μ₂ respectively. The upper half of the lens L₃ has a refractive index of μ₁  and the lower half has μ₁, A point object’ O is imaged at O₁ by the lens L₁ and at O₂ by the lens placed in the same position. If L₃ is placed in the same place

1. There will be an image at O
2. There will be an image at O₂
3. The only image will form somewhere between Oj and
4. The only image will form away from O₂

Answer: 1 And 2

Question 52. Radii of curvature of a thin concavo-convex lens are R and 2R respectively. The refractive index of the material of the lens is μ. When the lens is placed in the air the focal length

1. Depends on the direction of the incident ray from where it comes
2. Will remain the same and does not depend on which direction the incident ray comes
3. Will be \(\frac{R}{\mu-1}\)
4. Will be \(\frac{2 R}{\mu-1}\)

Answer: 2 And 4

Question 53. A luminous object is placed at D = 100 cm away from a screen. A conjugate focus is obtained due to two positions of a convex lens of 21 cm focal length, placed in between the object and screen. The gap between the two positions of the lens (A) and (B) is d. For the positions A and B respectively linear magnifications are m₁ and m₂.

1. d = 40 cm
2. d = 42 cm
3. \(m_1=\frac{7}{3}, m_2=\frac{3}{7}\)
4. \(m_1=\frac{7}{5}, m_2=\frac{5}{7}\)

Answer: 1 And 3

Question 54. The distance between a source of light and a screen is 1 m. By placing a convex lens between them an image is cast on the screen. The lens is shifted through a distance of 40 cm along the line joining the source and the screen and again an image is formed on the screen.

1. The focal length of the lens is

1. 40cm
2. 12cm
3. 21cm
4. 25cm

Answer: 3. 21cm

2. If the length of the images are 0.66 cm and 1.51 cm respectively the length of the object is

1. 0.998 cm
2. 1.3 cm
3. 0.667 cm
4. 0.889 cm

Answer: 2. 1.3 cm

Spherical lens multiple choice questions 

Question 55. A convex lens of power 5 D is placed on a plane mirror. A pin is placed above 30 cm straight from the lens.

1. The distance of the image from the lens at which the image is formed is

1. 30 cm
2. 40 cm
3. 50 cm
4. 60 cm

Answer: 4. 60 cm

2. The distance of the pin from the lens so that its image will coincide with the pin is

1. 20 cm
2. 30 cm
3. 40 cm
4. 50 cm

Answer: 1. 20 cm

Question 56. Two convex lenses of focal lengths 3cm and 4 cm are placed 8 cm apart from each other. An object of height 1 cm is placed at a distance of 4 cm from the lens’s smaller focal length.

1. The position of the final image formed by the two lenses

1. 8 cm away from the right side of the second lens
2. 2 cm away from the left side of the second lens
3. 2 cm away from the right side of the second lens
4. 1 cm away from the left side of the second lens

Answer: 3. 2 cm away from the right side of the second lens

2. The size of the image will be

1. 3 cm
2. 2 cm
3. 2.5 cm
4. 18 cm

Answer: 4. 18 cm

Question 57. A convex lens forms a five-times magnified real image of an object. When the object is further shifted by 6 cm away from the lens the magnification of the real image becomes double

1. The focal length of the lens is

1. 20 cm
2. 15 cm
3. 25 cm
4. 18 cm

Answer: 1. 20 cm

Spherical lens multiple choice questions 

2. The distance of the object from the lens in the first case

1. 18 cm
2. 24 cm
3. 20 cm
4. 32 cm

Answer: 2. 24 cm

Question 58. The refractive index of the material of a double equi convex lens is 1.5. If the radius of curvature of the lens is R, then its focal length is

1. Zero
2. Infinite
3. 2R
4. R

Answer: 4. R

If R is the radius of curvature of one surface of the lens, the radius of curvature of the other surface =-R

Now

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

⇒ \((1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)\)

⇒ \(0.5 \times \frac{2}{R}=\frac{1}{R}\)

f = R

Conceptual MCQs on Image Formation by Lenses

Question 59. A luminous object is separated from a screen by distance d is A convex lens is placed between the object and the object and the screen such that it forms a distinct image on the screen. The maximum possible focal length of this convex lens is.

1. 4d
2. 2d
3. \(\frac{d}{2}\)
4. \(\frac{d}{2}\)

Question 60. An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by 45 to get a sharp image again. The magnitude of the focal length of the concave lens is (in cm).

1. 72
2. 60
3. 36
4. 20

Answer: 4. 20

Here, u₁ = -30cm, f₁ = 10 cm, v₁ = ?

For the first case,

⇒ \(\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\)

⇒ \(\frac{1}{v_1}+\frac{1}{30}=\frac{1}{10}\)

Or, \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\)

⇒ \(\frac{1}{f}=\frac{1}{60}+\frac{1}{30}=\frac{1}{20}\)

Where , u₂= -30 cm, v₂= 15 + 45= 60 cm

∴ \(\frac{1}{f_2}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\)

f₂ = 20 cm

WBCHSE physics refraction at spherical surfaces MCQs 

Question 61. An object is located 4 m from the first of two thin converging lenses of focal lengths 2 m and lm respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of

1. 8.0 m
2. 7.5 m
3. 6.0 m
4. 6.5 m

Answer: 2. 7.5 m

For the first convex lens, the object is placed at a distance of 2/(2 x 2 m = 4 m). The image for the first lens would be formed at a distance of 4 m in the opposite direction.

But the second lens is at a distance of 3 m, consequently, the image of the first lens would be at a distance of m behind the second lens.

Hence, the effective object distance for the second lens is u = +1m.

The focal length of the second lens, f= + 1 m

If the image distance is v,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}\)

Or, v = \(\frac{u f}{u+f}=\frac{1 \times 1}{1+1}\)

= 0. 5 m

∴ Distance of the final image from the source

= 4 + 3 + 0.5 = 7.5 m

Question 62. A point object is held above a thin biconvex lens at its focus. The focal length is 0.1m and the lens rests on a horizontal thin plane mirror. The final image will be formed at

1. Infinite distance above the lens
2. 0.1m above the centre of the lens
3. Infinite distance below the lens
4. 0. 1 m below the centre of the lens

Answer: 2. 0.1m above the centre of the lens

Here a point object is placed at focus (F) of the convex lens. Thus, a beam of diverging rays emerges from it and becomes parallel after refraction from the lens and is incidentally on the plain mirror. As a result, reflected travels the same path and again refraction from the lens meets at the focus of the lens.

WBCHSE physics refraction at spherical surfaces MCQs 

Question 63. Two identical biconvex lenses, each of focal length/ are placed side by side by side in contact with each other with a layer of water in between them as shown in the figure. If the refractive index of the material of the lenses is greater than that of water, howthecombinedfocallength F related to f

1. F>f
2. \(\frac{f}{2}<F<f\)
3. \(F<\frac{f}{2}\)
4. F = f

Answer: 2. \(\frac{f}{2}<F<f\)

If the focal length of the lens formed by water be f_w then,

⇒ \(\frac{1}{f_w}=\left(\mu_w-1\right)\left(-\frac{1}{R}-\frac{1}{R}\right)\)

⇒  \(\left(\mu_w-1\right)\left(-\frac{2}{R}\right)\)

If the focal length of each lens is f then,

⇒  \(\frac{1}{f_l}=\left(\mu_l-1\right)\left\{\frac{1}{R}-\left(-\frac{1}{R}\right)\right\}=\left(\mu_l-1\right) \frac{2}{R}\) …………… (1)

⇒  \(\frac{1}{f_{\mathrm{eq}}}=\frac{1}{f_w}+\frac{1}{f_l}+\frac{1}{f_l}\)

= \(\left(\mu_w-1\right)\left(-\frac{2}{R}\right)+\frac{2}{f_l}\)

= \(-\frac{\mu_w-1}{\mu_l-1}\left(\frac{1}{f_l}\right)+\frac{2}{f_l}\) Using equation 1

= \(\frac{1}{f_l}\left\{2-\frac{\mu_w-1}{\mu_l-1}\right\}\) ………….. (2)

Since \(\frac{\mu_w-1}{\mu_l-1}<1\)

From eq. (2) we get \(\frac{1}{f_{\text {eq }}}<\frac{2}{f_l}\)

Therefore, \(\frac{1}{f_l}<\frac{1}{f_{\text {eq }}}<\frac{2}{f_l}\)

Or, \(f_l>f_{\text {eq }}>\frac{f_l}{2}\)

Given f_l = f and f_eq = F]

Or, \(\frac{f}{2}<F<f\)

Real-Life Scenarios Involving Refraction Questions

Question 64. A diverging lens with a magnitude of focal length 25cm is I placed at a distance of 15cm from a converging lens of a magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final image formed is 1 =

1. Real and at a distance of40cm from the convergent lens
2. Virtual and at a distance of40cm from the convergent lens
3. Reak and at a distance of 40cm from the convergent lens
4. Real and at a distance of cm from the convergent lens

Answer: 1. Real and at a distance of40cm from the convergent lens

For converging lens, u = -(15 + 25) = -40cm

Since u = 2f₂[f₂ = 20 cm ]

Therefore, v = u = 40crn

So, the final Image Is real and Is formed at a distance of 40cm from the converging lens,

Question 65. Two Identical thin plano-convex glass lenses (refractive index 1.5) each having a radius of curvature of 20 cm are placed with their convex surfaces In contact at the center. The Interveningspace Is filled with oil of refractive Index1,7. The focal length of the combination is

1. – 20 cm
2. – 25 cm
3. – 50 cm
4. 50 cm

Answer: 3. – 50 cm

We know \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

For the lens on the left,

⇒ \(\frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{-20}\right)=\frac{1}{40}\)

For the lens on the right,

⇒ \(\frac{1}{f_2}=(1.5-1)\left(\frac{1}{+20}-\frac{1}{\infty}\right)=\frac{1}{40}\)

For the oil in the Intervening space

⇒ \(\frac{1}{f_3}=(1.7-1)\left(\frac{1}{-20}-\frac{1}{20}\right)=-\frac{7}{100}\)

Hence, if the focal length of the combination is F, then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{40}+\frac{1}{40}-\frac{7}{100}=-\frac{1}{50}\)

I. e F = – 50 cm

WBCHSE physics refraction at spherical surfaces MCQs 

Refraction Of Light At Spherical Surface Lens Synopsis

1. Usually the transparent refracting medium is bounded by two spherical surfaces or a spherical surface and a plane surface. is. called a lens

• If the refractive index for the material of a lens is greater than the refractive index of its surrounding medium, then a convex lens behaves as a converging lens and a concave lens behaves as a diverging lens
• If the refractive index for the material of a lens is less than the refractive index of its surrounding medium, then a convex lens behaves as a diverging lens and a concave lens behaves as a converging lens

2. If the thickness of a lens is much less than the radii of curvature of two surfaces then that lens is called a thin lens

3. The point situated on the principal axis of the lens and within the lens through which rays oflight emerge without deviation after refraction, is called the optical centre of the
lens

4. A lens has two foci:

1. The first principal focus,
2. Second principal focus. Usually, the principal focus of a lens means the second principal focus.

5. The distance between the principal focus and the optical centre of a lens is called the focal length (f) of the lens

6. The ratio of the length of the image to that of the object is called linear magnification of the image.

7. The ratio of the area of the image to that of the two-dimensional object is called areal magnification of the image

8. The ratio of the length of the image to that of the object along the principal axis is called longitudinal or axial magnification.

9. A convex lens always forms a real image for a virtual object and that image always lies within the focus.

10. If the object distance for a virtual object in front of a concave lens be less than its focal length then the image is always real.

11. A pair of points is such that if an object is placed at one of them, its image is formed at the other by a fixed lens, then that pair of points are called conjugate foci of that lens

12. If the surrounding medium of a lens is denser than air then the focal length of a lens increases.

13. An image of an object is formed by a combination of a number of co-axial lenses. Now, keeping the position of the object and the image unaltered a single lens is used instead of a combination.

14. If the magnification of the image thus produced by the lens is equal to that in the case of the combination of lenses then this single lens is called the equivalent lens of the combination of lenses.

15. The degree of convergence or divergence of parallel rays of light incident on a lens gives the measure of the power of the lens.

16. The general form of refraction of light at the common spherical surface, when light rays moving from a medium of refractive index fix are refracted in the medium of refractive index μ₂, is

⇒ \(\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\)

Where, u = object distance; V = image distance and R = radius of curvature of spherical surface

17. General equation of a lens \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

This is also called the conjugate foci relation

18. Linear magnification

m = \(\frac{\text { length or height of the image }}{\text { length or height of the object }}\)

= \(\frac{v}{u}=\frac{f-v}{f}=\frac{f}{u+f}\)

19. Areal magnification

m’ = \(\frac{\text { area of the image }\left(A^{\prime}\right)}{\text { area of the object }(A)}=m^2\)

20. Longitudinal or axial magnification

m” = \(\frac{\text { length of the image along the principal axis }}{\text { length of the object along the principal axis }}\)

= \(\frac{d v}{d u}=-m^2\)

= – m²

21. Newton’s equation, xy = f²

Where x = u- f and y = v- f

22. Lens maker’s formula \(\frac{1}{f}=\left({ }_a \mu_b-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Where, _aμ_b = refractive index of the material of the lens concerning the surrounding medium a, rx = radius of curvature of the first refracting surface of the lens, r₂ radius of curvature of the second refracting surface of the lens.

23. Two thin coaxial lenses have focal lengths f₁ and f₂ respectively, the distance between their optical centres if the equivalent focal length of this combination of lenses is F then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{a}{f_1 f_2}\)

If a = 0, i.e., when the lenses are in contact then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\)

24. A convex lens is placed in between an object and a screen which are kept at a fixed distance D. In this case

If D > 4f, then for two different positions of the lens two real images will be formed on the screen.

If D = 4f, then for a fixed position of the lens a real image will be formed on the screen.

If D < 4f, then no image will be formed on the screen for any position of the lens.

If D > 4f, then f= \(\frac{D^2-x^2}{4 D}\) where x = distance between the two positions of the lens

If D > 4f, then u₁ = v₂ and v₁= u₂, where u₁ and v₁ are the object and image distance respectively for the first position of the lens and u₂ and v₂ are the object and image distance respectively for the second position of the lens.

1f D > 4f, then d = \(\sqrt{d_1 d_2}\), where d- the size of the object, dy and d2 = size of the two images for two different positions of the lens

Power of a lens, P = + \(+\frac{100}{f}\) dioptre {wheref is in cm]

= + \(\frac{1}{f}\) dioptre {wheref is in m]

If two lenses have power P₁ and P₂ then when these two lenses are kept in contact with each other, the power of their combination, P = P₁+P₂

WBCHSE Class 12 Physics Semiconductor Electronics Long Questions And Answers

Question 1. When a semiconductor is irradiated by light of a definite wavelength, its resistance decreases. Explain why.
Answer:

• In semiconductors, the greater the number produced, the greater its electrical conductivity and hence lesser its resistance.
• The energy of photons of light of definite wavelength is fixed.
• If the energy of photon incident oh a semiconductor is equal to the difference in the energy level between the conduction band and valence band,
• An electron transits from the valence band to the conduction band by generating a hole in the die valence band.
• Due to an increase in the electrical conductivity of a semiconductor resistance decreases

Question 2. What happens if the amount of reverse bias in a p-n junction diode is gradually increased?
Answer:

• If the reverse bias is increased gradually, the majority of carrier electrons of the TL -region and the majority of carrier holes of the p -region move away from the junction of the diode gradually.
• As a result, the thickness of the depletion layer on the two sides of the junction goes on increasing.
• Hence, although the applied potential difference is increased, the current being only due to minority carriers, is negligible.
• If the reverse bias voltage across the z.p-n junction diode is very high, the minority charge carriers get accelerated.
• Due to their high speed, the car. knock out electrons from the covalent bonds and in turn produce a large reverse current.
• This phenomenon is termed a breakdown

Read And Learn More WBCHSE Class 12 Physics Long Question And Answers

Question 3. Ax absolute zero temperature, the conductivities of the current p-n junction both the insulators and the semiconductors are zero. At any higher temperature, the semiconductor may have some conductivity but for insulator is zero. FT plain why
Answer:

• At absolute zero temperature, there is no existence of charge carriers in semiconductors or insulators.
• So, their electrical conductivities are zero.
• At a temperature higher than absolute zero, some electron-hole pairs are generated due to thermal energy, and they act as charge carriers in semiconductors.
• Some electrons or holes of the valence band, in p-type and p-type semiconductors respectively, can cross the narrow band gap and reach the conduction band.
• Hence, semiconductors possess some electrical conductivities.
• But at that temperature, no charge carriers are generated inside insulators and hence their thermal conductivities remain zero.

Question 4. You are given two bars of the same resistance; one of them is a conductor and the other is a semiconductor. How will you distinguish them experimentally?
Answer:

• At first, both the rods are connected separately with the same source of electricity at room temperature, and currents passing through them are measured.
• After that, the temperature of both the rods is increased.
• Then they are again connected with the same source of electricity and currents through them are measured.
• The bar through which the flow of current increases due to temperature rise is a semiconductor.

Question 5. In a transistor, the base is made very thin. Explain the reason. reverse
Answer:

• The number density of majority carriers in the base region is low as it is lightly doped.
• When the emitter-base junction is forward-biased, the majority of charge carriers crossing the emitter-base junction reach the base and result in electron-hole recombination in the base region.
• Since it is thin and lightly doped, only a small amount of electron-hole recombination takes place.
• The rest of the majority of charge carriers cross the barrier and reach the collector under the influence of the reverse bias and produce an appreciable current in the collector circuit.

Long Answer Questions on Transistors and Their Applications

Question 6. Why is a transistor called a temperature-sensitive device?
Answer:

• A transistor has free electrons and holes as charge tiers.
• When the emitter-base junction is forward-biased and the collector-base junction is reverse-biased, the charge carriers carry current through the transistor in the external circuit.
• When temperature increases, theÿelectrons gain sufficient energy to break the covalent bonds and thus produce a large amount of current
• .If the transistor is continued to operate at this temperature a strong current would flow through it.
• As a result, the transistor would get heated excessively and ultimately break down.
• Thus the operations of the transistor are restricted by its temperature

Question 7. In a transistor, the forward bias is always small compared to the reverse bias. Why?
Answer:

• In a transistor, the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
• If the forward bias is made large, the velocity of the majority of carriers entering the collector region through the base region would be very large.
• In addition to this, the number of these carriers is also very large.
• So, the heat produced would be large enough to break up the covalent bonds resulting in permanent damage of the i transistor.
• But the reverse bias applied to the collector is larger than this forward bias, this would restrict the majority of carriers coming from the emitter from reaching the collector with such high speed and thus saving the device from damage

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Question 8. Explain why the input resistance of a transistor is low while the output resistance is high.
Answer:

• While using a transistor the emitter-base junction is always forward biased and the collector-base junction is biased.
• For this reason, a small change in emitter voltage produces a large emitter current.
• This means a small variation of the input voltage produces a large variation in the emitter rent. This implies that the input resistance of the transistor is small.
• On the other hand, the collector being reverse-biased, collects all the charge carriers that enter through the base.; So a very large change in the collector voltage produces a very small change in the collector current.
• This implies that the output resistance of the transistor is very high

Question 9. The gain of a common-emitter amplifier is given by A_v= ~g_mR_L. Does it mean that if we keep on increasing- R_L indefinitely, the gain of the amplifier will also Increase? Explain your answer
Answer:

No, the gain of the amplifier will not increase indefinitely on increasing R_L.

We know that, V_CE = V_CC– I_CR_L

Where, V_CE → Collector-emitter voltage

V_CC → Output reverse battery voltage,

I_C → Current through output circuit.

Thus, when R_L is increased V_CE decreases and ultimately becomes less than the base voltage when both the junctions get forward biased and the collector current I_C gets saturated. Once Ic is saturated there will be no increase in the gain.

Practice Long Answer Questions on Logic Gates

Question 10. Under normal use of transistors, the emitter is forward-biased while the collector is reverse-biased. Can either of these biases be changed? Explain.
Answer:

1. If the emitter is reversed biased, no majority charge carrier would reach the base. So, if both the emitter and collector are reverse-biased, there would be no current through the transistor.
2. When both the emitter and collector are forward-biased, the majority of carriers would enter the base from both of these. So two separate circuits, i.e., one emitter-base and the other collector-base circuit will be formed and the purpose of the transistor will not be served.
3. If the emitter is reverse biased and the collector is forward biased, then the transistor will perform in the opposite direction, i.e., the emitter will become the collector and the collector will become the emitter. The base current will increase and the collector current

Question 11.

1. Draw the output characteristic curves of a n – p -n  transistor in a CE configuration and find the output resistance from it.
2. Explain, with a circuit diagram, the action of a transistor as a switch.

Answer: 

For an n-p-n transistor in CE configuration, the graphs of output current I_C concerning output voltage V_CE when the base current I_B is kept fixed at different values. These are called output characteristic curves of the n-p-n transistor. If two points A and B are taken on the same graph, the change in collector-emitter voltage = Δ V_CE and the change in collector current = ΔI_c for these two points.

∴ Output Resistance  = \(\frac{\Delta V_{C E}}{\Delta I_C}\)

Question 12. Derive an expression for the voltage gain of the amplifier and hence show that the output voltage is in the opposite phase with the input voltage
Answer:

Let us consider the input voltage V_i = 0.

ApplyingKirchhoff’s lawto the outputloop we get,

V_CC = V_CE+I_CR_L

In theinput loop, V_BB = V_BE+I_BR_B

When V_i isnot zero, we get,

V_BE+ V_i = V_BE+ I_BR_B +ΔI_B( R_B+r_i)

The change in V_BE can be related to the input resistance r_i and the change in I_B. Hence,

V_i = ΔI_B (R_B+ r_i) = RΔI_B

The change in I_B causes a change in Ic. We define a parameter /?

β = \(\frac{\Delta I_C}{\Delta I_B}=\frac{I_C}{I_B}\) …………………..  (2)

Again, the change in the I_C due to a change in I_B causes a change in V_CE and the voltage drop across the resistor RL because V_CC is fixed.

These changes can be given by equation (1) as,

ΔV_CC = ΔV_CE  + R_LΔI_C = 0

Or, ΔV_CE =  -R ΔI_C

The change in V_CE is the output voltage V₀. We get

V0 = ΔV_CE =  – βR_L ΔI_B

The voltage gain of the amplifier is

A_V = \(\frac{V_o}{V_i}=\frac{\Delta V_{C E}}{R \Delta I_B}=-\frac{\beta R_L}{R}\)

The negative sign represents that the output voltage is in the opposite phase to the input voltage.

Conceptual Long Answer Questions on Intrinsic and Extrinsic Semiconductors

Question 13. Draw V – I characteristics of a p- n junction

1. Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
2. Why does the reverse current show a sudden increase at the critical voltage?

Answer:

1. Up to a certain reverse bias potential, the small current p-n junction is due only to the saturated current produced by the minority carriers. It is independent of the applied potential.

2. Either,

1. When the reverse bias potential is fairly high, the high velocity attained by the minority carriers becomes sufficient to break the crystal bonds and produce a large number of electron-hole pairs. For this increase in the number of charge carriers (avalanche breakdown), the reverse current shows a sudden increase.
2. When the dopings, of the p-n junction are sufficiently high, even a comparatively small reverse potential can produce a high electric field across the junction. As a result, a large number of charge carriers become free (Zener breakdown). So, the reverse current shows a sudden increase.

The semiconductor device used in this case is called a Zener diode.

Question 14. Show how these characteristics can be used to determine output resistance.

The collector current (I_C) of a transistor is a function of the base current (JB) and the collector-emitter voltage (V). The output resistance of a transistor is r₀ = \(\frac{\Delta V_{C E}}{\Delta I_C}\)  when IB is kept constant On the linear graph in the. the active region of the output characteristics, two convenient points are chosen by a line representing a fixed I_B. The coordinates of the points, (V_CE, I_C) and (V’_CE, I’_C) Then r₀ is calculated from the relation \(r_o=\frac{v_{C E}^{\prime}-v_{C E}}{I_C^{\prime}-I_C}\)

Question 15. For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1kΩ
Answer:

Voltage amplification factor,

A_v = \(\frac{V_o}{V_i}=\beta \frac{R_C}{R_B}\)

∴ V_i = \(\frac{V_o R_B}{\beta R_C}\)

Now, V₀ = 2 V, R_B = 1 kΩ = 1000 Ω

β = 100, R_C = 2kΩ = 2000 Ω

V = \(\frac{2 \times 1000}{100 \times 2000}\) = 0.01 V

V = \(=\frac{V_l}{R_B}=\frac{0.01}{1000}\) = 10^-5 A = 10 μA

Question 16. The amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 V, calculate the output AC signal.
Answer:

Netvoltage gain, A = A₁ × A₂

∴ A = 10 × 20 = 200

Again A = \(\frac{V_o}{V_i}\)

Question 17. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

The energy of the photon of wavelength λ Is

E = hc/ λ

∴ E = \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9}}\)

= 0.206 eV

The band gap of the semiconductor = 2.8 eV

Since the band gap energy is more than, the energy of the light photon,it cannot be detected

Important Definitions and Explanations in Semiconductor Electronics

Question 18. In an intrinsic semiconductor, the energy gap E_g is 1.2 eV. j Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by n_i = \(n_0 \exp \left(-\frac{E_g}{2 k_B T}\right)\) where n_i is a constant
Answer:

Let K and K be the conductivity of the material at 600  and 300 K respectively.

⇒ \(\frac{K_1}{K_2}=\frac{n_1}{n_2}=\frac{n_0 e^{-\frac{E_g}{2 k_B T_1}}}{n_0 e^{-\frac{E_g}{2 k_B T_2}}}=e^{\frac{E_g}{2 k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}\)

Here, \(\frac{E_g}{2 k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=\frac{1.2}{2 \times 8.6 \times 10^{-5}}\left(\frac{1}{300}-\frac{1}{600}\right)\)

= 11.279

∴ \(\frac{K_1}{K_2}=e^{11.6279}\)

= 1.12 × 10⁵

Question 19. The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m3 of arsenic and 5 × 10²⁰ atoms per m³ of indium. Calculate the number of electrons and holes. Given that n_C

= 1.5 × 10¹⁶m^-3. Is the material n-type or p-type?

n_e = 5 × 10²²  – 5  × 10²⁰

= 4.95 × 10²² m^-3

The number density of hole

n_h = \(\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}\)

= 4.55 ×  10⁹ m^-3

Since n_e>>n_h, the material is of n-type.

Examples of Applications of Semiconductors in Technology

Question 20. For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2kΩ is 2 V. Given the current amplification factor of the transistors 100, find the input signal voltage and base current, if the base resistance 1kΩ.

Given R_C = 2kΩ ,V_CC =  2V , β = 100 , R_β = 1 kΩ

The current amplification factor β= \(\frac{I_C}{I_B}\)

I_C = \(\frac{V_{C C}}{R_C}=\frac{2}{2 \times 10^3}\)

= 1mA

I_B= \(\frac{I_C}{\beta}=\frac{1}{100}\)

= 0.01 mA

= 10 μA

∴  \(\frac{V_o}{V_i}=\frac{\beta R_C}{R_i}\)

Or, \(\frac{2}{V_i}=100 \times \frac{2000}{1000}\)

Or, V_i = 0.01 V

WBCHSE Class 12 Physics MCQs

Semiconductor Electronics Multiple Choice Questions

Question 1. In an n-type silicon, which of Ibe following statements is true?

1. Electrons are majority carriers and trivalent atoms are the dopants
2. Electrons are minority carriers and pentavalent atoms are dopants
3. Holes are minority carriers and pentavalent atoms dopants (oi holes are majority carriers and trivalent atoms are dopants
4. Holes are majority carriers and trivalent atoms are dopants

Answer: 3. Holes are minority carriers and pentavalent atoms dopants (oi holes are majority carriers and trivalent atoms are dopants

Question 2. In an unbiased p-n junction, boles diffuse from the p-region to the n -region because

1. Free electrons In the n -region attract them f v they move across the junction by the potential difference
2. They move across the junction by the potential difference
3. Hole concentration In p -the region is higher as compared To the n-region
4. All the above

Answer: 3. Hole concentration in p – the region is more as compared To n-region

Question 3. Carbon, silicon, and germanium have four valence electrons each.’ These are characterisedi>y valence and conduction bands separated by energy band gap equal to (£g)c, (£g)s. and (£g)Ge> Which of the following statements is true?

1. (E_g)S_i<(E_g)_GeE<(E_g)_C
2. (E_g)_C<(E_g)_Ge<(E_g)S_i
3. (E_g)_C>(E_g)S_i>(E_g)_Ge
4. (E_g)_C= (E_g)S_i= (E_g)_Ge

Answer: 3.(E_g)_C>(E_g)S_i>(E_g)_Ge

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. In V₀ is the potential barrier across a p-n junction, when no battery is connected across the junction

1. 1 and 3 both correspond to the forward bias of the junction
2. 3 corresponds to the forward bias of the junction and 1 corresponds to the reverse bias of the junction
3. 1 corresponds to forward bias and 3 corresponds to reverse bias of junction
4. 3 and 1 both correspond to the reverse bias of the junction

Answer: 2. 3 corresponds to the forward bias of the junction and 1 corresponds to the reverse bias of the junction

WBBSE Class 12 Semiconductor Electronics MCQs

Question 5. Assuming the diodes to be ideal

1. D₁ is forward-biased and D₂ is reverse biased, hence current flows from A to B
2. D₂ is forward biased and D₁ is reverse biased, hence n current flows from B to A and vice versa
3. D₁ and D₂ are both forward-biased and hence current flows from A to B
4. D₁ and D₂, are both reverse biased and hence no current flows from A to B and vice versa

Answer: 2. D₂ is forward biased and D₁ is reverse biased, hence n current flows from B to A and vice versa

Question 6. A 220 V AC supply is connected between points A and B will be the potential difference V across the capacitor

1. 220V
2. 110V
3. 0V
4. 222√2V

Answer: 4. 222√2V

WBCHSE class 12 physics MCQs 

Question 7. In the circuit, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

1. 1.3 V0
2. 2.3 V
3. 0
4. 0.5 V

Answer: 2. 2.3 V

Question 8. The output of the given circuit in

1. Would be zero
2. Would be like a half-wave rectifier with positive cycles in the output
3. Would be like a half-wave rectifier with negative cycles in the output
4. Would be like that of a full-wave rectifier

Answer: 3. Would be like a half-wave rectifier with negative cycles in the output

Short MCQs on p-n Junctions – Brief multiple 

Question 9. Consider an n-p-n transistor with its base-emitter junction forward-biased and collector-base junction reverse-biased. Which of the following statements is true?

1. Electrons cross over from emitter to collector
2. Holes move from base to collector
3. Electrons move from emitter to base
4. Electrons from the emitter move out of the base without going to the collector

Answer: 1 And 3

Question 10. The breakdown in a reverse-biased p-n junction is more likely to occur due to the

1. The large velocity of the minority charge carriers if the doping concentration is small
2. The large velocity of the minority carriers if the doping concentration is large
3. The strong electric field in a depletion region if the doping concentration is small
4. Strong electric field in the depletion region if the doping concentration is large

Answer: 1 And 4

Question 11. In an n-p-n transistor circuit, the collector current is 10 mA. If 95% of the electrons emitted reach the collector, which of the following statements is true?

1. The emitter current will be 8 mA
2. The emitter current will be 10.53 mA
3. The base current will be 0.53 mA
4. The base current will be 2 mA

Answer: 2 And 3

Question 12. If the potential difference is applied, the value of the electric current

1. Becomes infinite for an insulator kept at OK
2. Becomes zero for a semiconductor kept at 0 K
3. Becomes finite for a metal kept at 0 K
4. Becomes infinite in a forward-biased p-n junction diode kept at 300 K

Answer: 3. Becomes finite for a metal kept at 0 K

Semiconductor electronics class 12 MCQs 

Question 13. Two pieces of copper and germanium are cooled from room temperature to 77 K. As a result,

1. Resistance of both the pieces will increase
2. Resistance of both pieces will decrease
3. Resistance of copper will decrease but that of germanium will increase
4. Resistance of copper will increase but that of germanium will decrease

Answer: 3. Resistance of copper will decrease but that of germanium will increase

Question 14. For Intrinsic semiconductors, the energy gap of the forbidden zone Is approximately

1. 0.01 eV
2. 0.1 eV
3. 1 eV
4. 10 eV

Answer: 3. 1 eV

Question 15. In a semiconducting material, the mobilities of electrons and holes are pe and ph respectively. Which of the following is time?

1. μ_e>μ_h
2. μ_e<μ_h
3. μ_e= μ_h
4. μ_e>< 0 , μ_h>0

Answer: 1.μ_e>μ_h

Question 16. In some substances, the charge can flow at ordinary temperatures, but not at very low temperatures. These are called

1. Conductors
2. Insulators
3. Semiconductors
4. Dielectric

Answer: 3. Semiconductors

Question 17. If a small amount of antimony is added to the germanium crystal

1. It becomes a P-type semiconductor
2. The antimony becomes an acceptor atom
3. There will be more free electrons than holes in the semiconductor
4. Its resistance is increased

Answer: 3. There will be more free electrons than holes in the semiconductor

Semiconductor electronics class 12 MCQs 

Question 18. Pure S_i at 500K has an equal number of electron (nf) hole (n_h) concentrations of 1.5 × 10¹⁶.m^-3 Doping by indium increases n_h to 4.5 × 10²².m^-3. The doped semiconductor is of concentration

1. p-type basing electron concentration n_e= 5 × 10⁹.m^-3
2. n-type having electron concentration n_e= 5 × 10²².m^-3
3. p-type having electron concentration n_e= 2. 5 × 10¹⁰.m^-3
4. n-type having electron concentration n_e= 2.5 × 10²³m^-3

Answer: 1. p-type basing electron concentration n_e= 5 × 10⁹.m^-3

Question 19. A rectifier converts

1. Mechanical energy in electrical
2. Liglit energy into electrical
3. ac to dc
4. dc to ac

Answer: 3. ac to dc

Question 20. If a p-n junction diode is not connected in a circuit, then

1. The potential is the same everywhere
2. The potential of p -end is more than n -end
3. An electric field acts from the N-end junction
4. An electric field acts from the p -end to n -end at the junction

Answer: 3. An electric field acts from the N-end junction

Question 21. In the circuit, The forward bias resistances of both the diodes are 50, and reverse bias resistances are infinite. If the battery voltage is 6 V, the current through 100fl resistance in unit A is

1. Zero
2. 0.02
3. 0.03
4. 0.036

Answer: 2. 0.02

Practice MCQs on Logic Gates in Electronics

Question 22. Two identical p-n junctions are connected in series with a battery in three different ways. The potential drop across the two p-n junctions will be equal

1. In the circuits 1 and 2
2. In the circuits 2 and 3
3. In the circuits 3 and 1
4. In the circuit 1 only

Answer: 2. In the circuits 2 and 3

Semiconductor electronics class 12 MCQs 

Question 23. When a p-n diode is reverse-biased, then

1. No current flows 20 V 15V 1kft
2. The depletion region is increased
3. The depletion region is reduced
4. The height of the potential barrier is reduced

Answer: 2. The depletion region is increased

Question 24. The current through the given circuit 

1. \(\frac{3}{10}\) A
2. \(\frac{1}{10}\) A
3. \(\frac{3}{50}\) A
4. \(\frac{3}{10}\) A

Answer: 3. \(\frac{3}{50}\) A

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
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Question 25. The impurity atoms that should be added to germanium to make it n-type of semiconductor is

1. Iodine
2. Indium
3. Arsenic
4. Aluminium

Answer: 3. Arsenic

Question 26. The circuit has two oppositely connected ideal diodes in parallel. What is the current in the circuit

1. 1.33A
2. 1.71A
3. 2.00A
4. 2.31A

Answer: 3. 2.00A

Question 27. When a p-n junction is reverse-biased, then

1. No current flows
2. The depletion layer is increased
3. The depletion layer is decreased
4. The height of potential barriers is reduced

Answer: 3. The depletion layer is increased

Class 12 physics semiconductor questions 

Question 28. A Zener diode having a breakdown voltage equal to 15Y is used in a voltage regulator circuit. The current through the diode is

1. 5 mA
2. 10 mA
3. 15 mA
4. 20 mA

Answer: 5mA

Question 29. If the rate of dropping in the emitter, base, and collector of a transistor is De, Db, and De respectively, then

1. D_e= D_b = D_c
2. D_e < D_b = D_c
3. D_e > D_b > D_c
4. D_e > D_c > D_b

Answer: 4. D_e > D_c > D_b

Question 30. In the active region of a transistor, the biasing at the emitter-base junction and collector-base junction are respectively

1. Forward, Forward
2. Forward, Reverse
3. Reverse, Forward
4. Reverse, Reverse

Answer: 2. Forward, Reverse

Important Definitions Related to Semiconductor Electronics MCQs

Question 31. A transistor is used as

1. A rectifier
2. An amplifier
3. An oscillator
4. A source of electrons and holes

Answer:  2. An amplifier

Question 32. In an n-p-n transistor circuit. collector current Is 10 mA. If 90% A of electrons from the emitter enter into the collector then

1. Emitter current = 11.11 mA
2. Emitter current
3. = 9mA
4. Base current = 2.1 mA
5. Base current = 0.9 mA

Answer: 1. Emitter current = 11.11 mA

Class 12 physics semiconductor questions 

Question 33. For a transistor, if ΔI_C/ΔI_E = 0.96, then the current gain = β is

1. 6
2. 12
3. 24
4. 48

Answer: 3. 24

Question 34. A transistor is operated in a common-emitter configuration at Vc = 2V such that a change in the base current from 100 A to 300 A produces a change in the collector current from 10 mA to 20 mA. the current gain is

1. 50
2. 75
3. 100
4. 25

Answer: 1. 50

Question 35. If the open-loop gain feedback ratio of a positive feedback oscillator is A are r respectively, then the closed-loop gain A_f is

1. 1
2. <A
3. <A
4. Infinity

Answer:  4. Infinity

Question 36. The depletion region of the p-n junction completely vanishes, if.

1. A suitable forward bias is applied
2. a suitable reverse bias is applied
3. a light of suitable frequency is made incident from outside
4. The Zener effect takes place

Answer:  1,3,4

Class 12 physics semiconductor questions 

Question 37. If a pentavalent element is doped in an intrinsic semiconductor, then

1. a p-type semiconductor is formed
2. The effective energy gap of the forbidden zone decreases
3. The energy states of free electrons in  the pentavalent element come nearer to the conduction band
4. The energy states of free electrons in the pentavalent element come down near the valence band

Answer: 2,3

Question 38. If a p-n junction is reverse-biased, then

1. A potential barrier is created at the junction
2. The effective resistance reaches an infinite value
3. A small current is obtained due to the diffusion of the majority of carrier electrons and holes
4. Despite increasing the reverse bias, the reverse current remains almost unchanged

Answer:  1,2,3

Question 39. In case of a transfer

1. The emitter region is heavily doped
2. In CE configuration, if the base-emitter and the collector-emitter both junctions are forward biased, then the transistor acts in the saturation region
3. To use as an amplifier, it is operated in the active region
4. To use as a switch, it is operated in either cut region or saturation region

Answer: 1,2,3,4

Question 40. In case of a Zener diode,

1. To use as a voltage regulator, it is kept in reverse bias.
2. Power consumption in reverse bias is higher than that of an ordinary semiconductor diode
3. On forward bias, _ it behaves as an ordinary semiconductor diode is
4. If Zeiifer voltage is applied in reverse bias, then a high potential barrier is created at its junction

Answer:  1,2,3

Question 41. In a feedback oscillator

1. Effective amplification of the amplifier used in this oscillator is infinity
2. The output AC signal frequency is only determined by the feedback ratio
3. The output AC signal frequency is controlled by a tank circuit in the feedback device
4. The output AC signal is obtained without any externally applied input

Answer: 1,3,4

Question 42. In an n-type semiconductor, If n and p am dm WHvmtratlon of electrons and holes respectively and n( Is die concentration, of electron-hole pairs In a pure semi doctor, then

1. n = p
2. n>p-
3. n<p
4. np = n²₁

Answer:  1,3,4

Class 12 physics semiconductor questions 

Question 43.  A common-emitter DC circuit of the n-p-n transistor is shown. V_BB and V_CC are applied emf of the external source at the base end and collector end respectively. Generally, if we need to amplify an AC signal, it is to be applied at the base end. Hence base end is called the input end. R_B and I_B are input resistance and input DC respectively.

⇒ In most of the cases, input resistance R_B is induced due to the internal resistance of the transistor. The amplified signal of the transistor is obtained at the collector end. Thus the col¬lector end is called the output end. Here, R_L is external load resistance and I_C is the output DC

The terminal voltage across RL, V₀ = I_CR_L is the output voltage; V_i = I_BR_B is the input voltage. In the case of only DC biasing of the transistor, the current gain iff = voltage gain β =\(\frac{I_C}{I_B}\), power gain =. \(=\frac{P_o}{P_i}\)

In a given circuit, V_CC – 8V, the terminal voltage across load resistance =0.8V, load resistance = 800 Ω, the internal resistance of the transistor = 200 Ω, and β_dc – 250.

 1. The value of base current (in μA)

1. 0.8
2. 2.5
3. 4.0
4. 8.0

Answer:  3. 4.0

 2. The value of collector current (in mA)

1. 1.0
2. 2.5
3. 8.0
4. 10.0

Answer:  1. 1.0

3. The collector-emitter voltage (in volt)

1. 2.5
2. 7.2
3. 8.0
4. 8.8

Answer:  2. 7.2

Semiconductor multiple choice questions 

4. Voltage gain

1. 250
2. 1000
3. 2000
4. 4000

Answer:  2. 1000

5. Power gain

1. 250
2. 2500
3. 25000
4. 250000

Answer: 4. 250000

Real-Life Scenarios Involving Semiconductor Questions

Question 44. To maintain a constant potential difference across a regulative load resistance RL, it is not connected directly with an unregulated voltage source Vf. Hence a reverse biased Zener diode may be connected parallel to the load resistance. If the range of the current through the load resistance in which the value ofthe terminal voltage is well maintained within maximum value VQ and minimum value V, then percentage load regulation

r = \(\frac{V_0-V}{V_0} \times 100\)

Each Zener diode has a voltage rating and a watt rating. According to this, the current through the Zener diode has

a specific value above which the Zener diode will be burnt Hence to bypass this damage, an additional resistance R is to be connected in the circuit

In a given circuit, the maximum voltage of the external unregulated source is 15V, the rating of the Zener diode is 6 V-0.25 W and the maximum and minimum voltages across the load resistance are 5.95 V and 5.90 V respectively.

1. Maximum safe current through the Zener diode is approximately (in mA)

1. 1.5
2. 9.0
3. 24.0
4. 41.67

Answer: 4. 41.67

Semiconductor multiple choice questions 

2. The minimum value of the resistance R is approximately

1. 360
2. 216
3. 166
4. 83

Answer: 2. 216

3. Percentage load regulation

1. 0.84
2. 4.2
3. 5.95
4. 8.4

Answer: 1. 0. 84

Question 45. The impurity atom which when added to germanium makes it an n-type semiconductor

1. Boron
2. Indium
3. Arsenic
4. Aluminium

Answer: 3. Arsenic

Semiconductor multiple choice questions 

Question 46. What will happenifthe amount of reverse biasing a p-n junction diode is gradually increased?

1. The thickness of the depletion region will increase
2. The flow of current due to majority carriers will increase
3. The thickness of the depletion region will decrease
4. The flow of current due to majority carriers will decrease

Answer: 1. Thickness of the depletion region will increase

Question 47. For a transistor if β = 100, then α will be

1. 0.99
2. 1.01
3. 100
4. 0.0

Answer: 1. 0.99

β = \(\frac{\alpha}{1-\alpha}\)

Or, \(\frac{\beta}{1}=\frac{\alpha}{1-\alpha}\)

= \(\frac{\beta}{1+\beta}=\frac{\alpha}{(1-\alpha)+\alpha}\)

∴ α = \(\frac{\beta}{1+\beta}=\frac{100}{1+100}\)

= \(\frac{100}{101}\)

= 0.99

Question 48. The lengths, radii, and specific resistances of two conducting wires are each in the ratio of 1: 3. If the resistance of the thinner wire is 10Ω, then the resistance of the other wire will be

1. 40 Ω
2. 20 Ω
3. 10 Ω
4. 5 Ω

Answer: 3. 10Ω

We know, R = \(\rho \frac{l}{A}=\frac{\rho l}{\pi r^2}\)

∴ \(\frac{R_1}{R_2}=\frac{\rho_1}{\rho_2} \cdot \frac{l_1}{l_2} \cdot\left(\frac{r_2}{r_1}\right)^2\)

= \(=\frac{1}{3} \times \frac{1}{3} \cdot\left(\frac{3}{1}\right)^2\)

= 1

R₂=  R₁ =  10

WBCHSE physics semiconductor MCQs 

Question 49. Innormal transistor operation

1. Emitter-base junction and collector-base junction are both in reverse bias
2. Emitter-base junction is in forward bias and collector base junction is in reverse bias
3. Emitter-base junction and collector-base junction both are in forward bias
4. Emitter-base junction is in reverse bias and the collector-base junction is in forward bias

Answer: 2. Emitter-base junction is in forward bias and the collector-base junction is in reverse bias

In the common-base (CB) mode of a transistor, the emitterbase junction is kept at forward bias keeping the base grounded and the collector-base junction is kept at reverse bias to get the proper circuit action.

Question 50. In a transistor output characteristics commonly used in the common-emitter configuration, the base current I_B, the collector current I_C and the collector-emitter voltage V_CE have values of the following orders of the magnitude in the active region

1. I_B and I_C are both in μA and V_CE in V
2. I_B is in μA, IQ in mA, and V_CE in V
3. I_B is in mA, I_C in μA, and V_CE in mV
4. I_B is in mA, I_C is in mA and V_CE in mV

Answer: 2. I_B is in μA, IQ in mA, and V_CE in V

Question 51. In the circuit shown assume the diode to be ideal. When Vf increases from2Vto 6V, the changes in the current is(In mA)

1. Zero
2. 20
3. 80/3
4. 40

Answer: 2. 20

When V_i. increases from 2V to 3V, the diode remains in reverse bias, so current I = 0

When V_i increases from 3V to 6V, the current increases From 0 to; so,I = \(\frac{(6-3) V}{150 \Omega}\)= 0.02A

= 20mA

Question 52. If the band gap between the valence band and conduction band in a material is 5.0 eV, then the material

1. Semiconductor
2. Good conductors
3. Superconductor
4. Insulator

Answer: 4. Insulator

WBCHSE physics semiconductor MCQs 

Question 53. Assume that each diode. 1.70 has a forward bias resistance of 50Ω and an infinite reverse bias resistance. The current through the resistance 150Ω is

1. 0.66 A
2. 0.05 A
3. Zero
4. 0.04 A

Answer: 4. 0.04 A

The uppermost diode in the circuit is forward-biased. But the diode beneath it is reverse biased, hence would not conduct any current

The current through the closed circuit

= \(\frac{10 \mathrm{~V}}{(50+50+50) \Omega}\)

= \(\frac{10}{250}\)

= 0.04A

Question 54. In the case of a bipolar transistor β = 45. The potential drop T across the collector resistance of 1 kΩ is 5 V. The base current is approximately

1. 22 μA
2. 55 μA
3. 11μA
4. 45 μA

Answer: 3. 11 μA

Here β = 45

∴ \(\frac{I_C}{I_B}\) = 45

Now, I_C ×  1 × 10³ = 5 or, I_C = 5 × 10^-3

∴ \(I_B=\frac{I_C}{45}\)

= \(\frac{5 \times 10^{-3}}{45}\)

= 0.111 × 10^-3

= 111 μA

WBCHSE physics semiconductor MCQs 

Question 55. A zener diode having breakdown voltage 5.6 Vis connected in reverse bias with a battery ofemf10 V and a resistance of 100 fl in series. The current flowing through the more zen is

1. 88 mA
2. 4.4 mA
3. 0.88 mA
4. 44 mA

Answer: 4. 44 mA

Her, I= \(\frac{10-5.6}{100}\)

= 0.044 A

= 44 × 10^-3

= 44 mA

Question 56. When a semiconductor device is connected in series with a battery and a resistance, a current is found to flow in the circuit. If, however, the polarity of the battery is reversed, practically no current flows in the circuit. The device may be

1. A p-type semiconductor
2. A n-type semiconductor
3. An intrinsic semiconductor
4. A p-n junction

Answer: 4. A p-n junction

Question 57. What will be the current flowing through the 6kil resistor in the circuit shown, where the breakdown voltage of the Zener is 6 V?

1. \(\frac{2}{3}\)mA
2. 1 mA
3. 10 mA
4. \(\frac{3}{2}\)mA

Answer: 1. \(\frac{2}{3}\)mA

As the breakdown voltage of Zener is 6v, the potential drop in the 4 kΩ resistor is 6V.

The potential drop in 6 kΩ resistor is 4 V

∴ The current flowing through the 6kΩ resistor

= \(\frac{4}{6 \times 10^3}\) A

= \(\frac{2}{3}\)mA

Question 58. The forward-biased diode connection is

Answer: 1

In the case of forward bias of a p-n junction diode, the potential ofp -side is greater than n -side.

Examples of Applications of Semiconductor Devices

Question 59. Identify the semiconductor devices whose characteristics are given below, in the order (1), (2), (3), (4)

1. Simple diode, zener diode, solar cell, light-dependent resistance
2. Zener diode, simple diode, light-dependent resistance, solar cell
3. Solar cell, light-dependent resistance, zener diode, simple diode
4. Zener diode, solar cell, simple diode, light-dependent resistance

Answer: 1. Simple diode, zener diode, solar cell, light-dependent resistance

Question 60. For a common emitter configuration,if a and /? have their usual meanings, the incorrect relationship between cr and is

1. \(\frac{1}{\alpha}=\frac{1}{\beta}+1\)
2. \(\alpha=\frac{\beta}{1-\beta}\)
3. \(\alpha=\frac{\beta}{1+\beta}\)
4. \(\alpha=\frac{\beta^2}{1+\beta^2}\)

Answer: 3. \(\alpha=\frac{\beta}{1+\beta}\)

Question 61. In a common-emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be

1. 45°
2. 90°
3. 135°
4. 180°

Answer: 4. 180v

Question 62. The reading of the ammeter for a silicon diode in the given circuit is

1. 11.5 mA
2. 13.5 mA
3. 0
4. 15mA

Answer: 1. 11.5 mA

For silicon diode, the stopping potential is 0.7 V

I = \(\frac{V-V_{\text {diode }}}{R}\)

= \(\frac{3-0.7}{200} \times 1000\)

= 11.5 mA

WBCHSE physics semiconductor MCQs 

Question 63. The barrier potential of a p-n junction depends on

1. Type of semiconductor material

2. The amount of doping and

3. Temperature. Which one of the following is correct?

1.  (1) and (2) only
2. (2) only
3. (2) and (3) only
4. (1), (2), and (3)

Answer: 4. (1), (2), and (3)

The barrier potential V of-n junction diode,

• Is proportional to temperature T increases with the Increase of density of the donor atom and acceptor atom In the two parts of the diode.
• Decreases with the increase of number density (nÿ) of the thermal electron and hole of the pure semiconductor. This
• H depends on the material of the semiconductor.

Question 64. The given graph represents V-I characteristics for a semiconductor device. Which ofthe following statements Is correct?

1. It Is a V-I characteristic for solar cells where point A represents open circuit voltage and point B short circuit current
2. It is for a solar cell and points A and B represent open circuit voltage and current respectively
3. It Is for a photodiode and points A and B represent open circuit voltage and current respectively
4. It is for LED and points A and B represent open circuit voltage and short circuit current respectively

Answer: 1. It Is a V-I characteristic for solar cells where point A represents open circuit voltage and point B short circuit current

Question 65. If p-n junction, a square input signal of 10 V is applied, as shown,

Then the output across RL will be

Answer: 4

Only positive voltage can pass through a p-n junction diode

Question 66. Consider the junction diode as ideal. The value of current flowing through AB Is

1. 10^-2 A
2. 10^-1 A
3. 10^-3 A
4. 0

Answer: 1. 10^-3 A

The p-n junction diode is forward-biased

I = \(\frac{V}{R}=\frac{4-(-6)}{1000}\)

= 10^-2 A

Question 67. A n-p-n transistor is connected to a common emitter configuration in a given amplifier. A load resistance of 800 ft is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 ft, the voltage gain and the power gain of the amplifier will respectively be

1. 3.69, 3.84
2. 4,4
3. 4,369
4. 4,3.84

Answer: 4. 4,3.84

Here β =  0.96

R_L = 800

V₀ = 0.8V

R_B = 192Ω

∴ Voltage grain = \(\beta \frac{R_L}{R_n}=0.96 \times \frac{800}{192}\)

= 4

∴ Power gain = \(\beta^2 \frac{R_L}{R_B}=(0.96)^2 \times \frac{800}{192}\)

= 3.84

Question 68. Two sides of a semiconductor germanium crystals A and B are doped with arsenic and indium, respectively. They are connected to a battery. 

The correct graph between current and voltage for the arrangement is

Answer: 1.

Arsenic is a pentavalent sample and indium is a trivalent sample. As side A is doped with arsenic is n-type and as side B is doped with indium it is p-type. Therefore, the circuit is reverse-biased. But the given electric cell is a fixed source. If it is considered as an ideal source then there will be no change in the electromotive force.

Question 69. A common-emitter amplifier circuit is shown in the figure below. For the transistor used in the circuit the current amplification factor, β_dc = 100. Other parameters are mentioned.

We find that

1. V_BE=  . + 18.2 V, V_BC = -3.45 V, and the amplifier is  working
2. V_BE = + 18.5 V, V_BC =  +2.85 V, and the amplifier is not working
3. V_BE =. + 120.7 V, V_BC = + 3.75 V=  and amplifier Is not working
4. V_BE=  . + 21.5 V, V_BC = -2.75 V, and the amplifier is  working

Answer: 3. V_BE =. + 120.7 V, V_BC = + 3.75 V=  and amplifier Is not working

V_CE =  V_CC – I_C R_C =  24- (1.5 × 10^-3) × (4.7 × 10³)

= 16.95 V

V_BE V_CC – I_B R_B =  \(V_{C C}-\frac{I_C}{\beta} R_B\)

= 24 – \(\frac{1.5 \times 10^{-3}}{100} \times\left(220 \times 10^9\right)\)

= 20.7 V

The p – side of the n-p-n transistor i.e the base connected with +20 .7V, means it is kept in forward.

Again, the n-type collector is connected with +16.95 V, which means it is kept in reverse bias.

∴ This transistor is in the active region. So it acts as an amplifier

V_BC = V_BC + V_EC = V_BE – V_CE

20.7-16.95 = 3.75 V

WBCHSE physics semiconductor MCQs 

Question 70. In the circuit shown in the figure, the input voltage V. is 20 V, VB£ = 0 and VCE – 0. The values of IB, Ic, and /? are given by

1. I_B = 20μA, I_C= 5 mA, β = 250
2. I_B = 25μA, I_C = 5 mA, β = 200
3.  I_B = 40 μA, I_C= 5 mA, β = 250
4.  I_B = 40μM, I_C= 5 mA, β = 125

Answer: 4.  I_B = 40μM, Ic = 5 mA, β = 125

Applying KVL in loop ABEKD we get,

500 × l0³ I_B + V_BE-20 = 0

Or, \(=\frac{20}{5 \times 10^5}\)

= 40 μA

Applying KVL in loop FGHKEBC we get

⇒ \(4 \times 10^3 I_C+V_{C E}-20\) = 0

⇒ \(I_C=\frac{20}{4 \times 10^3}\)

= 5 mA

Or, I_C= βI_B

Or, β = \(\frac{I_C}{I_B}=\frac{5 \times 10^{-3}}{40 \times 10^{-6}}\)

= 125

Question 71. Inap-w junction diode, changeintemperaturedue to heating

1. Does not affect the resistance of the p-n junction
2. Affects only forward resistance
3. Affects only reverse resistance
4. Affects the overall V-I characteristics of the p-n junction

Answer: 4. Affects the overall V-I characteristics of the p-n junction

The number density of electrons and holes of a p-n junction diode increases due to heating. So, the characteristics graph of the p-n junction diode will change totally

WBCHSE Class 12 Physics Communication System Short Question And Answers

Question 1. Why is satellite used for TV transmission to far places?
Answer:

The ionosphere is unable to reflect the signals above 40 MHz. TV signals have a frequency range of 100 MHz to 200 MHz. So TV signals are not properly reflected by the ionosphere.

So satellites are used for TV transmission to far places. Also, a communication satellite has an amplifier and transmitter that work with solar energy. The satellite amplifies the weak TV signals sent to it from the ground stations and then transmits them to the different ground stations from where they are relayed

Question 2. Write down different modes by which electromagnetic waves, can propagate from transmitting to receiving antenna. Mention one important use of the microwave.
Answer:

Electromagnetic waves propagate through the atmosphere V from the transmitting antenna to the receiving antenna in the form of ground waves, space waves, and sky waves. One important use of microwaves is communication through artificial satellites.

Read And Learn More WBCHSE Class 12 Physics Short Question And Answers

Practice Short Questions on Signal Types

Question 3. Draw a neat diagram of the amplitude-modulated waveform. Write down the expression’ modulation index’ and show each term in the diagram
Answer:

Modulation index, β =  \(k \frac{v_0}{V_0}\)

k = is constant

Question 4. What is the importance of the modulation index?
Answer:

The degree of modulation or modulation index is an indication of the strength of the message signal, i.e., the greater the modulation index, the stronger and clearer will be the message signal.

Question 5. In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their functions.

Answer:

X is the IF block which converts the carrier frequency to a lower value and Y is the amplifier

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WBBSE Class 12 Communication System Short Q&A

Question 6. Write the function of the following in communication, and distinguish between the ‘sky wave’ and ‘space wave’ modes of the system:
Answer:

1. Transmitter: A device that transmits a suitably designed modulated signal to long distances in the form of electromagnetic wave

2. Modulator: A device that superimposes a data signal on a suitable carrier wave.

Question 7. Write two basic modes of communication.
Answer:

1. Point-to-point communication: This communication takes place over a link between a single transmitter and a receiver. Example—telephonic communication.
2. Broadcast: In Broadcast mode, there are a large number of receivers corresponding to a single transmitter. Example—radio, television

Question 8. Write the function of the following in the communication system:
Answer:

The following in the communication system:

1. Receiver
2. Demodulator

1. Receiver: A device that accepts, electromagnetic waves transmitted from distant stations, through an antenna and then sends it to the demodulator.

2. Demodulator: A device that separates, at the receiving end, a data signal from the modulated signal transmitted from distant stations

Short Answer Questions on Modulation Techniques

Question 9. Write the function of a transducer in communication
Answer:

A transducer is any device that converts one form of energy into another. In a communication system, the transducer in the receiver converts the signal into sound waves

Question 10. Why is the sky wave mode of propagation restricted to frequencies up to 40 MHz
Answer:

The sky wave mode of propagation is ionospheric propagation. In the frequency range from a few MHz up to 30MHz to 40 MHz, long-distance communication can be achieved by ionospheric reflection of radio waves back toward the earth. Hence, the sky wave mode of propagation is restricted to frequencies up to 40 MHz

WBCHSE Class 12 Physics Communication System Long Question And Answers

Question 1. Explain briefly the following terms used in communication systems:
Answer:

The following terms used in communication systems:

1. Transducer
2. Repeater
3. Amplification.

1. Transducer:

An energy transformation device. In communication systems, it is used to convert the electrical energy from the modulator into the energy carried by electromagnetic waves. Also, it produces the reverse transformation at the receiving end. These transducers are essential parts of these transmitting and receiving antennas.

2. Repeater:

Space wave communication is a highly efficient mode of long-distance communication. But space waves travel in straight lines and cannot cover a large distance due to the curvature ofthe earth’s surface. So intermediate repeaters or repeater stations are set up to receive the transmitted signal from one side and to send it to other required directions

3. Amplification:

In communication sterns, the generally feeble data signals need adequate amplification before they, after modulating a suitable carrier wave, are transmitted from an antenna

Read And Learn More WBCHSE Class 12 Physics Long Question And Answers

Question 2. In the block diagram of a simple modulator for obtaining an AM signal,  identify the boxes A and B. j Write their functions.

Answer:

A : Linear amplifier; it amplifies the modulated signal and sends it to the transmitting antenna.

B: Transmitting antenna; it transmits the modulated signal in the form of electromagnetic waves.

Question 3. Name the type of waves that are used for line of sight (LOS) communication. What is the range of their frequencies?
Answer:

A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for
satisfactory communication in LOS mode. (Radius of the earth 6.4 × 10⁶ m)

These waves are space waves. The frequency range is from about 30 MHz to 300 MHz and higher.

The maximum distance between the two towers, earth = 6.4 × 10⁶ m )

= \(\sqrt{2 R}(\sqrt{H}+\sqrt{h})\)

= \(=\sqrt{2 \times\left(6.4 \times 10^6\right)} \times(\sqrt{20}+\sqrt{45})\)

= 40000 m ‘

= 40 km

Question 4. Draw a block diagram of a simple modulator for obtaining an amplitude modulated signal A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal to have a modulation index of 75%?
Answer:

 Block diagram of a simple modulator for obtaining amplitude-modulated signal:

A_c = 12 V

μ = 75

= 0.75%

∴ μ = \(\frac{A_m}{A_c}\)

Or, A_m = μ A_c = 0.75 × 12 V

= 9V

∴ The peak voltage of the modulating signal should be 9 V

WBBSE Class 12 Communication System Long Answer Questions

Question 5. A signal of 5 kHz frequency is amplitude modulated carrier wave off-frequency 2 MHz. What are the frequencies ofthe sidebands produced?
Answer:

Upper sideband frequency:

= 5 kHz + 2 MHz =5 × 10^-3 MHz + 2 MHz

= 0.005 + 2 = 2.005 MHz

Lower sideband frequency:

= 2 MHz- 5 kHz = 2 MHz- 0.005 MHz

= 1.995 MHz

Question 6. Why is the baseband signal not transmitted directly? Give any two reasons.
Answer:

The baseband signal is not transmitted directly, because:

1. Its higher wavelength (low frequency) will require the size of the antenna or aerial used for transmission to be very high.
2. For linear antenna (length l) the power radiated is proportional to Q)2> and hence, the effective power radiated by a long wavelength baseband signal would be small.

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Question 7. State the factors which limit its range of propagation
Answer:

Factors that limit the range of propagation:

• At frequencies higher than 40 MHz, the space waves scatter more easily and the communication is limited to line-of-sight propagation.
• Because of the line-of-sight nature of propagation, direct waves get blocked at some point by the curvature of the Earth

Question 8. Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving a reason, the frequency range used in this mode of propagation.
Answer:

• The basic mode of communication used in satellite communication is point-to-point mode.
• Space wave propagation is used in satellite communication
• In space wave propagation, the range of frequency is between 54 MHz to 4.26 GHz since both the ground wave and sky wave propagation fail at such high frequencies

Question 9.

1. What is the line of sight communication?
2. Why is it not possible to use sky waves for the transmission of TV signals? Up to what distance can a signal be transmitted using an antenna ofheight h ?

Answer:

1. A space wave travels in a straight line from the transmitting antenna to the receiving antenna. This is known as line-of-sight communication.

2. TV signals lie in the frequency range of 100 MHz-220 MHz. Waves in this frequency range are not reflected in the ionosphere and hence cannot be transmitted via sky waves.  A signal can be transmitted using an antenna of height h up to a distance of 2Rh

Question 10. Explain any two factors that justify the need for modulating a low-frequency baseband signal.
Answer:

The two factors that justify the need for modulating a low-frequency baseband signal are:

Power radiated by the antenna is given by \(P \propto \frac{l^2}{\lambda^2}\). Hence, there is a need for higher frequency conversation for effective power transmission by the antenna.

To transmit a signal effectively, the size of the antenna should be at least of the size \(\frac{\lambda}{4}\), where λ is the wavelength of the signal to be transmitted. So if λ is small (or frequency is high) signal can be transmitted with reasonable antenna length.

Detailed Explanations of Communication System Principles

Question 11. The frequencies of two sidebands in an AM wave an 640 kHz and 660 kHz respectively. Find the frequencies of carrier and modulating signal. What, is the bandwidth required for amplitude modulation?
Answer:

Let the frequencies of carrier and modulating signal bi denoted by and f respectively

According to the problem

f_c + f_m = 660

f_c – f_m = 640

∴ f_c =  \(\frac{660+640}{2}\) = \(\frac{1300}{2}\)

= 659 kHz

∴ f_m  =  \(\frac{660-640}{2}\) = \(\frac{20}{2}\)

= 10 kHz

Hence, the bandwidth required for amplitude modulation = 660 – 640 = 20 kHz

Question 12.

1. Give three reasons why modulation of a message signal is necessary for long-distance transmission.
2. Show graphically an audio signal, a carrier wave, and an amplitude-modulated wave.

Answer:

Three reasons behind modulating message signals for long-distance transmission are:

1. The audio / audiovisual signals when converted into electromagnetic waves do not have sufficiently high energy to travel up to long distances, because of their lower frequency. Hence these signals must be modulated with high-frequency carrier waves, before beginning

2. For effective transmission by an antenna, its size should at least be of the order \(\frac{\lambda}{4}\), where λ is the wavelength of the signal to be sent For an electromagnetic wave of frequency 20 kHz, we need an antenna of size \(\frac{\lambda}{4}\)  i.e., 3.75 km high, which is practically impossible. Hence, these low-frequency signals are first converted into high frequencies so that transmission can be obtained with reasonable antenna Lengths

3. The power radiated by an antenna is related to its length l and wavelength A of the signal as \(P \propto \frac{l^2}{\lambda^2}\) The relation shows that for die same antenna length, power (P) radiated increases with decreasing wavelength or increasing frequency. Hence, there is a need for higher frequency conversion for effective power transmission by the antenna.

Question 13. A carrier wave of peak voltage 15 V is used to transmit a message signal Find the peak voltage of the modulating signal to have a modulation index of 60%
Answer:

Modulation index is given by μ = \(\frac{A_m}{A_c}\)

Where A_m is the die amplitude of the modulated wave, A_c is the amplitude of the carrier wave.

Here, A_c = 15 V, μ = 60% = 0.6

As,  μ =  \(\frac{A_m}{A_c}\)

Or, A_m= μ A_c

A_m= 0.6 × 15

= 9V

Question 14. What is an antenna? Find the length of a dipole antenna for
Answer:

The antenna is a device used in communication systems for sending electromagnetic waves in all directions from the transmitter and receiving them at the receiver. The wavelength of carrier wave,

λ = \(\frac{\text { velocity }}{\text { frequency }}\)

= \(\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{3 \times 10^8 \mathrm{~s}^{-1}}\)

= 1m

∴ The length of the dipole antenna should be about 1 m

Practice Long Answer Questions on Modulation Techniques

Question 15.  Due to economic reasons, only did upper side bards of an AM wave are transmitted; but at the receiving station there is a facility for generating the carrier.
Answer:

Let the two signals be represented by \(A_1 \cos \left(\omega_c+\omega_m\right) t\) and \(A_c \cos \omega_c t\)

Multiplying the waves, we get,

⇒ \(A_1 A_c \cos \left(\omega_c+\omega_m\right) t \cdot \cos \omega_c t\)

= \(\frac{A_1 A_c}{2} \cos \omega_m t+\frac{A_1 A_c}{2} \cos \left(2 \omega_c+\omega_m\right) t\)

The separation of the relationship indicates that the modulating signal \(\frac{A_c A_1}{2} \cos \omega_m t\)can be easily recovered at the receiving station

Question 16. Write down the Difference between amplitude modulation (AM) and Frequency(FM)
Answer: 

WBCHSE Class 12 Physics MCQs

Communication System Multiple Choice Questions

Question 1. Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves?

1. 10 kHz
2. 10 MHz
3. 1 GHz
4. 1000 GHz

Answer: 2. 10 MHz

Frequency 10 MHz will suffer from antenna problems. Frequency of 1 GHz and 1000 GHz have so high penetration power that the earth would absorb them

Question 2. Frequencies in the UHF range normally propagate using

1. Ground waves
2. Surface waves
3. Sky waves
4. Space waves

Answer: 4. Space waves

Question 3. Digital signals

1. Do not provide a continuous set of values,
2. Represent values as discrete steps,
3. Can utilize only binary systems, and
4. Can utilize decimal as well as binary systems

Question 4. Which ofthe above statements are true?

1. (1) and (2) only
2. (2) and (3) only
3. (1), (2), and (3) but not (4)
4. All of (1), (2), (3) and (4)

Answer: 3. (1), (2), and (3) but not (4)

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 5. Is it necessary for a transmitting antenna to be at the same Ac 6 height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m all How much service area can it cover if the receiving antenna is at the ground level?
Answer:

Two antennas don’t need to be of height

Required area = \(\pi d^2=\pi(\sqrt{2 \times h \times R})^2\)

= \(\frac{22}{7} \times 2 \times 81 \times 6.4 \times 10^6\)

= 3258 km²

WBBSE Class 12 Communication System MCQs

Question 6. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer:

μ = \(\frac{A_m}{A_c}\)

∴ A_m = μ A_c 

= \(\frac{75}{100} \times 12\)

= 9V

Question 7. For an amplitude-modulated wave, the maximum amplitude is found to be 10 V while minimum amplitude is found to be 2 V. Determine the modulation index H . What would be the value of μ if the minimum amplitude is zero?
Answer:

A_max = A_c+A_m and A_min= A_c– A_m

∴  A_c = \(\frac{A_{\max }+A_{\min }}{2}\) = \(\frac{10+2}{2}\)

= 6V

A_m = \(\frac{A_{\max }-A_{\min }}{2}\) = \(\frac{10-2}{2}\)

= 4V

∴ μ = \(\frac{A_m}{A_c}=\frac{4}{6}\)

= 0.67

If A_min= 0 , then A_m = A_c and μ = 1 .

WBCHSE class 12 physics MCQs

Question 8. Three waves A, B, and C of frequencies 1600 kHz, 5 MHz carrier wave of frequency (oc to get modulated wave (AM ). The frequency of the AM wave will and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?

1. A is transmitted via space waves while B and C are transmitted via sky waves
2. A is transmitted via ground wave, B via sky wave and C via space wave
3. B and C are transmitted via ground wave while A is transmitted via sky wave
4. B is transmitted via ground wave while A and C are transmitted via space wave

Answer: 1. A is transmitted via space waves while B and C are transmitted via sky waves

Question 9. A 100-long antenna is mounted on a 500 m tall building The complex can become a transmitting tower for waves with A

1. 400m
2. 25m
3. 150m
4. 2400m

Answer: 1. 400m

Question 10. A 1 kW signal is transmitted using a communication channel that provides attenuation at the rate of -2dBperkmrIf the communication channel has a total length of 5 km, the power of the signal received is [gain in dB = 10Iog(P₀/P_t)

1. 900 W
2. 100 W
3. 990W
4. 1010W

Answer: 2. 100 W

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Short Answer Questions on Communication Systems

Question 11. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1MHz, using amplitude modulation. The frequencies ofthe sidebands will be

1. 3001 kHZ and 2997 kHz
2. 1.003 MHz and 0.997 MHz
3. 1003 kHz and 1000 kHz
4. 1 MHz and 0.997 MHz

Answer: 2. 1.003 MHz and 0.997 MHz

Question 12. A message signal of frequency com is superposed on a carrier wave of frequency to get an amplitude-modulated wave (AM). The frequency of the AM wave will.

1. ω_m
2. ω_c
3. \(\frac{\omega_c+\omega_m}{2}\)
4. \(\frac{\omega_c-\omega_m}{2}\)

Answer: 3. \(\frac{\omega_c+\omega_m}{2}\)

Question 13. I-V characteristics of four devices are shown. Identify devices that can be used for modulation

1. (1) and (3)
2. Only (3)
3. (2) and some regions of (4)
4. All the devices can be used

Answer: 3.

Question 14. A male voice after modulation transmission sounds like that of a female to the receiver. The problem is due to

1. Poor selection of modulation index (selected 0 < m < 1)
2. Poor bandwidth selection of amplifiers
3. Poor selection of carrier frequency
4. Loss of energy in transmission

Answer: 2. Poor bandwidth selection of amplifiers

WBCHSE class 12 physics MCQs

Question 15. Identify the mathematical expression for the amplitude-modulated wave

1. \(A_c \sin \left[\left\{\omega_c+k_1 v_m(t)\right\} t+\phi\right]\)
2. \(A_c \sin \left\{\omega_c t+\phi+k_2 v_m(t)\right\}\)
3. \(\left\{A_c+k_2 v_m(t)\right\} \sin \left(\omega_c t+\phi\right)\)
4. \(A_c v_m(t) \sin \left(\omega_c t+\phi\right)\)

Answer: 3. \(\left\{A_c+k_2 v_m(t)\right\} \sin \left(\omega_c t+\phi\right)\)

Question 16. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation because

1. The size of the required antenna would be at least 5 km which is inconvenient
2. The audio signal cannot be transmitted through sky waves
3. The size of the required antenna would be at least 20 km which is inconvenient
4. Effective power transmitted would be very low if the size of the antenna is less than 5 km

Answer: 1,2 And 4

Question 17. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?

1. The sideband frequencies are 1506 kHz and 1494 kHz
2. The bandwidth required for amplitude modulation is 6 kHz
3. The bandwidth required for amplitude modulation is 3 MHz
4. The sideband frequencies are 1503 kHz and 1497 kHz

Answer: 2 And 4

Question 18. In amplitude modulation, the modulation index m, is kept less than or equal to 1 because

1. m > 1 will result in interference between frequency and message frequency, distortion
2. m> 1 will result in overlapping of both sidebands resulting in loss of information
3. m> 1 will result in a change in phase between the carrier signal and the message signal
4. m> 1 indicates amplitude of the message signal greater than the amplitude of the carrier signal resulting in distortion

Answer: 2 And 4

Practice Questions on Transmission Media

Question 19. A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of(assume the radius of the earth to be 6.4 × 10⁶ m )

1. 100 km
2. 24 km
3. 55km
4. 50 km

Answer: 2,3 And 4

Question 20. Out of the following, which is an essential element of a communication system?

1. Transistor
2. Transmitter
3. Computer
4. Both 1 and 3

Answer: 3. Computer

Question 21. The distortion in the transmission and processing of signals is called

1. Interference
2. Diffraction
3. Noise
4. Scattering

Answer: 2. Diffraction

Question 22. If the maximum frequency of an audio signal is f, then what would be the bandwidth of an Amplitude Modulated (AM) wave?

1. \(\frac{f}{2}\)
2. f
3. 2f
4. 4f

Answer: 3.2f

Communication system class 12 MCQs 

Question 23. If the frequencies of carrier waves for AM and FM are f_A and f_F respectively, then

1. f_A ≈ f_F
2. f_A < f_p
3. f_A>f_p
4. f_A>f_p

Answer: 2. f_A < f_p

Question 24. If the modulation frequency of an FM wave is f, then the modulation index will be directly proportional to:

1. \(\frac{1}{f}\)
2. f
3. \(\frac{1}{f^2}\)
4. f²

Answer: 1.\(\frac{1}{f}\)

Question 25. In amplitude modulation modulation index of a transmitted carrier wave is 0. The increase in power dissipation would be directly proportional to

1. β
2. β²
3. 1+β²
4. 1+β²/2

Answer: 2. β²

Question 26. The rate of energy dissipation in a carrier wave transmission is 10 kW. What would be the rate of energy dissipated if the wave is frequency modulated to a 10% level?

1. 10 kW
2. 10.05 kW
3. 10.1 kW
4. 10.5 kW

Answer: 1. 10 kW

Important Definitions in Communication Systems

Question 27. In an amplitude-modulated wave, the circular frequency of the carrier wave is ft and that of the data signal lies in the range between (ω’ -ω) Then the width of a single sideband is

1. Ω +ω’
2. Ω + (ω’ -ω)
3. 2ω’
4. (ω’ -ω)

Answer: 4.(ω’ -ω)

Question 28. A MODEM is used

1. To superimpose a data signal on a carrier wave
2. To retrieve a data signal from its mixture with a carrier wave
3. To amplify a data signal
4. To convert an analog signal to a digital signal and vice-versa

Answer: 4. To convert an analog signal to a digital signal and vice-versa

Question 29. The process in which the amplitude of the carrier wave is made proportional to the instantaneous amplitude ofthe signal wave is called

1. Amplitude modulation
2. Demodulation
3. Rectification
4. Amplification

Answer: 1. Amplitude modulation

Question 30. Radio waves of low frequencies cannot be transmitted to long distances. So these are. superposed on a high-frequency carrier signal. This process is known as

1. Amplification
2. Modulation
3. Rectification
4. Demodulation

Answer: 2. Modulation

Communication system class 12 MCQs 

Question 31. Long-distance transmission is not possible using ground waves due to which characteristics of electromagnetic waves?

1. Scattering
2. Diffraction
3. Interference
4. Polarisation

Answer: 4. Polarisation

Question 32. A transmitting antenna and a receiving antenna, both of, height H, are used for a space wave transmission. At; another place, a similar transmission uses a transmitting antenna of height 2H, but the receiving antenna is set very near the ground. The ratio of the radio horizons for these two cases is
Answer:

1. 1: 1
2. 2: 1
3. √2: 1
4. 1: √2

Answer:  3. √2: 1

Question 33. An example of employing electromagnetic wave, with a range of frequency from 30 MHz to very high 300 MHz, as carrier wave is

1. Radio transmission of short-distance
2. Transmission of FM radio
3. Television transmission
4. Radar system

Answer: 2 And 3

Examples of Communication System Applications

Question 34. In distant communication, the data signal is not transmitted directly without the help of a carrier wave. The reason is

1. An extremely large transmitting antenna is necessary
2. The transmitting antenna is to be installed at a very high altitude
3. The chance of mixing noise in data signal increased
4. Due to the overlapping of many data signals, the data at the receiving end becomes very unclear

Answer: 1 And 4

Question 35. If a properly modulated carrier wave is transmitted through an antenna, then rate of energy dissipated at the antenna

1. Remains the same for AM wave
2. Increases in AM wave
3. Remains the same for FM wave
4. Increases in FM wave

Answer: 2 And 3

Question 36. Long-distance communication is possible through space wave

1. Due to reflection at the ionosphere
2. Due to reflection at artificial satellite
3. Through successive reflections from the earth’s surface
4. By Installing many antennas one after another at a specific distance apart

Answer: 2 And 4

Question 37. Which of the following is used as carrier wave, in a communication system?

1. Ultraviolet wave
2. Infrared wave
3. Microwave
4. Radio wave

Answer: 3 and 4

Communication system class 12 MCQs 

Question 38. In long-distance communication, the similar potential wave produced from a carrier wave is V – V₀ sin Ω t. On the other side, the potential wave corresponding to data signal

Is, v = v₀ sin ω t,  v₀ << V₀ and ω <<Ω. Superimposing data signal on a carrier wave, generally two types of modulated waves are produced; they are:

1.  Amplitude Modulated (AM) wave:

V_AM = \(V_0\left(1+\beta_1 \sin \omega t\right) \sin \Omega t\)

Where β₁ =  \(k_1 \frac{v_0}{V_0}\) = Modulation index

k₁ = constant

2. Frequency Modulated (FM) wave:

V_FM = \(V_0 \sin \left(\Omega t-\beta_2 \cos \omega t\right)\)

Where, \(\beta_2=k_2 \frac{v_0}{\omega}\)

Mathematical analysis of two modulated waves shows that, for the AM wave, apart from Ω frequency, there are two sine wave components of frequencies Ω – ω and Ω + ω. As a result, the bandwidth of the waves becomes (Ω – ω)- (Ω + ω). = 2ω). On the other hand, in FM wave, the number of single wave components is infinite and their frequencies are Ω ±ω, Ω ±2ω, ……….

1000 Hz

Besides that, if the dissipated power for a transmitting antenna to transmit a carrier wave be P_C and the dissipated power for transmitting AM and FM waves be P_AM and P_FM, then

P_AM  = \(\left(1+\frac{\beta^2}{2}\right)\)

P_FM  = P_C

1. The unit of constant K₁

1. Does not have any unit
2. V.s^-1
3. V^-1. s
4. V^-1 . s^-1

Answer: 1. Does not have any unit

2. The unit of constant K₂

1. 1
2. V.s^-1
3. V^-1. s
4. V^-1 . s^-1

Answer:  4. V^-1 . s^-1

Conceptual Questions on Signal Processing

Question 39. What Would Be The Width OF SIdeband On Either side of the amplitude modulated (AM) wave, if there is a mixing of frequency of 1000 Hz to 1500 Hz in the data signal?

1. 500 Hz
2. 1500 Hz
3. 2000 Hz
4. 3000 Hz

Answer: 4. 3000 Hz

Question 40. What would be the width of the sideband on either side of the amplitude-modulated (AM) wave, if there is a mixing frequency of 1000 Hz to 1500 Hz in the data signal?

1. 500 Hz
2. 1000 Hz
3. 1500 Hz
4. 2000 Hz

Answer: 4. 2000 Hz

Question 41. In converting a carrier wave into an amplitude-modulated (AM) wave, the modulation index was 0.05. Then, what would be the percentage increase in dissipated power at the transmitting antenna

1. 2.5
2. 1.25
3. 0.25
4. 0.125

Answer: 4. 0.125

Class 12 physics communication questions 

Question 42. As radio waves of low frequencies cannot be transmitted to long distances, a high-frequency carrier signal is superposed on it. This process is called

1. Amplification
2. Rectification
3. Modulation
4. Oscillation

Answer: 3. Modulation

Question 43. Satellite communication is done with

1. Ground wave
2. Skywave
3. Space Wave
4. None of these

Answer: 3. Space Wave

Class 12 physics communication questions 

Question 44. A signal of 5 kHz frequency is amplitude modulated carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are

1. 2 MHz only
2. 2005 kHz and 1995 kHz
3. 2005 kHz, 2000 kHz and 1995 kHz
4. 2000 kHz and 1995 kHz

Answer: 3. 2005 kHz, 2000 kHz and 1995 kHz

The two sidebands are also present along with the main frequency ofthe carrier wave

Question 45. Choose the correct statement:

1. In amplitude modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
2. In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
3. In frequency modulation the amplitude of the high. frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
4. In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the frequency of the audio signal

Answer: 1. In amplitude modulation the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal