WBCHSE Class 12 Physics Reflection Of Light Question And Answers

WBCHSE Class 12 Physics  Reflection Of Light Long Questions And Answers

Question 1. What are the differences in reflection by

1. A plane mirror,
2. The wall of a building and 
3. A clean glass plate?

Answer:

1. Regular reflection of light takes place in a plane mirror and a bright and distinct image is formed.
2. Diffuse reflection takes place from the wall of a building due to its rough surface and no image is formed. The wall itself is seen with equal brightness.
3. If light is incident on a clean glass plate, a small portion of the incident light is reflected.
4. A major portion of the light enters the glass plate and by refraction is transmitted to air through the other side. So, a dim but distinct image is formed.

Question 2. At night the objects situated outside a brightly illuminated room are not visible through the glass windows. But if the lights inside the room are off, external objects are visible. Explain the reason.
Answer:

1. At nightlight, external objects enter a room through glass windows. If the bulb glows inside the room, reflected light from the glass window comes to the eyes of the observer.
2. The intensity of this reflected light is much greater than that of the light coming from outside. So, the external objects are not visible.
3. But if the room is made dark, the light coming from outside is sufficient to make the external objects visible.

Important Definitions Related to Reflection Q&A

Question 3. What type of mirror is used as a rear-view mirror in motor cars and other vehicles and why?
Answer:

Convex mirror is used as a rear-view mirror in motor cars and other vehicles. In a convex mirror, always erect and diminished virtual image is formed and the convex side of the mirror faces the object. As a result, the field of view is large. So, the driver can see a large number of vehicles approaching from behind.

Question 4. In the dial of a clock, lines are marked instead of numbers. Observing through a plane mirror the time appeared to be 7.25. What was the actual time?
Answer:

In the mirror, the image of the dial is laterally inverted. The hour hand actually between 4 and 5 appeared in the mirror between 7 and 8, and the minute hand actually at 7 appeared at 5. So, the actual time was 4.35.

WBBSE Class 12 Reflection of Light Q&A

Question 5. Can a plane mirror form a real image?
Answer:

A plane mirror can form a real image of a virtual object. As converging rays AB, CD, and PO would meet at P’ in the absence of the mirror P’ acts as the virtual object. But the rays after reflection meet at P and form a real image.

Question 6. Why are paraboloidal mirrors used in car headlights and searchlights instead of using spherical concave mirrors?
Answer:

Parallel rays of high intensity even at a large distance are required for motor car headlights and searchlights. We know that if a source of light is placed at the focus of a concave mirror only paraxial rays travel parallel ahead after reflection in the mirror.

But the other rays do not travel parallel. So, the rays spread after traversing only a little distance, and the intensity at a large distance becomes very low.

But, if a source of light is placed at the focus of a paraboloidal mirror, all the reflected rays advance parallel to the principal axis. So, the intensity remains high even at a large distance.

For this reason, paraboloidal mirrors are used in car headlights and searchlights instead of spherical concave mirrors.

Question 7. Based on the relation among u, and f of a spherical mirror determine the nature, position, and size of the image of an object formed by a plane mirror.
Answer:

The radius of curvature of a plane mirror, r = ∞o; so, its focal length, f =\(\frac{r}{2}\)  = ∞

From the relation of a spherical mirror,

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{1}{\infty}\) = 0

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

Or, v= -u

The negative sign indicates that the image is formed behind the mirror as the object is in front of the mirror and it is virtual.

Again the values of u and v are equal; so, object distance from the mirror = image distance.

Magnification of the image, m == 1, ie, the size of the object and image are equal.

Short Questions on Laws of Reflection

Question 8. Two persons of the same height are standing, one Inside a shop and another outside the shop, on either side of a glass window. The second man sees his image behind the glass window due to reflection in the glass and his image appears larger than the other person. What is the type of glass on the window?
Answer:

The outside of the glass window acts as a concave mirror. We know that if an object is placed within the focus of a concave mirror, a virtual, erect, and magnified image is formed behind the mirror.

In the case of a convex mirror, the image is diminished in size, and in the case of a plane mirror the image is of the same size as the object. So, for the man standing outside, the glass window behaves like a concave mirror.

Question 9. If an object of height I is placed perpendicularly on the principal axis of a spherical mirror at a distance u from it, prove that the height of Its image,
Answer:

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

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

Or, \(\frac{u-f}{f u}\)

Or,  \(\frac{v}{u}=\frac{f}{u-f}\)

Linear magnification, m= \(\frac{l^{\prime}}{l}=\frac{v}{u}=\frac{f}{u-f}\) [considering mod value of m only]

Or, l’ = \(l\left(\frac{f}{u-f}\right)\)

Question 10. Show that in a spherical mirror both the object and the image are situated on the same side of the focus. Answer:

In the case of a spherical mirror, if we take object distance and image distance from the focus as x and y respectively then, according to Newton’s equation,

xy = f² [f=focal length of the mirror]

f²  first always positive. Therefore xy is always positive. So, x and y must be of the same sign. Now x and y will be of the same sign if the object and the image are situated on the same side of the focus.

Question 11. Two concave mirrors are placed facing each other and have the same center of curvature. Where will the images of the point source kept at this common center of curvature be formed?
Answer:

The rays of light from a point source may fall on either of the concave mirrors and after reflection from the respective mirrors, all the reflected rays will meet at the common center of curvature. Therefore, the images will coincide at that point.

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Question 12. Show that a convex mirror always forms a virtual image of diminished size as compared to the object. Answer:

According to the equation of the spherical mirror,

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

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

Or, \(v=\frac{u f}{u-f}\) ………………….. (1)

Now for a convex mirror ƒ is considered as positive. Again, as the object is real, u is taken as negative.

Following this sign convention, from equation (1) we get,

v= \(\frac{(-u) \times f}{-u-f}=\frac{u f}{u+f}\) ………………………(2)

As v is positive, so for any position of the object in front of a convex mirror, the image will be formed behind the mirror i.e., the image will always be virtual.

Again, magnification,

m =  \(\frac{v}{u}\) [considering the magnitude of m only]

= \(\frac{f}{u+f}\) [from equation:(2)]

U+f>f. So, m <1

i.e., the image formed by a convex mirror is diminished in size as compared to the object mirror.

Practice Questions on Image Formation by Mirrors

Question 13. Explain under what conditions a real image of a virtual object and a virtual image of a virtual object is formed in a convex mirror.
Answer:

According to the equation of a spherical mirror,

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

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

Or, v=  \(\frac{u f}{u-f}\)

Now in the case of a convex mirror, f>0, and for a virtual object, u> 0.

Now, if f>u then v<0, i.e., if the virtual object is situated Now differentiating concerning u, between the pole and the focus, a real image will be formed.

And if f<u then v>0, i.e., if the virtual object is situated between the focus and infinity, a virtual image will be formed.

Question 14. An object is approaching a convex mirror from a long distance. Will the image move with the same velocity as the object? In which direction will the image move?
Answer:

In the case of a spherical mirror,

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

Differentiating concerning what we get,

⇒ \(-\frac{1}{v^2} \frac{d v}{d t}-\frac{1}{u^2} \frac{d u}{d t}\) = 0

Or, \(\frac{d v}{d t}=-\frac{v^2}{u^2} \frac{d u}{d t}\)

Here, \(\frac{d v}{d t}\)  – velocity of image \(\frac{d u}{d t}\) – velocity of object

The velocity of the image

= \(-\frac{v^2}{u^2} \times \text { velocity of the object }\)

= \(-\left(\frac{v}{u}\right)^2 \times \text { velocity of the object }\)

In the case of convex mirror, m = \(\left|\frac{v}{u}\right|\)  <1

So, the velocity of the image < velocity of the object. The negative sign of equation (1) indicates that the image is moving.  So, the image will move towards the mirror with less velocity than the object.

Question 15. The image of a candle formed by a concave mirror is cast on a screen. If some parts of the mirror are covered then how will the image be affected?
Answer:

The parts of the mirror that are covered do not take part in image formation. Therefore, the brightness of the image is diminished Still the full image will be formed no part of the image will be missing.

Conceptual Questions on Total Internal Reflection

Question 16. A small linear object of length I is placed along the principal axis of a concave mirror. The nearest point of the object is at a distance d from the mirror. If the focal length of the mirror is ƒ then, prove that the size (length) of the image will be l \(\left(\frac{f}{d-f}\right)^2\) 
Answer:

Equation of a concave mirror,

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

Now differentiating concerning u,

⇒ \(-\frac{1}{v^2} \frac{d v}{d u}-\frac{1}{u^2}\) = 0

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

Therefore, the magnitude of axial magnification of the image

m = \(=\left|\frac{d v}{d u}\right|=\left(\frac{v}{u}\right)^2\)

From the equation of mirror, we get,  \(\frac{u}{v}+1\) = \(\frac{u}{f}\)

Or, \(\frac{u}{v}=\frac{u-f}{f}=\frac{d-f}{f}\) [ given u = d]

∴ m = \(\left(\frac{v}{u}\right)^2=\left(\frac{f}{d-f}\right)^2\)

∴ Length of the image = l ×m = l\(\left(\frac{f}{d-f}\right)^2\)

Question 17. The focal length of a concave mirror is f. What will be the magnification of an object placed at a distance x from the principal focus (given, object distance>f)? For what value of x will the size of the object and the image be the same? 
Answer:

Object distance concerning the pole, u = f+x

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

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

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

=  \(\frac{f+x}{f}-1=\frac{x}{f}\)

Magnification of the image, m= \(=\frac{v}{u}=\frac{f}{x}\) [ take mod value of m].

If the size of the object and the image is the same then, m = 11
i.e., \(\frac{f}{x}=1\) = 1 or, x= f

Real-Life Scenarios Involving Reflection Questions

Question 18. What will be the magnification of the image of an object placed in the mid-point between the focus and the pole of a concave mirror?
Answer:

Equation of a concave mirror, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

∴ \(\frac{1}{v}+\frac{2}{f}=\frac{1}{f}\)

Since u = \(\frac{f}{2}\)

Or, \(\frac{1}{v}=\frac{1}{f}-\frac{2}{f}=-\frac{1}{f}\) Or, v = -f

∴  The image will be formed  at a distance of f behind the mirror

Magnificatiojn, m = \(m=\frac{v}{u}=\frac{f}{\frac{f}{2}}\)

= 2. [  considering the mod value of m]

Question 19. What is the minimum distance between the object and its real image formed by a concave mirror and when is it possible?
Answer:

The minimum distance between the object and its real image is zero when u = v = 2f; i.e., the object is placed at the
center of curvature of the concave mirror.

Question 20. What sort of a reflector is used with a street bulb to light up streets in a better way?
Answer:

A convex reflector is used. This is because the area of vision is larger for a convex mirror and this helps in illuminating a larger area.

Question 21. A cube is placed in front of a large concave mirror. Will the image of the cube be a cube?
Answer:

The height of the image of the front side of the cube facing the concave mirror will not be equal to that of the back side of it. the image of the cube will not be a cube.

Question 22. An object is placed at a distance of 20 cm in front of a concave mirror of a focal length of 10 cm. Determine the position of the image. What is the ratio of the size of the image to the size of the object?
Answer:

Here, u-20 cm, f = -10 cm according to the equation of spherical mirror,

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

⇒ \(\frac{1}{v}+\frac{1}{-20}=-\frac{1}{10}\)

⇒ \(\frac{1}{v}=\frac{1}{20}-\frac{1}{10}\)

⇒ \(\frac{1-2}{20}=-\frac{1}{20}\)

v= – 20 cm

∴ The image will be formed in front of the concave mirror at a distance of 20 cm.

Now, \(\text { Now, } \frac{\text { Size of the image }(I)}{\text { Size of the object }(O)}\)

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

= \(-\frac{-20}{-20}\)

= -1

The negative sign shows that the image is real. Hence the ratio of the size of the image to the size of the object is 1: 1.

Question 23. The inside of optical instruments like the camera and telescope are blackened. Explain.
Answer:

When light is incident on an object of black color, the light is not reflected and most of the light is absorbed by the object. If the inside of an optical instrument is not blackened, then the light will be reflected from the parts inside the instrument and it will overlap with the image.

Hence the image will become blurred. Thus if the inside of the optical instrument is blackened, then there is no unwanted reflection of light and the instrument can function properly.

Examples of Applications of Reflection in Daily Life

Question 24. Light from an object is incident on a concave mirror and forms a real image of the object. If both the object and the mirror are immersed in water, then does the position of the image change?
Answer:

The relation between object distance u, image distance v, and the radius of curvature of the mirror r for a concave mirror is,

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{2}{r}\)

According to this relation, the image distance remains constant if the object distance is constant. Since u and r are independent of the medium, the relation is the same for any medium. Thus if the object and the mirror are immersed in water, then the position of the image will not change.

Question 25. Can a plane mirror form an inverted image of an object?
Answer:

No, a plane mirror always forms a virtual and erect image of an object. So the image formed by a plane mirror is never inverted. It should be noted that the distance of the image from the mirror is the same as the object distance and the image is laterally inverted.