WBCHSE Class 12 Physics Refraction Of Light At Spherical Surface Lens Short Answer Questions

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.