Multiple Choice Questions on Lenses

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Here are a few questions from optics involving the lens maker’s equation and the concept of magnification in the case of lenses:

(1) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R/2, the refractive index of the material of the lens is

(a) 1.2

(b) 1.33

(c) √2

(d) √3

(e) 2

The refractive index n is related to the focal length f and the radii of curvature R₁ and R₂ by the lens maker’s equation,

1/f = (n – 1) (1/R₁ – 1/R₂)

By Cartesian sign convention, R₁ is positive and R₂ is negative for a biconvex lens so that the above equation becomes

1/f = (n – 1) (1/R₁ + 1/R₂)

Substituting for f, R₁ and R₂ we obtain

1/(R/2) = (n – 1) (2/R)

Or, 2/R = (n – 1) (2/R) from which n = 2

(2) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R, what will be its focal length in a liquid of refractive index 1.5?

(a) R/2

(b) R

(c) 2R

(d) 4R

(e) Infinite

Many of you might have noted that an equiconvex lens of refractive index 1.5 has focal length equal in magnitude to the radius of its faces.

Some of you might have noted the converse of the above fact: If the refractive index is 1.5 in the case of an equiconvex (or equiconcave) lens, the magnitude of its focal length will be the same as that of the radius of curvature of its faces

[You can check this by substituting n = 1.5 and R₁ = R₂ = R in the lens maker’s equation].

Since the refractive index of the material of the lens is 1.5, its focal length in a liquid of refractive index 1.5 will be infinite.

(3) A converging lens produces a real, magnified and well defined image of a small illuminated square on a screen. The area of the image is A₁. When the lens is moved towards the screen without disturbing the object and the screen, the area of the well defined image obtained on the screen is A₂. What is the side of the square object?

(a) (√A₁ + √A₂)/2

(b) [(A₁ + A₂)/2]^1/2

(c) (A₁A₂)^1/2

(d) (A₁A₂)^1/4

(e) [√A₁ +√A₂]^1/4

The two positions (of the lens) for which well defined images of the square are obtained, are the conjugate positions and hence we have

a =√(a₁a₂) where a₁ and a₂ are the sides of the images in the two cases and a is the side of the square object.

[The linear magnification in the first case is v/u = a₁/a.

Since the object distance u and the image distance v are interchanged in conjugate positions, we have, in the second case,

u/v = a₂/a.

From the above expressions, 1 = a₁a₂/a² from which a =√(a₁a₂)].

But a₁ = √A₁ and a₂ = √A₂

Therefore, a =√(√A₁√A₂) = (A₁A₂)^1/4

You will find similar useful multiple choice questions on optics here as well as here.