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.

Questions from Optics- Cartesian Sign Convention

The sign convention adopted widely in Optics is the Cartesian convention. The ray incident on the curved surface is to be considered as proceeding in the positive X-direction and you have to measure all distances from the pole, which is supposed to be the origin.
Occasionally, you may be given a ray diagram in which the incident ray may be proceeding from right to left. To avoid confusion, imagine that the direction of the incident ray is still in the positive X-direction. You can even redraw the diagram to make the incident ray proceed from left to right if you want. The signs given to the distances are as in the Cartesian coordinate system: pole to right positive and pole to left negative. Distances measured (from pole) upwards are positive and those measured downwards are negative but you will mostly encounter problems with leftward and rightward measurements.
While solving problems, you should apply the signs to all known quantities. The unknown quantities are left as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you finally arrive at as the answer. For instance, if the distance of an image is obtained as negative, you will immediately understand that the image is on the same side of the curved surface as the object is.
Let us consider the following M.C.Q.:
A convex lens made of crown glass of refractive index 1.5 has focal length 20 cm. What will be its focal length within water (refractive index 4/3)?
(a) 10 cm (b) 20 cm (c) 40 cm (d) 60 cm (e) 80 cm
Writing lens maker’s equation 1/f = [(μ₂/μ₁) -1] [(1/R₁) - (1/R₂)] for the two cases, we have,
1/20 = [1.5-1] [(1/R₁) - (1/R₂)] for the first case, where f = +20cm (for convex lens), μ₂ = 1.5 and μ₁ = 1 (for air) and
1/f = [1.125 – 1] [(1/R₁)-(1/R₂)] for the second case where μ₂= 1.5 and μ₁ = 4/3 (for water).
Note that we have ignored the signs of the unknown quantities f, R₁ and R₂. On dividing the first equation by the second, we obtain f = 80 cm[option (e)].
The result shows that within water, the lens is still a converging lens. If the liquid has refractive index more than that of the lens, the lens will become diverging within the liquid. Let us consider such a case:
A converging lens of focal length 20cm is made of glass of refractive index 1.5. How will this lens behave if it is immersed in a liquid of refractive index 1.6?
(a) As a converging lens of focal length 80cm (b) As a diverging lens of focal length 80cm (c) As a diverging lens of focal length 120cm (d) As a diverging lens of focal length 160cm (e) As a converging lens of focal length 120cm
The correct option to this question is (d). You can check this by writing the lens maker’s equation for the two cases as in the previous question. The first equation is unchanged: 1/20 = [1.5-1] [(1/R) - (1/R₂)]. The second equation is
1/f = [0.9375 – 1] [(1/R₁) - (1/R₂)] since μ₂/μ₁) = 1.5/1.6 = 0.9375.
Dividing the first equation by the second, we obtain f = -160cm. As we get a negative value for the focal length, the lens behaves as a diverging lens within the given liquid [option (d)].
Let us discuss one more question:
The plane face of a planoconvex lens is silvered. If the refractive index of the material of the lens is μ, and the radius of curvature of its curved face is R, then the system behaves like a concave mirror of radius of curvature
(a) Rμ (b) R/μ (c) R/ (μ-1) (d) R/2(μ-1) (e) R(μ-1)
When one surface of a lens is silvered, the rays of light entering through the un-silvered surface are refracted first, reflected by the silvered surface next and finally are once more refracted. The effective focal length (F) of the system is given by
1/F = 1/f + 1/fm + 1/f where ‘f’ is the focal length of the un-silvered lens and fm is the focal length of the silvered surface which makes a mirror. In the present problem since the silvered surface is plane, fm is infinity so that the above equation becomes
1/F = 1/f + 0 + 1/f = 2/f. Therefore, F = f/2.
But we have for the original un-silvered lens, 1/f = (μ – 1) [(1/R₁) - (1/R₂)] where R₁ = R and R₂ = infinity. Therefore, f = R/(μ – 1)
Since F = f/2, the system has a focal length of R/2(μ – 1).The radius of curvature of a concave mirror is twice its focal length. Therefore, the system behaves like a concave mirror of radius of curvature R/(μ – 1). [Option (c)]. You will find multiple choice questions (with solution) on refraction at plane surfaces here as well as here