Optics: Karnataka CET Questions (MCQ) on Lenses

Today we will discuss two questions (MCQ) involving lenses. The first one appeared in Karnataka CET 2011 question paper and the second one appeared in Karnataka CET 2010 question paper.

(1) Two thin lenses have a combined power of + 9 D. When they are separated by a distance of 20 cm, their equivalent power becomes + 27/5 D. Their individual powers (in dioptre) are ……..

(1) 4, 5

(2) 3, 6

(3) 2, 7

(4) 1, 8

We have P₁ + P₂ = 9 where P₁ and P₂ are the powers of the two lenses.

[Note that combined power means 1/f₁ + 1/f ₂ which is the effective power when the lenses are in contact. Power is the reciprocal of focal length and when the lenses are in contact, we have 1/F = 1/f₁ + 1/f ₂ where f₁ and f₂ are the focal lengths of the lenses and F is the combined focal length].

When the lenses are separate by a distance d, the power becomes P₁ + P₂ – d P₁P₂

[This follows from the expression, 1/F = 1/f₁ + 1/f ₂ – d/f₁f₂].

Therefore, we have

P₁ + P₂ – d P₁P₂ = 27/5

Or, 9 – (0.2× P₁P₂) = 27/5

[You have to substitute the separation in metre since the power is the reciprocal of focal length in metre].

From the above equation, 18/5 = 0.2× P₁P₂

Or, P₁P₂ = 18

Evidently the values given in option (2) satisfy the above condition so that we have P₁ = 3 and P₂ = 6

(2) The focal length of a plano-convex lens is ‘f’ and its refractive index is 1.5. It is kept over a glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes 2f. Then the refractive index of the liquid is ……

(1) 1.5

(2) 2

(3) 1.25

(4) 1.33

In the case of a lens placed in air or vacuum, lens maker’s equation is

1/f = (n – 1)(1/R₁ – 1/R₂) where n is the refractive index of the lens.

[General form of lens maker’s equation is 1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are the radii of curvature of its faces, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed].

In the case of the plano-convex lens, the equation becomes

1/f = (n – 1)(1/R₁ – 1/∞) = (n – 1)(1/R₁)

Since n = 1.5, R₁ = 0.5f ……..(i)

The focal length of the combination (F) of the plano convex lens (of focal length f) and the plano-concave liquid lens (of focal length f_ℓ)is given by

1/F = 1/f + 1/f_ℓ

Therefore, 1/2f = 1/f + 1/f_ℓ

This gives f_ℓ = –2f………(ii)

But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (n_ℓ) of the material (liquid) as

1/f_ℓ = (n_ℓ – 1)(–1/R₁ – 1/∞) = (n_ℓ – 1)(– 1/R₁)

[Note that the radius of the concave surface, which the liquid lens presents to the incident light, has to be negative in accordance with the Cartesian sign convention. Click on the label ‘Cartesian sign convention’ below this post to learn more about this convention].

Therefore, 1/(–2f) = – (n_ℓ – 1)/0.5f , on substituting for R₁ and f_ℓ from equations (i) and (ii).

This gives 2 n_ℓ = 2.5 so that n_ℓ = 1.25