Question

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm, and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be

• A. -55 cm

Answer 1

Answer: (A)

The focal length of the liquid-glass combination is approximately -55 cm.

Answer 2

We can “build up” the combined refraction from the two surfaces that are in contact. (Here we adopt the following sign‐convention: distances measured in the direction of ray propagation are positive; a spherical surface has positive radius if its center of curvature is to the right of the vertex.)

The lens is made of glass (n₍glass₎ = 1.5). Its upper (concave) surface is in contact with a liquid of n₍liq₎ = 1.3 and has radius R₁ = 30 cm, while its lower (convex) surface is in air (n₍air₎ = 1) with R₂ = 20 cm.

For two “refracting surfaces in contact” the net power (in units of cm⁻¹) is obtained by “adding” the powers of the individual surfaces. For a spherical surface the power is given by

  Power = (n₂ – n₁)/R

(with the appropriate sign for R). Thus we have

  Power of 1st surface = (1.5 – 1.3)/30 = 0.2/30 = 0.00667 cm⁻¹
  Power of 2nd surface = (n_air – n_glass)/R₂ = (1 – 1.5)/20 = (–0.5)/20 = –0.025 cm⁻¹

So, the net power is

  Φ = 0.00667 – 0.025 = –0.01833 cm⁻¹.

When the final medium is air (n = 1) the effective focal length (with the proper sign) is given by

  f = 1/Φ = 1/(–0.01833) ≈ –54.55 cm.

A negative focal length means that, as seen from the incident (liquid) side the combination is diverging. (If one wishes to express the answer to two‐significant–figures one may write –55 cm.)

Core Explanation:

1. For the first interface (liquid→glass):
     Power₁ = (1.5 – 1.3)/30 = 0.00667 cm⁻¹.

2. For the second interface (glass→air):
     Power₂ = (1 – 1.5)/20 = –0.025 cm⁻¹.

3. Net power:
     Φ = 0.00667 – 0.025 = –0.01833 cm⁻¹
     Effective focal length: f = 1/Φ ≈ –54.55 cm.