Question

A convex lens of glass (n = 1.5) has focal length 0.2 m The lens is immersed in water of refractive index 1.33. The change in the power of convex lens is _________ . 

(A) 3.72 D (B) 4.62 D (C) 6.44 D (D) 1.86 D

Answer 1

The correct option is (A) 3.72 D.

Explanation:

 (1/f) = (n – 1) [(1 / R₁) – (1 / R₂)]

Here   (1 / f_a) (a_{(n)g – 1}) [(1 / R₁) – (1 / R₂)] ‑‑‑‑‑‑‑‑‑‑‑‑‑‑ in air .....(1)

(1 / f_w) (w_{(n)g – 1}) [(1 / R₁) – (1 / R₂)] ‑‑‑‑‑‑‑‑‑‑‑‑‑‑ in water ......(2)

from (1),  (1 / 0.2) = (1.5 – 1) [(1 / R₁) – (1 / R₂)]

(1 / R₁) – (1 / R₂) = 10 ......(3)

from (2) (1 / f_w) = {(1.5 / 1.33) – 1} [(1 / R₁) – (1 / R₂)]

using (3) we get (1 / f_w) = 0.127 × 10 = 1.27

as power = {1 / (focal length)} hence

P_a = (1 / f_a) = (1 / 0.2) = 5D

P_w = (1 / f_w) = 1.27 D

Hence change in power = P_a – P_w = 5 – 1.27 = 3.73 D