Question

A thin convergent glass lens $ \mu =1.5 $ has a power of $ +5.0\,D $ . When this lens is immersed in a liquid of refractive index $ \mu _{1} $ it acts as a diverging lens of focal length $ 100\,cm $ . The value of $ \mu _{l} $ should be

• A. $ \frac{3}{2} $
• B. $ \frac{4}{3} $
• C. $ \frac{5}{3} $
• D. $ 2 $

Answer 1

Answer: (C)

When the lens in air, we have
$ P_a =\frac{1}{f_a}=\frac{\mu_g-\mu_a}{\mu_a}\bigg[\frac{1}{R_1}-\frac{1}{R_2}\bigg] $
When the lens is in liquid, we have
$ P_1 =\frac{1}{f_l}=\frac{\mu_g-\mu_l}{\mu_l}\bigg[\frac{1}{R_1}-\frac{1}{R_2}\bigg] $
Given, $ P_a = 5, P_1 = - 1,$
$ \mu_a = 1, \mu_g = 1.5 $
On soiving, we get,
$ \mu_l = \frac{5}{3} $