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)

$\frac{5}{3}$

Answer 2

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, $\mu_l = \frac{5}{3}$