Question

A convex lens of glass $(\mu =1.5)$ has a focal length of $8cm$ when placed in air. What is the focal length of lens when it is immersed in water $\left(\mu =\frac{4}{3}\right)?$

• A. 4 cm
• B. 8 cm
• C. 16 cm
• D. 32 cm

Answer 1

Answer: (D)

Given,
focal length $f=8cm$
${\mu }_g=1.5$
$\therefore \frac{1}{f}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\frac{1}{8}=(1.5-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\frac{1}{8}=(0.5)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\left[\frac{1}{R_1}-\frac{1}{R_2}\right]=\frac{1}{4}$
Lens when immersed in water,
Lens when immersed in water,
$\frac{1}{f_w}=\left(\frac{{\mu }_g}{{\mu }_w}-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$=\left[\frac{1.5}{1.33}-1\right]\frac{1}{4}$
$=\frac{1}{8}\times \frac{1}{4}$
$=\frac{1}{f_W}=\frac{1}{32}$
$\therefore f_w=32$