Question

A convex lens made of glass has focal length $0.15\, m$ in air. If the refractive index of glass is $3/2$, and that of water is $4/3$ the focal length of lens when immersed in water is

• A. 0.15 m
• B. 0.30 m
• C. 0.6 m
• D. 0.45 m

Answer 1

Answer: (C)

0.6 m

Answer 2

Given, $f_{a}=0.15 \,m , \mu_{g}=\frac{3}{2}, \mu_{w}=\frac{4}{3}$
According to Lens maker's formula
$\frac{1}{f} =(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \,\,\,\,\,\,$ where $\mu=\frac{\mu_{L}}{\mu_{M}}$
$\frac{1}{f_{a}} =\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$=\left(\frac{(3 / 2)}{1}-1\right) C\,\,\,\,\,\,\,\,$ where $C=\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
or $\,\,\,\frac{1}{f_{a}}=\frac{C}{2}\,\,\,\,\,\,\,\,...(i)$
Also, $\frac{1}{f_{w}}=\left(\frac{\mu_{g}}{\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\left(\frac{(3 / 2)}{(4 / 3)}-1\right) C$
or $\,\,\,\,\,\frac{1}{f_{w}}=\frac{C}{8}\,\,\,\,\,\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$\frac{f_{w}}{f_{a}}=\frac{C}{2} \times \frac{8}{C}=4$
or$\,\,\,\,\,\,f_{w}=4 f_{a}$
$=4 \times 0.15=0.6\, m$