Question

A thin lens of refractive index 1.5 has a focal length 15 cm in air. When the lens is placed in a medium of refractive index 4/3, what is the new focal length?

• A. 60 cm
• B. 120 cm
• C. 70 cm
• D. 50 cm

Answer 1

The correct option is A 60 cm

For the lens in air

$\frac{1}{f_a}=(a\mu g-1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{15}=(1.5-1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{2}{15}=\frac{1}{R_1}-\frac{1}{R_2}$

When the lens is immersed in water

$\frac{1}{f_w}=(\frac{a\mu g}{a\mu w}-1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{f_w}=(\frac{1.5}{\frac{4}{3}}-1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{f_w}[\frac{4.5}{4}-1][\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{f_w}=\frac{1}{8}\times \frac{2}{15}$

$\Rightarrow \frac{1}{f_w}=\frac{1}{60}$

$\Rightarrow f_w=60cm$.