Question 301: Ray Optics

Question

A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm.
(Given refractive index of water $$=\frac{4}{3}$$)

Accepted Answer

From lens makers formula
$$
\frac{1}{f}=\left(\frac{\mu_{	ext {lens }}}{\mu_{	ext {mrdium }}}-1ight)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}ight]
$$
when in air
$$
\frac{1}{18}=\left(\frac{1.5}{1}-1ight)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}ight]\quad...(1)
$$
$\mu_{	ext {lense }}=1.5, \mu_{	ext {air }}=1$.
when in water
$$
\frac{1}{f}=\left(\frac{1.5}{4 / 3}-1ight)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}ight]\quad...(2)
$$
from (1) & (2)
$f=72$
Change in focal length $=72-18 = 54$

Ray Optics

A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm. (Given refractive index of water $=\frac{4}{3}$)

Choose the correct answer: