JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 20)
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 $$=\frac{4}{3}$$)
Answer
54
Explanation
From lens makers formula
$$ \frac{1}{f}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {mrdium }}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] $$
when in air
$$ \frac{1}{18}=\left(\frac{1.5}{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(1) $$
$\mu_{\text {lense }}=1.5, \mu_{\text {air }}=1$.
when in water
$$ \frac{1}{f}=\left(\frac{1.5}{4 / 3}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(2) $$
from (1) & (2)
$f=72$
Change in focal length $=72-18 = 54$
$$ \frac{1}{f}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {mrdium }}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] $$
when in air
$$ \frac{1}{18}=\left(\frac{1.5}{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(1) $$
$\mu_{\text {lense }}=1.5, \mu_{\text {air }}=1$.
when in water
$$ \frac{1}{f}=\left(\frac{1.5}{4 / 3}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(2) $$
from (1) & (2)
$f=72$
Change in focal length $=72-18 = 54$
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