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 (integer)
54
Solution
From lens makers formula
<br/><br/>
$$
\frac{1}{f}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {mrdium }}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]
$$
<br/><br/>
when in air
<br/><br/>
$$
\frac{1}{18}=\left(\frac{1.5}{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(1)
$$
<br/><br/>
$\mu_{\text {lense }}=1.5, \mu_{\text {air }}=1$. <br/><br/>when in water
<br/><br/>
$$
\frac{1}{f}=\left(\frac{1.5}{4 / 3}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(2)
$$
<br/><br/>
from (1) & (2)
<br/><br/>
$f=72$
<br/><br/>
Change in focal length $=72-18 = 54$
About this question
Subject: Physics · Chapter: Optics · Topic: Lenses and Optical Instruments
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