A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

Geometrical Optics PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q66 with Solution

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Correct answer: C

For a thin lens in a surrounding medium of refractive index $n_m$ :

\frac{1}{f}=\left(\frac{n_l}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

where $n_l$ is refractive index of the lens.

So,

In air ( $n_m=1$ ):
$\frac{1}{f_{air}}=(n_l-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
In water ( $n_m=1.28$ ):
$\frac{1}{f_w}=\left(\frac{n_l}{1.28}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Given:

$n_l=1.6, \qquad f_{air}=12\text{ cm}$

First,

\frac{n_l}{1.28}-1=\frac{1.6}{1.28}-1=1.25-1=0.25

Also,

Hence,

\frac{1/f_w}{1/f_{air}}=\frac{0.25}{0.6}

So,

\frac{f_w}{f_{air}}=\frac{0.6}{0.25}=2.4

Therefore,

f_w=2.4\times 12=28.8\text{ cm}

28.8\text{ cm}=288\text{ mm}

\boxed{288\text{ mm}}

So the correct option is C.