A biconvex lens of refractive index 1.5 has a focal length of 20 ~cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

Geometrical Optics PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q70 with Solution

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Correct answer: D

Use the lens maker formula in a medium

For a thin lens in a medium of refractive index $n_m$ , the focal length is

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

where $n_l$ is the refractive index of the lens.

First use the given focal length in air

In air, $n_m=1$ , so

\frac{1}{f_{air}}=(n_l-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Given:

$n_l=1.5$
$f_{air}=20\,\text{cm}$

So,

\frac{1}{20}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\frac{1}{20}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Hence,

\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{10}

Now immerse the lens in liquid of refractive index $1.6$

Now,

\frac{1}{f_{liq}}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Calculate the refractive factor:

\frac{1.5}{1.6}=0.9375

So,

0.9375-1=-0.0625

Thus,

\frac{1}{f_{liq}}=(-0.0625)\left(\frac{1}{10}\right)=-\frac{1}{160}

Therefore,

f_{liq}=-160\,\text{cm}

Interpretation

Since the liquid has refractive index greater than that of the lens ( $1.6>1.5$ ), the biconvex lens behaves like a diverging lens in that medium, so the focal length is negative.

Option check

A: $-16\,\text{cm}$ ❌
B: $+16\,\text{cm}$ ❌
C: $+160\,\text{cm}$ ❌
D: $-160\,\text{cm}$ ✅

Hence the correct answer is D.

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