The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 ~cm . The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be x: 1 . The value of x is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q74 with Solution

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Correct answer: 4

For a thin lens in a surrounding medium,

P = \left(\frac{n_\ell}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

where:

For a biconvex lens with equal radii:

R_1 = +20\text{ cm}, \qquad R_2 = -20\text{ cm}

Hence,

\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{20}-\left(-\frac{1}{20}\right)=\frac{2}{20}=\frac{1}{10}\text{ cm}^{-1}

In air, $n_m = 1$ and $n_\ell = 1.8$ . So,

P_{\text{air}} = (1.8-1)\left(\frac{1}{10}\right)=0.8\cdot \frac{1}{10}=0.08

In liquid, $n_m = 1.5$ . Thus,

P_{\text{liq}}=\left(\frac{1.8}{1.5}-1\right)\left(\frac{1}{10}\right)

Now,

\frac{1.8}{1.5}=1.2

So,

P_{\text{liq}}=(1.2-1)\left(\frac{1}{10}\right)=0.2\cdot \frac{1}{10}=0.02

\frac{P_{\text{air}}}{P_{\text{liq}}}=\frac{0.08}{0.02}=4

Therefore,

P_{\text{air}}:P_{\text{liq}} = 4:1

So, the required value is

x=4

Stored correct answer = $4$

This matches our derived answer.