A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f1 , then the ratio f_1 \over f is closest to the integer :

Geometrical Optics PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q53 with Solution

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Correct answer: C

For a thin lens in a surrounding medium,

\frac{1}{f} = \left(\frac{n_{\text{lens}}}{n_{\text{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Let

In air,

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

Given $f=16\text{ cm}$ , so

\frac{1}{16} = (1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\frac{1}{16} = 0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Thus,

\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{8}

In liquid,

\frac{1}{f_1} = \left(\frac{1.5}{1.42}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Now,

\frac{1.5}{1.42}-1 = \frac{1.5-1.42}{1.42} = \frac{0.08}{1.42}

So,

\frac{1}{f_1} = \frac{0.08}{1.42}\cdot \frac{1}{8}

Instead of calculating $f_1$ directly, take ratio:

\frac{\frac{1}{f_1}}{\frac{1}{f}} = \frac{\left(\frac{1.5}{1.42}-1\right)}{(1.5-1)}

Hence,

\frac{f_1}{f} = \frac{1.5-1}{\frac{1.5}{1.42}-1}

\frac{f_1}{f} = \frac{0.5}{\frac{0.08}{1.42}} = 0.5\times \frac{1.42}{0.08}

\frac{f_1}{f} = \frac{0.71}{0.08} = 8.875

\frac{f_1}{f} \approx 8.875

Closest integer is

9

So the correct option is C.