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Ray Optics

Section: Physics

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Question : 1 of 17

Marks: +1, -0

The ratio of the focal lengths of a convex lens when kept in air and when it is immersed in a liquid is

1: 2

. If the refractive index of the material of the lens is 1.5 , then the refractive index of the liquid is

[AP EAPCET 24 May 2025 S1]

$1.20$
$1.30$
$1.25$
$1.35$

Solution:

By Lens maker's formula, For a equviconvex lens,

\; \frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{2}{R} \right)

In air,

n_1 = 1, n_2 = \; \frac{3}{2}

\therefore \;\; \frac{1}{f_{\;\text{air }\;}} = \left( \frac{3}{2} - 1 \right) \left( \frac{2}{R} \right)

\; \frac{1}{f_{\;\text{air }\;}} \; = \; \frac{1}{R}

In liquid,

n_1 = x, n_2 = \; \frac{3}{2}

\; \frac{1}{f_l} = \left( \; \frac{3}{2x} - 1 \right) \left( \; \frac{2}{R} \right)

Given,

\; \frac{f_{\;\text{air }\;}}{f_{\;\text{liquid }\;}} = \; \frac{1}{2}

So, from Eq. (i) and (ii)

\; \frac{f_{\;\text{air }\;}}{f_{\;\text{liquid }\;}} = \frac{R \times 2 \left( \frac{3}{2x} - 1 \right)}{R \times 1} = \; \frac{1}{2}

\Rightarrow 2 \left( \; \frac{3}{2x} - 1 \right) = \; \frac{1}{2}

\Rightarrow \; \frac{3}{2x} - 1 = \; \frac{1}{4}

\Rightarrow \; \frac{3}{2x} = \; \frac{5}{4} \; \text{ or } \; x = \; \frac{6}{5} = 1:20

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