JEE Main 2025 Jan 29 Shift 2, Physics Q13: Thin Lens and Mirrors

JEE Main2025Jan 29, Shift 2Physics

A convex lens mode of glass (refractive index = $1.5$ ) has focal length $24$ cm in air. When it is totally immersed in water (refractive index = $1.33$ ), its focal length changes to

A
$72$ cm
B
$96$ cm
C
$24$ cm
D
$48$ cm

Solution

Lensmaker formula with the surrounding medium:

$\frac{1}{f} = (\frac{μ _{lens}}{μ _{medium}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .

In air: $\frac{1}{24} = (1.5 - 1) \cdot (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.5 \cdot S$ , where $S$ is the curvature factor. So $S = \frac{1}{12}$ .

In water, taking $μ_{water} = 4/3$ , the relative index becomes $\frac{1.5}{4/3} = \frac{9}{8}$ , and

$\frac{1}{f ^{'}} = (\frac{9}{8} - 1) S = \frac{1}{8} \cdot \frac{1}{12} = \frac{1}{96}$ .

$∴ f^{'} = 96$ cm.

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JEE Main 2025 Jan 29 Shift 2, Physics Q13: Thin Lens and Mirrors

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