JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Geometrical Optics

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✓ Correct
(C)24 cm
(D)48 cm

Answer: (B)

Solution

Using the lens maker's formula, in air:

\frac{1}{f} = (\frac{μ _{l e n s}}{μ _{m e d i u m}} - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]

\frac{1}{24} = (1.5 - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}] ...(i)

Immersed in water:

\frac{1}{f ^{'}} = (\frac{1.5}{1.33} - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}] = (\frac{1.5 \times 3}{4} - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}] = \frac{0.5}{4} [\frac{1}{R _{1}} - \frac{1}{R _{2}}] ...(ii)

Dividing (i) by (ii):

\frac{f ^{'}}{24} = \frac{0.5}{0.5/4} = 4 ⟹ f^{'} = 96 cm

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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