JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Geometrical Optics

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A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water.

(Take refractive index of water as 1.28)

(A)355 mm
(B)288 mm✓ Correct
(C)555 mm
(D)655 mm

Answer: (B)

Solution

The lens maker's formula fixes the power through the ratio of lens index to surrounding index. In air,

\frac{1}{12} = (1.6 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) \Rightarrow (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = \frac{1}{7.2} cm^{- 1} .

In water the relative index is

1.6/1.28 = 1.25

, so

\frac{1}{f _{w}} = (1.25 - 1) \times \frac{1}{7.2} = \frac{0.25}{7.2} cm^{- 1}, f_{w} = \frac{7.2}{0.25} = 28.8 cm .

The focal length of the lens in water is

288

mm.

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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