JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Geometrical Optics

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The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm . The refractive index of the lens material is

(A)1.2
(B)1.4
(C)1.5✓ Correct
(D)1.8

Answer: (C)

Solution

The lens maker's formula gives

\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

. For the biconvex lens

R_{1} = + 10 cm

and

R_{2} = - 15 cm

, so

\frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) = (μ - 1) \frac{1}{6} .

Hence

μ - 1 = \frac{6}{12} = \frac{1}{2}

, and the refractive index of the lens material is

1.5

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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