JEE Main 2025 Apr 3 Shift 1, Physics Q18: Thin Lens and Mirrors

JEE Main2025Apr 3, Shift 1Physics

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
D
1.8

Solution

Lensmaker's equation: $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) .$

For a biconvex lens with $R_{1} = + 10$ cm and $R_{2} = - 15$ cm:

$\frac{1}{12} = (μ - 1) (\frac{1}{10} - \frac{1}{- 15}) = (μ - 1) \cdot \frac{3 + 2}{30} = (μ - 1) \cdot \frac{1}{6} .$

$\Rightarrow μ - 1 = \frac{6}{12} = \frac{1}{2} \Rightarrow μ = 1.5.$

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JEE Main 2025 Apr 3 Shift 1, Physics Q18: Thin Lens and Mirrors

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