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Physics · Geometrical Optics

JEE Main 2026 — 22 January, Morning Shift — Question 46

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm . Refractive index of glass is 1.5 . The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is ____\_\_\_\_ cm .

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

μ2v−μ1u=μ2−μ1R⇒1.5v−1∞=1.5−150\frac{\mu_{2}}{\mathrm{v}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}} \Rightarrow \frac{1.5}{\mathrm{v}}-\frac{1}{\infty}=\frac{1.5-1}{50} V=150 cm\mathrm{V}=150 \mathrm{~cm} x→\mathrm{x} \rightarrow measure from center x=V−Rx=V-R =150−50=100 cm=150-50=100 \mathrm{~cm}

Solution figure

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Refraction at Curved Surface and Glass Sphere

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