Start free

Physics · Geometrical Optics

JEE Main 2025 — 3 April, Evening Shift — Question 68

Light from a point source in air falls on a spherical glass surface (refractive index, μ=1.5\mu=1.5 and radius of curvature =50 cm=50 \mathrm{~cm} ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is \qquad m.

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

V=200 cmV=200 \mathrm{~cm}

μ2v−μ1u=μ2−μ1R\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}

1.5200−1u=1.5−150=1100\frac{1.5}{200}-\frac{1}{u}=\frac{1.5-1}{50}=\frac{1}{100}

1u=152000−1100=15−202000=−52000\frac{1}{u}=\frac{15}{2000}-\frac{1}{100}=\frac{15-20}{2000}=\frac{-5}{2000}

u=−400 cm=−4 mu=-400 \mathrm{~cm}=-4 \mathrm{~m}

Solution figure

Answer key and solution verified before publishing.

Practise Geometrical Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Refraction at Curved Surface and Glass Sphere

More Geometrical Optics questions from this paper

Light from a point source in air falls on a spherical glass surface… | JEE Main 2025 PYQ with Solution · DhiX AI