Physics · Geometrical Optics

JEE Main 2026 — 6 April, Morning Shift — Question 12

A spherical interface lens of radius R separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of 4R in front of spherical interface. The magnitude of the magnification of point source image is ______.

Option A:
1.66
Correct
Option B:
2.33
Option C:
2.66
Option D:
1.33

Answer: A

Step-by-step solution

Using refraction formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}$ . Here $n_1=1, n_2=1.4, u=-4R, R$ positive? Since center of curvature on the right? Assuming convex interface, $R$ positive. $\frac{1.4}{v} - \frac{1}{-4R} = \frac{0.4}{R} \Rightarrow \frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R} = \frac{2}{5R}$ . Solve: $\frac{1.4}{v} = \frac{2}{5R} - \frac{1}{4R} = \frac{8-5}{20R} = \frac{3}{20R} \Rightarrow v = \frac{1.4 \times 20R}{3} = \frac{28R}{3}$ . Magnification $m = \frac{n_1 v}{n_2 u} = \frac{1 \times (28R/3)}{1.4 \times (-4R)} = \frac{28/3}{-5.6} = -\frac{28}{16.8} = -1.666$ . Magnitude = 1.66.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Geometrical Optics
Topic: Refraction at Curved Surface and Glass Sphere