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Physics Question 41 – JEE-MAIN 2026

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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 _______.

Identify the given parameters: refractive indices of the two media, object distance, and radius of curvature.

● Video Walkthrough
Step 1: Apply Spherical Refraction Formula✦ Active

Given refractive indices n1=1 and n2=1.4. The object is placed at a distance 4R in front of the spherical interface, so the object distance is u=−4R. The interface is convex from the left, so the radius of curvature R is positive. The formula for refraction at a spherical surface is:

n2v−n1u=n2−n1R

Substitute the given values:

1.4v−1−4R=1.4−1R
Step 2: Calculate Image Distance○ Expand

Simplify the equation to find the image distance v:

1.4v+14R=0.4R

Rearrange and solve for v:

1.4v=0.4R−14R=1.64R−14R=0.64R

Therefore,

v=1.4×4R0.6=5.6R0.6=56R6=28R3
💡 Teacher's Secret Hint

Ensure correct algebraic manipulation and common denominators when combining terms.

Step 3: Calculate Magnitude of Magnification○ Expand

The transverse magnification m for refraction at a spherical surface is given by:

m=n1vn2u

Substitute the values of n1, n2, u, and v:

m=(1)(28R3)(1.4)(−4R)=28R3−5.6R=283×(−5.6)=28−16.8=−280168=−53

The magnitude of the magnification is |m|=|−53|=53≈1.666... which rounds to 1.66.

💡 Teacher's Secret Hint

Remember that magnification can be negative, indicating an inverted image. The question asks for the magnitude.

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