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JEE Main 2026 · 6 Apr · Shift 1, Physics question 41

A spherical interface lens of radius RR 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 4R4R in front of spherical interface. The magnitude of the magnification of point source image is ________.

A point source p on the axis, 4R to the left of a spherical surface that bulges towards it; the centre of curvature is to the right (radius R). Medium n₁ = 1 on the left, n₂ = 1.4 on the right.
Choose one option
Answer and worked solution

Official answer: A

Refraction at a curved surface. u=−4Ru=-4R, and the centre of curvature is on the far side, so the radius is +R+R:
1.4v−1−4R=0.4R ⇒ 1.4v=0.15R ⇒ v=283R.\frac{1.4}v-\frac1{-4R}=\frac{0.4}R\ \Rightarrow\ \frac{1.4}v=\frac{0.15}R\ \Rightarrow\ v=\frac{28}3R.

Magnification.
m=n1vn2u=28R/31.4(−4R)=−53,m=\frac{n_1v}{n_2u}=\frac{28R/3}{1.4(-4R)}=-\frac53,
with magnitude about 1.66.

(A)

Topics:Refraction at spherical surfaces
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