JEE Main 2026 Apr 6 Shift 1, Physics Q16: Refraction at Plane and Spherical surfaces

JEE Main2026Apr 6, Shift 1Physics

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 $4 R$ in front of spherical interface. The magnitude of the magnification of point source image is _________.

A
$1.66$
B
$2.33$
C
$2.66$
D
$1.33$

Solution

Use the single-surface refraction formula with sign convention (light moves left to right, distances measured from pole, $u = - 4 R$ , $R$ positive for convex toward incoming light):

$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ .

$\frac{1.4}{v} - \frac{1}{- 4 R} = \frac{0.4}{R}$ .

$\frac{1.4}{v} = \frac{0.4}{R} - \frac{1}{4 R} = \frac{1.6 - 1}{4 R} = \frac{0.6}{4 R} = \frac{0.15}{R}$ .

$v = \frac{1.4 R}{0.15} = \frac{28 R}{3}$ .

Lateral magnification: $m = \frac{n _{1} v}{n _{2} u} = \frac{1 \cdot ( 28 R /3 )}{1.4 \cdot ( - 4 R )} = \frac{28}{3 \cdot ( - 5.6 )} = - \frac{28}{16.8} \approx - 1.67$ .

$∣ m ∣ \approx 1.66$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q16: Refraction at Plane and Spherical surfaces

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