JEE Main 2026 Apr 2 Shift 1, Physics Q16: Refraction at Plane and Spherical surfaces

JEE Main2026Apr 2, Shift 1Physics

Refer the figure given below. $μ_{1}$ and $μ_{2}$ are refractive indices of air and lens material. The height of image will be _____ cm.

A
$1$
B
$0.5$
C
$1.2$
D
$0.25$

Solution

Apply refraction at a single spherical surface with sign convention ( $u = - 40$ cm, $R = - 20$ cm based on the geometry of the figure):

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

$\frac{1.54}{v} - \frac{1}{- 40} = \frac{1.54 - 1}{- 20} = - \frac{0.54}{20}$

$\frac{1.54}{v} = - \frac{1}{40} - \frac{0.54}{20} = - \frac{2.08}{40} \Rightarrow v \approx - 29.61$ cm

Lateral magnification for a single refracting surface: $m = \frac{μ _{1} v}{μ _{2} u} = \frac{1 \times ( - 29.61 )}{1.54 \times ( - 40 )} \approx 0.48$ .

$h_{i} = m \times h_{o} = 0.48 \times 2 \approx 0.96 \approx 1$ cm.

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JEE Main 2026 Apr 2 Shift 1, Physics Q16: Refraction at Plane and Spherical surfaces

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