JEE Main 2025 Jan 29 Shift 2, Physics Q9: Refraction at Plane and Spherical surfaces

JEE Main2025Jan 29, Shift 2Physics

Two concave refracting surfaces of equal radii of curvature and refractive index $1.5$ face each other in air as shown in figure. A point object $O$ is placed midway, between $P$ and $B$ . The separation between the images of $O$ , formed by each refracting surface is :

A
$0.214 R$
B
$0.114 R$
C
$0.411 R$
D
$0.124 R$

Solution

Use the single-surface refraction formula $\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$ for each surface, treating each surface as concave (centre of curvature on the object side, so $R$ is taken as $- R$ relative to the light direction).

Surface B (right): object distance $∣ u ∣ = R /2$ , $μ_{1} = 1$ , $μ_{2} = 1.5$ , surface radius $- R$ .

$\frac{1.5}{v _{B}} + \frac{1}{R /2} = - \frac{0.5}{R} \Rightarrow \frac{1.5}{v _{B}} = - \frac{2.5}{R} \Rightarrow v_{B} = - 0.6 R$ .

Surface A (left): object distance $∣ u ∣ = 3 R /2$ , same indices and radius convention.

$\frac{1.5}{v _{A}} + \frac{2}{3 R} = - \frac{0.5}{R} \Rightarrow \frac{1.5}{v _{A}} = - \frac{7}{6 R} \Rightarrow v_{A} = - \frac{9 R}{7}$ .

Both images are virtual, on the same side as the object. With the two vertices separated by $2 R$ , the separation between the images works out to

$2 R - (0.6 R + \frac{9 R}{7}) = 2 R - \frac{13.2 R}{7} \approx 0.114 R$ .

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JEE Main 2025 Jan 29 Shift 2, Physics Q9: Refraction at Plane and Spherical surfaces

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