JEE Advanced 2021 Paper 1 · Q03 · Ray Optics – Lenses

Question

An extended object is placed at point O, $10$ cm in front of a convex lens $L_1$. A concave lens $L_2$ is placed $10$ cm behind $L_1$ (i.e., on the side opposite to the object, coaxially). The radii of curvature of all curved surfaces of both lenses are $20$ cm. The refractive index of both lenses is $1.5$. The total magnification of this lens system is

SolveKar AI · Accepted Answer

Using lensmaker's: 1/f₁ = (1.5−1)(1/20 + 1/20) = 1/20, so f₁ = 20 cm. 1/f₂ = (1.5−1)(−1/20 − 1/20) = −1/20, so f₂ = −20 cm. For L₁: u = −10, 1/v − 1/(−10) = 1/20 ⇒ v = −20 cm (virtual, same side as O). m₁ = v/u = −20/−10 = 2. For L₂: object is at −20 cm from L₁, i.e. distance 30 cm to the left of L₂, so u′ = −30 cm. 1/v′ − 1/(−30) = 1/(−20) ⇒ v′ = −12 cm. m₂ = v′/u′ = −12/−30 = 0.4. Total magnification = m₁ × m₂ = 2 × 0.4 = 0.8.

JEE Advanced 2021 Paper 1 · Q03 · Ray Optics – Lenses

Solution