JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Geometrical Optics

A thin biconvex lens is prepared from the glass (

μ = 1.5

) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of ____ cm.

(A)10✓ Correct
(B)12.5
(C)13
(D)13.5

Answer: (A)

Solution

For the equiconvex lens,

\frac{1}{f _{L}} = (μ - 1) (\frac{1}{R} + \frac{1}{R}) = (0.5) \frac{2}{20} = \frac{1}{20}

, so

f_{L} = 20 cm

. The silvered surface behaves as a concave mirror of focal length

f_{m} = \frac{R}{2} = 10 cm

. Light crosses the lens, reflects at the mirror and crosses the lens again, so the system acts as a single mirror whose power is

P = 2 P_{L} + P_{m}

, giving

\frac{1}{F} = \frac{2}{f _{L}} + \frac{1}{f _{m}} = \frac{2}{20} + \frac{1}{10} = \frac{1}{5}

and

F = 5 cm

. Object and image coincide when the object sits at the centre of curvature of this equivalent mirror, that is at

2 F

. The object should be placed at

10 cm

from the lens.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Geometrical Optics

Solution

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