JEE Main 2026 Apr 8 Shift 2, Physics Q5: Thin Lens and Mirrors

JEE Main2026Apr 8, Shift 2Physics

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
$10$
B
$12.5$
C
$13$
D
$13.5$

Solution

Silvering the left curved surface turns the system into an equivalent mirror whose power equals the sum of contributions from one lens refraction (entering from the right), one mirror reflection (at the silvered surface), and a second lens refraction (exiting on the right).

Lens focal length: $\frac{1}{f _{L}} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.5 (\frac{1}{20} - \frac{1}{- 20}) = \frac{1}{20}$ , so $f_{L} = 20$ cm.

Mirror focal length of the silvered concave surface (radius $20$ cm, concave toward the incident light): $f_{M} = R /2 = 10$ cm.

Equivalent mirror power: $P_{eq} = 2 P_{L} + P_{M} ⟹ \frac{1}{f _{eq}} = \frac{2}{f _{L}} + \frac{1}{f _{M}} = \frac{2}{20} + \frac{1}{10} = \frac{1}{5}$ .

So $f_{eq} = 5$ cm (concave-mirror-like).

For image and object to coincide, the object must lie at the centre of curvature of the equivalent mirror, i.e. at $2 f_{eq} = 10$ cm.

Concept behind this question

Thin Lens and MirrorsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 8 Shift 2, Physics Q5: Thin Lens and Mirrors

Concept behind this question

More previous year questions on Thin Lens and Mirrors

Practice more Physics