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JEE MAIN 2026 · Optics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 30

Updated
Single correct+4 / -1Answer matches NTA’s final key

A thin biconvex lens is prepared from the glass (μ=1.5\mu = 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.

Choose one

Solution

Solution

Focal length of the lens. 1f=(1.5−1)(120+120)=120\dfrac1f = (1.5 - 1)\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = \dfrac{1}{20}, so f=20f = 20 cm.

The silvered surface acts as a concave mirror of radius 20 cm: fm=10f_m = 10 cm.

The combination is light passing through the lens, reflecting, and passing back through. Its power (in the concave-mirror sense) adds as

P=2Plens+Pmirror=220+110=15 cm−1.P = 2P_{\text{lens}} + P_{\text{mirror}} = \frac{2}{20} + \frac{1}{10} = \frac15\ \text{cm}^{-1}.

So the system behaves as a concave mirror of focal length F=5F = 5 cm. An image forms on the object itself when the object is at the centre of curvature, 2F=102F = 10 cm.

Answer: 10 cm.

Answer checked against the official answer key. The solution is our own working.

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