JEE Main 2026 Apr 5 Shift 2, Physics Q16: Thin Lens and Mirrors

JEE Main2026Apr 5, Shift 2Physics

An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is _____.

A
virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB
B
real, formed at 7.5 cm right of lens Q, with a size same as that of AB
C
formed at infinity
D
real, formed at 7 cm right of lens Q, with a size smaller than that of AB

Solution

Lens P: Object at $u = - 15$ cm, $f = + 10$ cm. $\frac{1}{v} - \frac{1}{- 15} = \frac{1}{10} \Rightarrow v = + 30$ cm.

The image from P forms 30 cm right of P, which is 15 cm right of Q (i.e., 15 cm beyond Q on the far side). For lens Q this image acts as a virtual object at $u_{Q} = + 15$ cm.

Lens Q: $\frac{1}{v ^{'}} - \frac{1}{15} = \frac{1}{15} \Rightarrow v^{'} = + 7.5$ cm. Image is real, 7.5 cm right of Q.

Magnification: $m_{P} = \frac{v}{u} = \frac{30}{- 15} = - 2$ , and $m_{Q} = \frac{7.5}{15} = 0.5$ . Net $m = - 2 \times 0.5 = - 1$ , so the image is the same size as AB and inverted.

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JEE Main 2026 Apr 5 Shift 2, Physics Q16: Thin Lens and Mirrors

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