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

Answer: (b). Lens P: u = -15, f = 10: v = 30 cm, m₁ = 30/-15 = -2. This image is 15 cm beyond Q: virtual object, u = +15, f = 15: 1/v = 1/15 + 1/15 ⇒ v = 7.5 cm. m₂ = 7.5/15 = 0.5; total m = -2 × 0.5 = -1. So the answer is (b) real, 7.5 cm right of Q, same size.

An object AB is placed 15 cm on the left of a convex lens P of focal… | JEE Main 2026 (5 Apr, Shift 2)

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Combination of Lenses

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 ______.

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Answer: (b)

Lens P: $u = -15$ , $f = 10$ : $v = 30\text{ cm}$ , $m_1 = \frac{30}{-15} = -2$ .
This image is 15 cm beyond Q: virtual object, $u = +15$ , $f = 15$ : $\frac{1}{v} = \frac{1}{15} + \frac{1}{15} \Rightarrow v = 7.5\text{ cm}$ .
$m_2 = \frac{7.5}{15} = 0.5$ ; total $m = -2 \times 0.5 = -1$ .
So the answer is (b) real, 7.5 cm right of Q, same size.

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