Question:

A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m. A plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final image formed by the system is :

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Light passes through the lens, reflects from the plane mirror and passes through the lens again; apply the lens formula twice and treat the mirror image as the object for the second pass.
Updated On: Nov 5, 2026
  • 2.6 m from the mirror, real
  • 1 m from the mirror, real
  • 2.6 m from the mirror, virtual
  • 1 m from the mirror, virtual
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The Correct Option is A

Solution and Explanation

First refraction: \(u = -1\) m, \(f = 0.5\) m. \(\frac1v = \frac1f + \frac1u = 2 - 1 = 1\Rightarrow v = 1\) m (behind the lens).
This image is 1 m from the lens, i.e. 1 m in front of the mirror (mirror at 2 m). The plane mirror forms an image 1 m behind it, i.e. 3 m from the lens.
This image acts as a (virtual) object for the lens on the second pass, at distance \(u = -3\) m.
\(\frac1v = \frac{1}{0.5} - \frac13 = 2 - 0.333 = 1.667\Rightarrow v = 0.6\) m on the side of the lens away from the mirror.
Distance of this final image from the mirror \(= 2 + 0.6 = 2.6\) m. The rays actually converge, so the image is real.
Answer: (1) 2.6 m from the mirror, real
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