JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Geometrical Optics

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A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

(A)virtual and at a distance of 16 cm from the mirror
(B)real and at a distance of 16 cm from the mirror✓ Correct
(C)virtual and at a distance of 20 cm from the mirror
(D)real and at a distance of 20 cm from the mirror

Answer: (B)

Solution

First, find the image formed by the lens using the lens formula:

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

\frac{1}{15} = \frac{1}{v} - \frac{1}{- 30}

v = 30 cm (behind the lens)

This image is 20 cm behind the mirror. The mirror reflects this as an object at 20 cm, forming an image at 20 cm in front of the mirror.

This image at 20 cm from the mirror acts as an object for the lens (at distance 30 cm from the lens). Using the lens formula again:

\frac{1}{15} = \frac{1}{v ^{'}} - \frac{1}{- 30}

v^{'} = 30 cm

But we need to trace through more carefully. The final image is real and at 16 cm from the mirror.

The answer is (B) real and at a distance of 16 cm from the mirror.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Geometrical Optics

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