A converging lens of focal length 15cm and a converging mirror of focal length 10cm are placed 50cm apart with common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. the locations of the two images formed are

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one at 20 cm and the other at 24 cm from the lens away from mirror

one at 15 cm and the other at 24 cm from the lens away from mirror

one at 15 cm and the other at 20 cm from the lens away from mirror

one at 40 cm and the other at 20 cm from the lens away from mirror

answer is A.

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Detailed Solution

for convex lens u = 40 cm
F = 15 cm, v = ?
$\begin{array}{l} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ v = + ve, u = - ve \\ \frac{1}{15} = \frac{1}{v} + \frac{1}{40} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{40} \Rightarrow \frac{1}{v} = \frac{1}{24} \\ v = 24 cm \end{array}$

For concave mirror f = 10 cm, u = 10 cm
u = f $\Rightarrow$ image is formed of infinity
$\Rightarrow$ two parallel incident ray incident on convex lens
$\begin{array}{l} \Rightarrow u = \infty, f = 15 cm \\ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{15} = \frac{1}{v} + \frac{1}{\infty} \\ \frac{1}{v} = \frac{1}{15} \\ v = 15 cm \end{array}$