A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is

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Real and at a distance of 40 cm from convergent lens

Virtual and at a distance of 40 cm from convergent lens.

Real and at a distance of 40 cm from the divergent lens.

Real and at a distance of 6 cm from the convergent lens.

answer is A.

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Detailed Solution

$\frac{1}{f_{1}} = \frac{1}{V_{1}} - \frac{1}{u_{1}} \Rightarrow \Rightarrow \frac{- 1}{25} = \frac{1}{V_{1}} - \frac{1}{\infty}$

$\Rightarrow V_{1} = - 25 cm$ .

Image from concave will be formed at its focus will act as the object for convex lens.

Object distance u=25+15=40cm.

$\frac{1}{f_{2}} = \frac{1}{V_{2}} - \frac{1}{u_{2}}$

$\Rightarrow \frac{1}{20} = \frac{1}{V_{2}} + \frac{1}{40} \Rightarrow V_{2} = 40 cm$

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