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

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Real at a distance of 40cm from converging lens.

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

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

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

answer is A.

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Detailed Solution

Here $f_{1} = - 25 c m, f_{2} = 20 c m$

For diverging lens $v = - 25 c m$

Fro converging lens $u = - (15 + 25) = - 40 c m$

$\frac{1}{v} - (- \frac{1}{40}) = \frac{1}{20}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{40}$

$v = 40 c m$

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