JEE Main 2017 · 2 Apr · Shift 1 · Physics Q22

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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 :

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 and worked solution

Official answer: A

The diverging lens forms a virtual image of the parallel beam 25 cm in front of it, i.e. $25+15=40$ cm from the converging lens. For that lens, $u=-40$ cm and $f=20$ cm:
$\frac1v=\frac1{20}-\frac1{40}\implies v=40\text{ cm},$
a real image 40 cm beyond the converging lens.

Topics:Combination of lenses and mirrors Thin lenses and lens maker's formula

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