Q 12-09-219JEE MainJEE Main 2017 (2 Apr)Medium
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:
Answer: (B) Real and at a distance of $40$ cm from convergent lens
Parallel light through the diverging lens forms a virtual image at its focus, $25$ cm in front of it. This image is $25 + 15 = 40$ cm in front of the converging lens and acts as its object ($u = -40$ cm, $f = +20$ cm):
$$\frac1v = \frac1f + \frac1u = \frac1{20} - \frac1{40} = \frac1{40} \;\Rightarrow\; v = +40\ \text{cm}$$
The final image is real, $40$ cm from the converging lens.
Solution by Sreeraj P, M.Sc Physics