A diverging lens with magnitude of focal length 25 cm 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: (a) real and at a distance of 6 cm from the convergent lens. (b) real and at a distance of 40 cm from convergent lens. (c) virtual and at a distance of 40 cm from convergent lens. (d) real and at a distance of 40 cm from the divergent lens.

Answer: (b). Diverging lens ( f = -25) makes the parallel beam appear to come from 25 cm before it. For the converging lens this object is 25 + 15 = 40 cm away: u = -40, f = 20. 1/v = 1/20 - 1/40 = 1/40 ⇒ v = +40 cm (real) So the answer is (b) real, 40 cm from the convergent lens.

A diverging lens with magnitude of focal length 25 cm is placed at a… | JEE Main 2017

Q65JEE Main 2017Medium

Combination of Lenses

A diverging lens with magnitude of focal length 25 cm 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:

Show answer and solution

Answer: (b)

Diverging lens ( $f = -25$ ) makes the parallel beam appear to come from 25 cm before it.
For the converging lens this object is $25 + 15 = 40\text{ cm}$ away: $u = -40$ , $f = 20$ .
$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \Rightarrow v = +40\text{ cm}$ (real)
So the answer is (b) real, 40 cm from the convergent lens.

Report a mistake

Combination of Lenses question from JEE Main 2017

More questions on Combination of Lenses