Optics Ques 57
- 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
(2017 Main)
(a) virtual and at a distance of $40 $ $ cm$ from convergent lens
(b) real and at a distance of $40 $ $ cm$ from the divergent lens
(c) real and at a distance of $6 $ $ cm$ from the convergent lens
(d) real and at a distance of $40 $ $ cm$ from convergent lens
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Answer:
Correct Answer: 57.(b)
Solution:
Here, $f _1=-25 $ $cm, f _2=20 $ $cm$
For diverging lens, $v=-25 $ $cm$
For converging lens, $u=-(15+25)=-40 $ $cm$
$\therefore \quad \frac{1}{v}-\frac{1}{-40}=\frac{1}{+20} \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40} \Rightarrow v=40 $ $cm$
BETA
