Optics 3 Question 10
10. 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: 10. (b)
Solution:
Here, $f _1=-25 cm, f _2=20 cm$
For diverging lens, $v=-25 cm$
For converging lens, $u=-(15+25)=-40 cm$
$\therefore \quad \frac{1}{v}-\frac{1}{-40}=\frac{1}{+20} \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40} \Rightarrow v=40 cm$
