Optics Ques 185
- A biconvex lens of focal length $15$ $ cm$ is in front of a plane mirror. The distance between the lens and the mirror is $10 $ $cm$. A small object is kept at a distance of $30$ $ cm$ from the lens. The final image is
(2010)
(a) virtual and at a distance of $16 $ $cm$ from the mirror
(b) real and at a distance of $16 $ $cm$ from the mirror
(c) virtual and at a distance of $20 $ $cm$ from the mirror
(d) real and at a distance of $20 $ $cm$ from the mirror
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Answer:
Correct Answer: 185.(b)
Solution:
- Object is placed at distance $2 f$ from the lens. So first image $I _1$ will be formed at distance $2 f$ on other side. This image $I _1$ will behave like a virtual object for mirror. The second image $I _2$ will be formed at distance $20$ $ cm$ in front of the mirror, or at distance $10 $ $ cm$ to the left hand side of the lens.
Now applying lens formula
$ \begin{aligned} & \frac{1}{v}-\frac{1}{u} =\frac{1}{f} \\ \therefore & \quad \frac{1}{v}-\frac{1}{+10} =\frac{1}{+15} \\ \text { or } & v =6 cm \end{aligned} $
Therefore, the final image is at distance $16 $ $cm$ from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.
BETA
