Optics 7 Question 12
12. 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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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
$$ \begin{aligned} & \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \therefore & \frac{1}{v}-\frac{1}{+10} & =\frac{1}{+15} \\ \text { or } & v & =6 cm \end{aligned} $$
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.
