Question

A biconvex lens of focal length $15cm$ is in front of a plane mirror.The distance between lens and the mirror is $10cm.$ A small object is kept at a distance of $30cm$ from the lens. The final image is:

• A. virtual and at a distance of $16cm$ from the mirror
• B. real and at a distance of $16cm$ from the mirror
• C. virtual and at a distance of $20cm$ from the mirror
• D. real and at a distance of $20cm$ from the mirror

Answer 1

Answer: (B)

real and at a distance of $16cm$ from the mirror

$\begin{array}{c}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}-\frac{1}{-30}\\ v=+30\\ \frac{1}{15}=\frac{1}{v}-\frac{1}{10}\Rightarrow \frac{1}{15}+\frac{1}{10}=\frac{1}{v}\\ v=\frac{150}{25}=+6cm\end{array}$

The final image ${\hat{I}}_3$ is $6cm$ (real) from

the lens.

$10+6=16cm$

from the mirror ( real)

So option $B$ is correct.