Question

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

• 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

Answer 1

The Correct Option is B

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
$\hspace25mm \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\therefore \hspace15mm \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15}$
or $\hspace25mm$ v = 6 cm
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.
$\therefore$ The correct option is (b).