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 $16cm$ 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. None of the above

Answer 1

Answer: (B)

Object is placed at distance 2 $f$ from the lens. So first image $I$ 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
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$


$\therefore \frac{1}{v}-\frac{1}{+ 10}=\frac{1}{+ 15}$
$or \, v=16 \, cm$
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.