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. Real and at a distance of $20cm$ from the mirror

Answer 1

Answer: (B)

Image after refraction from lens,

$\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{15}$

$v_1=30cm$

Hence the image is produced 20 cm behind the mirror.

The mirror reflects the image to produce its image at 20 cm in front of the mirror. This image acts as object for the lens.

$u_2=20-10=10cm$

$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f}$

Thus $v_2=6cm$

This is the distance of image from lens.

Therefore the distance of image from mirror= $6cm+10cm=16cm$