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. real and at a distance of 16 cm from the mirror
• B. virtual 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

Answer: (A)

We get the image distance due to the lens using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

or

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

or

$v=30cm$ in front of the convex mirror, which means $20$cm behind the mirror. Assuming mirror to reflect from both the sides and this image will act as virtual object for mirror and second image, $I_2$ will be formed at 20 cm in front of mirror. That is $10$ cm in front of lens.

Understanding the convention used while taking +ve and -ve sign for image and object distance, the direction along which ray of light travels is taken as +ve.

So, here, the ray of light is coming from behind the mirror towards lens, for the second image, $I_2$ with respect to lens,

$u=+10cm;f=15cm$

So, applying lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

or

$\frac{1}{v}-\frac{1}{10}=\frac{1}{15}$

or

$v=6cm$ from lens (in front of lens)

So this image is at a distance of 16 (10 + 6) cm in front of mirror.