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

Answer: (B)

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

Answer 2

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).