Question

A plane mirror is placed at a distance of $50\text{cm}$ from a concave mirror of focal length $16\text{cm}$. Where should a short object be placed between the mirrors and facing both the mirrors so that its virtual image in the plane mirror coincides with the real image in concave mirror?

• A. $30\text{cm}$ from the concave mirror
• B. $20\text{cm}$ from the plane mirror
• C. $15\text{cm}$ from the concave mirror
• D. $30\text{cm}$ from the plane mirror

Answer 1

The correct option is D $30\text{cm}$ from the plane mirror


Let, the distance between object $\text{AB}$ and the plane mirror be $x$ cm.

$\therefore$ distance between object $\text{AB}$ and concave mirror be $(50-x)\text{cm}$.

From the figure,

$AM=A^{\prime }M=x$
${OA}^{\prime }=OM+A^{\prime }M=(50+x)\text{cm}$

Object distance for concave mirror, $u=-(50-x)$

Image distance for concave mirror, $v=-(50+x)$

Applying the mirror formula,

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

$\Rightarrow \frac{1}{-(50+x)}+\frac{1}{-(50-x)}=\frac{1}{-16}$

$\Rightarrow \frac{50-x+50+x}{{50}^2-x^2}=\frac{1}{16}$

$\Rightarrow 100\times 16=2500-x^2$

$\Rightarrow x^2=900$

$\therefore x=30\text{cm}$

Hence, option (d) is the correct answer.

Why this question: To make the student understand the image formation by combination Spherical and Plane mirrors