Question

When an object is at distance $u_1$ and $u_2$ from a concave mirror, real and virtual images are formed respectively having same magnification. The focal length of the concave mirror is

• A. $u_1-u_2$
• B. $\frac{u_1-u_2}{2}$
• C. $\frac{u_1+u_2}{2}$
• D. $u_1+u_2$

Answer 1

The correct option is C $\frac{u_1+u_2}{2}$

Let $f$ be the focal length of the mirror.

Using mirror equation , magnification $\left(m\right)$ from a concave mirror is given by

$m=\frac{f}{f-u}$

For real image, object distance $u_1>f$

For a concave mirror, a real object produces a real and inverted image.

$\therefore -m=\frac{f}{f-u_1}.....\left(1\right)$

For virtual image , object distance $u_2<f$

$\therefore +m=\frac{f}{f-u_2}.....\left(2\right)$

From Eqs.$\left(1\right)$ and $\left(2\right)$

$\frac{f}{f-u_2}=-\frac{f}{f-u_1}\Rightarrow f=\frac{u_1+u_2}{2}$

Hence, option (c) is the correct answer.