Question

When an object is at distances of $u_1$ and $u_2$ from the poles of a concave mirror, images of the same size are formed. The focal length of the mirror is:

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

Answer 1

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

The lens formula is given as

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

Therefore when the object is placed at a distance $u_1$ and image is formed at a distance $v_1$,

$\therefore \frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}$

$\Rightarrow \frac{1}{v_1}=\frac{1}{f}-\frac{1}{u_1}$

$\Rightarrow \frac{1}{v_1}=\frac{u_1-f}{u_{1}f}$

$\Rightarrow \frac{v_1}{u_1}=\frac{f}{u_1-f}=m_1$ (magnification) ----------$\left(1\right)$

When the object is placed at a distance $u_2$ for a virtual image the image distance becomes $-v_2$,

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

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

$\Rightarrow \frac{1}{v_2}=\frac{f-u_2}{u_{2}f}$

$\Rightarrow \frac{v_2}{u_2}=\frac{f}{f-u_2}=m_2$ (magnification) ----------$\left(2\right)$

It is given that the images are of the same size.

Hence, $m_1=m_2$

$\frac{f}{u_1-f}=\frac{f}{f-u_2}$

$\Rightarrow u_1-f=f-u_2$

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

Hence the correct answer is option (C).