Question

The magnification of the image when an object is placed at a distance x from the principle focus of a mirror of focal length f is?

• A. $\frac{x}{f}$
• B. $1+f/x$
• C. $\frac{f}{x}$
• D. $1-\frac{f}{x}$

Answer 1

The correct option is C $\frac{f}{x}$

$\text{Step 1: Sign Convention and position of Object [Refer fig.]}$

Pole of the mirror is taken as origin, and all distances are taken from pole.

Sign convention $\to$ Direction of incident ray is +ve, i.e. rays coming from object to mirror, rightwards here.

As Object$\left(O\right)$ is leftwards, So object distance $u=-(x+f)$

As focus is also to the left of the pole, So focal length $=-f$

$\text{Step 2: From mirror formula}$

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

$\frac{1}{-f}=\frac{1}{v}+\frac{1}{-(x+f)}$

$\frac{1}{v}=\frac{1}{x+f}-\frac{1}{f}$

$\frac{1}{v}=\frac{f-x-f}{f(x+f)}$

$v=-\frac{f(x+f)}{x}$

$\text{Step 2: Magnification of mirror}$

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

$m=\frac{\frac{-f(x+f)}{x}}{-(x+f)}$ $=-\frac{f}{x}$

$\therefore |m|=\frac{f}{x}$

Hence, Option $C$ is correct