Question

An object is moving along the principal axis of a concave mirror of focal length $8\text{cm}$. At a distance of $4\text{cm}$ from the mirror, the image formed is twice the height of the object. Then, the magnification produced for object distance $20\text{cm}$ will be

• A. $0.57$
• B. $0.67$
• C. $0.47$
• D. $0.77$

Answer 1

The correct option is B $0.67$

From given data,
$u=-4\text{cm}$

Since the object distance is within the focal length of concave mirror, image formed will be virtual.
$\therefore m=2$

Now using the formula of magnification,
$m=\frac{-v}{u}\Rightarrow 2=\frac{-v}{-4}$
$\Rightarrow v=8\text{cm}$

Now, for $u=-20\text{cm}$, image formed will be real.

Using mirror formula,

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

$\Rightarrow \frac{1}{v}+\frac{1}{-20}=\frac{1}{-8}$

$\Rightarrow \frac{1}{v}=\frac{-3}{40}$

$\Rightarrow v=\frac{-40}{3}\text{cm}$

Then, magnification $m=\frac{-v}{u}$

$\Rightarrow m=-\left(\frac{\frac{-40}{3}}{-20}\right)=-0.67$

or $|m|=0.67$

Hence, option $\left(b\right)$ is the correct answer.

Why this question? In order to solve this question, you should have a solid understanding of the nature/magnification of image formed by a concave mirror at different object distances.