Q144 / 180 · COMEDK 2015 · Physics
A concave mirror gives an image three times as large as its object placed at a distance of $20 \mathrm{~cm}$ from it. For the image to be real, the focal length should be
- A$10 \mathrm{~cm}$
- Correct answerB$15 \mathrm{~cm}$
- C$20 \mathrm{~cm}$
- D$30 \mathrm{~cm}$
Solution
Given, size of image formed by concave mirror,
$$
h_{2}=3 h_{1}
$$
where, $h_{1}$ is the size of object.
$$
m=\frac{h_{2}}{h_{1}}=\frac{3 h_{1}}{h_{1}}=3
$$
But for real image, magnification $m$ will be negative.
$\therefore \quad m=-3$
$\Rightarrow \quad-\frac{v}{u}=-3$
$\Rightarrow \quad v=3 u$
Here, $\quad u=-20 \mathrm{~cm}$,
$\therefore \quad v=3(-20)=-60 \mathrm{~cm}$
By using mirror formula,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$=\frac{1}{-20}+\frac{1}{-60}=-\frac{1}{15}$
$\Rightarrow \quad f=-15 \mathrm{~cm}$
Hence, the correct answer is option: 'B'.