Question 81: Ray Optics

Question

Distance between an object and three times magnified real image is 40 cm . The focal length of the mirror used is $\_\_\_\_$ cm .

Accepted Answer

Since the image is real, the mirror must be a concave mirror.
For a real image formed by a concave mirror, the magnification is negative.
$$ \mathrm{m}=-3 $$
We know that $m=-\frac{v}{u}$, where $m$ is the magnification of the image, $v$ is the image distance and $u$ is the object distance.
$$\Rightarrow $$ $$ -3=-\frac{v}{u} \Rightarrow v=3 u $$
The distance between the object and the image is 40 cm . In a concave mirror, for a real magnified image, both the object ( $u$ ) and the image ( $v$ ) are on the same side of the mirror.
The distance between them is $|\mathrm{v}|-|\mathrm{u}|$.
$$ |\mathrm{v}|-|\mathrm{u}|=40 \mathrm{~cm} $$
Substituting $\mathrm{v}=3 \mathrm{u}$,
$$ 3|\mathrm{u}|-|\mathrm{u}|=40 $$
$$\Rightarrow $$ $$ 2|\mathrm{u}|=40 \Rightarrow|\mathrm{u}|=20 \mathrm{~cm} $$
Using sign convention, the object distance is negative; $\mathrm{u}=-20 \mathrm{~cm}$.
Therefore, the image distance is :
$$ \mathrm{v}=3 	imes(-20)=-60 \mathrm{~cm} $$
Using the mirror formula :
$$ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} $$
Substituting the values :
$$ \frac{1}{f}=\frac{1}{-60}+\frac{1}{-20} $$
$$\Rightarrow $$ $$ \frac{1}{f}=\frac{-1-3}{60} \Rightarrow f=\frac{60}{-4}=-15 \mathrm{~cm} $$
Therefore, the focal length of the concave mirror is -15 cm . The negative sign indicates it is a concave mirror.
Hence, the correct option is (D)

Answer

-20

Answer

$-15 / 2$

Answer

-10

Answer

-15

Ray Optics

Distance between an object and three times magnified real image is 40 cm . The focal length of the mirror used is \_\_\_\_ cm .

Choose the correct answer: