JEE Physics 2024 Question with Solution

Given: The image is virtual and two-times magnified, so magnification is $m = +2$. The distance between the object and image is $15 \, \text{cm}$.

Find: The focal length $f$ of the curved mirror.

A virtual magnified image by a mirror is formed by a concave mirror. For mirrors,

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

Using $m = 2$,

$$2 = -\frac{v}{u}$$

so,

$$v = -2u$$

The distance between object and image is

$$|v-u| = 15$$

Substituting $v = -2u$,

$$|-2u-u| = 15$$ $$|3u| = 15$$

Thus,

$$u = -5 \, \text{cm}$$

Now,

$$v = -2u = 10 \, \text{cm}$$

Using the mirror formula,

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

Substitute $v = 10 \, \text{cm}$ and $u = -5 \, \text{cm}$:

$$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$$ $$\frac{1}{f} = \frac{1}{10} - \frac{2}{10} = -\frac{1}{10}$$

Hence,

$$f = -10 \, \text{cm}$$

Therefore, the focal length of the mirror is $-10 \, \text{cm}$. The correct option is C.

Given: Magnification $m=+2$ and distance between object and image $=15 \, \text{cm}$.

Find: The focal length $f$.

The solution explicitly states that the correct option is C and concludes that the focal length is $-10 \, \text{cm}$.

Its working uses

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

with

$$v=-2u$$

and then the mirror formula

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

to obtain

$$f=-10 \, \text{cm}$$

Although one alternate approach block on the page contains sign inconsistencies in intermediate steps, its final stated answer still matches $-10 \, \text{cm}$. Therefore, from the solution, the correct option is C.