JEE Physics 2026 Question with Solution

Given: Distance between the object and image is $40 \, \text{cm}$. The image is real and three times magnified, so magnification is $m = -3$.

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

For mirrors,

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

Since the image is real and three times magnified,

$$m = -3$$

Therefore,

$$v = -3u$$

The separation between object and image is

$$|v-u| = 40$$

Substituting $v = -3u$,

$$|-3u-u| = 40$$ $$4|u| = 40$$ $$u = -10 \, \text{cm}$$

Hence,

$$v = 30 \, \text{cm}$$

Now use the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ $$\frac{1}{f} = \frac{1}{30} - \frac{1}{10}$$ $$\frac{1}{f} = -\frac{1}{15}$$ $$f = -15 \, \text{cm}$$

Therefore, the focal length of the mirror is $-15 \, \text{cm}$, so the correct option is D.

Given: A real image is formed and its magnification is three times.

Find: The focal length of the mirror.

For a real image formed by a mirror, magnification is always negative. So take

$$m = -3$$

Using

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

we get

$$\frac{v}{u} = -3$$ $$v = -3u$$

The object-image distance is $40 \, \text{cm}$, so

$$|v-u| = 40$$

Substitute $v = -3u$:

$$|-3u-u| = 40$$ $$|-4u| = 40$$ $$4|u| = 40$$ $$|u| = 10$$

Using Cartesian sign convention for a real object in front of the mirror,

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

Then,

$$v = -3(-10) = 30 \, \text{cm}$$

Apply mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ $$\frac{1}{f} = \frac{1}{30} + \frac{1}{-10}$$ $$\frac{1}{f} = \frac{1}{30} - \frac{3}{30}$$ $$\frac{1}{f} = -\frac{2}{30} = -\frac{1}{15}$$ $$f = -15 \, \text{cm}$$

Therefore, the correct option is D.