JEE Physics 2025 Question with Solution

Given: Magnification of the image is $m = -3$. The distance between the object and image is $|v-u| = 20 \, \text{cm}$.

Find: The magnitude of the radius of curvature $R$ of the concave mirror.

For mirrors, the magnification relation is

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

Substituting $m = -3$,

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

so

$$v = 3u$$

Using the given distance between object and image,

$$|v-u| = 20$$ $$3u-u = 20$$ $$2u = 20$$ $$u = 10 \, \text{cm}$$

Hence,

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

Now apply the mirror formula,

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

Substituting the values,

$$\frac{1}{f} = \frac{1}{10} + \frac{1}{30}$$ $$\frac{1}{f} = \frac{3+1}{30} = \frac{4}{30} = \frac{2}{15}$$

Therefore,

$$f = \frac{15}{2} = 7.5 \, \text{cm}$$

For a spherical mirror,

$$R = 2f$$

Thus,

$$R = 2 \times 7.5 = 15 \, \text{cm}$$

Therefore, the magnitude of the radius of curvature is $15 \, \text{cm}$ and the correct option is C.

Given: A concave mirror forms an image with magnification $-3$, and the separation between object and image is $20 \, \text{cm}$.

Find: The magnitude of the radius of curvature $R$.

The key relations are:

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

and

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

Also,

$$f = \frac{R}{2}$$

From magnification,

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

which gives

$$v = 3u$$

Now use the object-image separation:

$$|v-u| = 20$$ $$|3u-u| = 20$$ $$2u = 20$$ $$u = 10 \, \text{cm}$$

Then,

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

Substitute into the mirror formula:

$$\frac{1}{f} = \frac{1}{10} + \frac{1}{30}$$ $$\frac{1}{f} = \frac{4}{30} = \frac{2}{15}$$ $$f = 7.5 \, \text{cm}$$

Finally,

$$R = 2f = 15 \, \text{cm}$$

Therefore, the radius of curvature is $15 \, \text{cm}$.