JEE Physics 2025 Question with Solution

Given: Magnification is $m=-\frac{1}{3}$ and the distance between the object and its image is $30 \, \text{cm}$.

Find: The magnitude of $x$ if the focal length is $\frac{x}{4} \, \text{cm}$.

For a mirror,

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

So,

$$-\frac{1}{3}=-\frac{v}{u}$$ $$v=\frac{u}{3}$$

The distance between object and image is

$$|v-u|=30$$

Substituting $v=\frac{u}{3}$,

$$\left|\frac{u}{3}-u\right|=30$$ $$\left|\frac{-2u}{3}\right|=30$$ $$\frac{2|u|}{3}=30$$ $$|u|=45 \, \text{cm}$$

Since the object is in front of the mirror,

$$u=-45 \, \text{cm}$$

Then,

$$v=\frac{-45}{3}=-15 \, \text{cm}$$

Using the mirror formula,

$$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$ $$\frac{1}{f}=\frac{1}{-45}+\frac{1}{-15}$$ $$\frac{1}{f}=-\left(\frac{1}{45}+\frac{3}{45}\right)=-\frac{4}{45}$$ $$f=-\frac{45}{4} \, \text{cm}$$

Comparing with $\frac{x}{4} \, \text{cm}$,

$$\frac{x}{4}=-\frac{45}{4}$$ $$x=-45$$

Therefore, the magnitude of $x$ is $45$.

Given: $m=-\frac{1}{3}$ and object-image distance is $30 \, \text{cm}$.

Find: Magnitude of $x$ when focal length is written as $\frac{x}{4} \, \text{cm}$.

From magnification,

$$m=-\frac{v}{u}=-\frac{1}{3}$$

Hence,

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

Now use the distance condition,

$$|v-u|=30$$ $$\left|\frac{u}{3}-u\right|=30$$ $$\left|\frac{-2u}{3}\right|=30$$ $$|u|=45 \, \text{cm}$$

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

$$u=-45 \, \text{cm}$$

and therefore,

$$v=-15 \, \text{cm}$$

Now apply

$$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$ $$\frac{1}{f}=\frac{1}{-45}+\frac{1}{-15}=-\frac{4}{45}$$

Thus,

$$f=-\frac{45}{4} \, \text{cm}$$

So,

$$\frac{x}{4}=-\frac{45}{4}$$

which gives

$$x=-45$$

Hence, $|x|=45$. The required numerical answer is $45$.