JEE Physics 2025 Question with Solution

Given: Object distance is $u_1 = -40 \, \text{cm}$ and magnification is $m_1 = \frac{1}{2}$.

Find: How the object must be moved so that magnification becomes $m_2 = \frac{1}{3}$.

Using the magnification relation for a spherical mirror,

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

For the first case,

$$\frac{1}{2} = \frac{f}{f-(-40)} = \frac{f}{f+40}$$ $$f + 40 = 2f$$ $$f = 40 \, \text{cm}$$

Now for magnification $\frac{1}{3}$,

$$\frac{1}{3} = \frac{40}{40-u}$$ $$40 - u = 120$$ $$u = -80 \, \text{cm}$$

So the object distance changes from $40 \, \text{cm}$ to $80 \, \text{cm}$, which means the object is moved $40 \, \text{cm}$ farther from the mirror.

Therefore, the correct option is A.

Given: Initially, the object is at $u_1 = 40 \, \text{cm}$ away from the mirror and the magnification is $m_1 = \frac{1}{2}$.

Find: The displacement needed to make the magnification $m_2 = \frac{1}{3}$.

For a spherical mirror,

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

and magnification is

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

So,

$$v = -mu$$

Substitute this into the mirror formula:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{-mu}$$ $$\frac{1}{f} = \frac{m-1}{mu}$$ $$f = \frac{mu}{m-1}$$

For the first condition,

$$f = \frac{\left(\frac{1}{2}\right)(40)}{\frac{1}{2}-1}$$ $$f = \frac{20}{-\frac{1}{2}} = -40 \, \text{cm}$$

For the second condition,

$$-40 = \frac{\left(\frac{1}{3}\right)u_2}{\frac{1}{3}-1}$$ $$-40 = \frac{\frac{1}{3}u_2}{-\frac{2}{3}} = -\frac{u_2}{2}$$ $$u_2 = 80 \, \text{cm}$$

Hence the object must go from $40 \, \text{cm}$ to $80 \, \text{cm}$, so it is moved $40 \, \text{cm}$ away from the mirror.

The solution also notes a discrepancy because one step states $u = -80 \, \text{cm}$ while the listed option text concludes "40 cm away from the mirror". The defensible final conclusion from the worked comparison is that the object is moved $40 \, \text{cm}$ away from the mirror, so the correct option is A.