JEE Main 2025 Apr 4 Shift 1, Physics Q9: Magnification

JEE Main2025Apr 4, Shift 1Physics

When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$ , the object is to be moved :

A
40 cm away from the mirror.
B
80 cm away from the mirror.
C
20 cm towards the mirror.
D
20 cm away from the mirror.

Solution

For a spherical mirror, magnification can be written $m = \frac{f}{f - u}$ (with the standard sign convention).

First condition: $m = \frac{1}{2}$ with $u = - 40$ cm gives $\frac{1}{2} = \frac{f}{f + 40}$ , so $f + 40 = 2 f$ , hence $f = 40$ cm (a convex mirror).

Second condition: $m = \frac{1}{3}$ , $\frac{1}{3} = \frac{40}{40 - u}$ , giving $40 - u = 120$ , so $u = - 80$ cm.

The object distance increases from 40 cm to 80 cm — the object must be moved 40 cm farther from the mirror.

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JEE Main 2025 Apr 4 Shift 1, Physics Q9: Magnification

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