JEE Main 2025 Apr 2 Shift 2, Physics Q14: Magnification

JEE Main2025Apr 2, Shift 2Physics

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of $12$ cm, at same distance of $18$ cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

A
$1/2$
B
$2$
C
$3$
D
$1/3$

Solution

Both mirrors have $∣ f ∣ = R /2 = 6$ cm. Use the magnification formula $m = \frac{f}{u - f}$ with the sign convention that gives the size ratio directly.

For the convex mirror, $f = + 6$ cm, $u = - 18$ cm (or equivalently using the form above with $u = 18$ cm distance): $∣ m_{convex} ∣ = \frac{6}{18 + 6} = \frac{6}{24} = \frac{1}{4}$ .

For the concave mirror, $f = - 6$ cm: $∣ m_{concave} ∣ = \frac{6}{18 - 6} = \frac{6}{12} = \frac{1}{2}$ .

Ratio of image sizes: $\frac{∣ m _{convex} ∣}{∣ m _{concave} ∣} = \frac{1/4}{1/2} = \frac{1}{2}$ .

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JEE Main 2025 Apr 2 Shift 2, Physics Q14: Magnification

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