JEE Main 2025 Jan 28 Shift 2, Physics Q13: Reflection at Plane and Spherical Surfaces

JEE Main2025Jan 28, Shift 2Physics

A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is $- 3$ , then the magnitude of the radius of curvature of the mirror is :

A
$3.75$ cm
B
$30$ cm
C
$7.5$ cm
D
$15$ cm

Solution

Magnification $m = - v / u = - 3$ implies $v = 3 u$ (with the sign convention that $u$ is negative for a real object in front of the mirror).

A real inverted magnified image is on the same side as the object, so the object-image distance is $∣ v ∣ - ∣ u ∣ = 3∣ u ∣ - ∣ u ∣ = 2∣ u ∣ = 20$ cm, giving $∣ u ∣ = 10$ cm, $∣ v ∣ = 30$ cm.

Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = - \frac{1}{30} - \frac{1}{10} = - \frac{4}{30}$ , so $∣ f ∣ = 7.5$ cm.

Radius of curvature $∣ R ∣ = 2∣ f ∣ = 15$ cm.

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JEE Main 2025 Jan 28 Shift 2, Physics Q13: Reflection at Plane and Spherical Surfaces

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