Physics · Geometrical Optics

JEE Main 2026 — 6 April, Evening Shift — Question 20

A concave mirror of focal length $10\mathrm{cm}$ forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is ______ cm.

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

Magnification $m = \pm 2$ . For real image (inverted) $m=-2$ : $v=-2u$ . Mirror formula: $1/v+1/u=1/f$ ⇒ $1/(-2u)+1/u = -1/10$ ⇒ $1/(2u) = -1/10$ ⇒ $u = -5$ cm. For virtual image (erect) $m=+2$ : $v=2u$ . Then $1/(2u)+1/u = -1/10$ ⇒ $3/(2u) = -1/10$ ⇒ $u = -15$ cm. Distance between positions = 15-5=10 cm.

Answer key and solution verified before publishing.

Practise Geometrical Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Geometrical Optics
Topic: Reflection of Light at Curved Surfaces and Spherical Mirrors

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is

Answer: 10

Step-by-step solution

Practise Geometrical Optics

More Geometrical Optics questions from this paper