JEE Main 2026 — Ray Optics Question with Solution

Given focal length of the concave mirror, $f=-10$ cm.

The image is double the size of the object, so the magnification can be $m=\pm 2$.

Using the magnification formula in terms of focal length and object distance:

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

Case 1: For a real image, $m=-2$

$-2=\frac{-10}{-10-u_1}$

$20+2u_1=-10$

$2u_1=-30\Rightarrow u_1=-15$ cm

Case 2: For a virtual image, $m=+2$

$2=\frac{-10}{-10-u_2}$

$-20-2u_2=-10$

$-2u_2=10\Rightarrow u_2=-5$ cm

The distance between the two positions of the object is:

$\Delta u=|u_1-u_2|=|-15-(-5)|=10$ cm

Answer: $10$