JEE Main 2026 Apr 6 Shift 2, Physics Q25: Thin Lens and Mirrors

JEE Main2026Apr 6, Shift 2Physics

A concave mirror of focal length 10 cm forms an image which is double the size of the object when the object is placed at two different positions. The distance between the two positions of the object is ____ cm.

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Solution

Concave mirror, $f = - 10$ cm (using sign convention). Magnification $m = - v / u = \pm 2$ .

Case 1 — real, inverted image ( $m = - 2$ , so $v = 2 u$ with object beyond $f$ ):

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ gives $\frac{1}{- 2 u} + \frac{1}{- u} = \frac{1}{- f}$ $\Rightarrow$ $u_{1} = \frac{3 f}{2} = 15$ cm.

Case 2 — virtual, erect image ( $m = + 2$ , $v = - 2 u$ ):

$\frac{1}{+ 2 u} + \frac{1}{- u} = \frac{1}{- f}$ $\Rightarrow$ $u_{2} = \frac{f}{2} = 5$ cm.

$Δ u = u_{1} - u_{2} = 15 - 5 = 10$ cm.

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JEE Main 2026 Apr 6 Shift 2, Physics Q25: Thin Lens and Mirrors

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