JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

Distance between an object and three times magnified real image is $40$ cm. The focal length of the mirror used is _____ cm.

A
$- 15/2$
B
$- 10$
C
$- 20$
D
$- 15$

Solution

For a mirror with a real image of a real object, both $u$ and $v$ are negative (taking the mirror at origin and the incident direction as positive). The magnification is $m = - \frac{v}{u} = - 3$ , so $v = 3 u$ (both negative).

Distance between object and image: $∣ v ∣ - ∣ u ∣ = 40 \Rightarrow 3∣ u ∣ - ∣ u ∣ = 40 \Rightarrow ∣ u ∣ = 20 cm$ . So $u = - 20$ , $v = - 60$ .

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{- 60} + \frac{1}{- 20} = - \frac{1}{60} - \frac{3}{60} = - \frac{4}{60} = - \frac{1}{15}$ .

$f = - 15 cm$ (concave mirror).

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