JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

A rod of length $10$ cm lies along the principle axis of a concave mirror of focal length $10$ cm as shown in figure. The length of the image is ________ cm.

A
$2.5$
B
$5$
C
$7.5$
D
$7$

Solution

The rod's two endpoints are at object distances $20$ cm and $30$ cm. Image each end with the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and sign convention $u, f < 0$ .

End A ( $u = - 20$ , $f = - 10$ ): $\frac{1}{v} = - \frac{1}{10} + \frac{1}{20} = - \frac{1}{20} \Rightarrow v_{A} = - 20$ cm.

End B ( $u = - 30$ , $f = - 10$ ): $\frac{1}{v} = - \frac{1}{10} + \frac{1}{30} = - \frac{1}{15} \Rightarrow v_{B} = - 15$ cm.

Image length $= ∣ v_{A} - v_{B} ∣ = ∣ - 20 - (- 15) ∣ = 5$ cm.

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