Question

A thin rod of $5 \, cm$ length is kept along the axis of a concave mirror of $10 \, cm$ focal length such that its image is real and magnified and one end touches the rod. Its magnification will be

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

End $A$ of the rod acts as an object for mirror and $A′$ will be its image so $u=2f-l=20-5=15 \, cm$
$\because $ $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ $\Rightarrow \frac{1}{- 10}=\frac{1}{v}-\frac{1}{15}\Rightarrow v=-30 \, cm.$
Now $m=\frac{\text{Length of image}}{\text{Length of object}}=\frac{\left(\right. 30 - 20 \left.\right)}{5}=2$