Question

A rod of length $10\text{cm}$ lies along the principal axis of a concave mirror of focal length $10\text{cm}$ in such a way that its end close to the pole is $20\text{cm}$ away from the mirror. The length of the image is

• A. $10\text{cm}$
• B. $15\text{cm}$
• C. $2.5\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is D $5\text{cm}$


The rod is lying axially along the principal axis, its two ends will acts as two point objects $O_1$ and $O_2$ for the mirror.
The Object distance for the end close to the pole is $20\text{cm}$
$\because R=2f=20\text{cm}$
Thus, image $I_1$ for object $O_1$ will form at object itself, because $O_1$ lies at centre of curvature.
$\Rightarrow |P{I}_1|=20\text{cm}$
For object $O_2$, let image is formed at $v_2$ (Taking positive along the incident ray)
$u=-30\text{cm}$
Using mirror formula,
$\frac{1}{v_2}+\frac{1}{(-30)}=\frac{1}{(-10)}$
$(\because f=-10\text{cm})$
$\Rightarrow \frac{1}{v_2}=\frac{1}{30}-\frac{1}{10}=\frac{-2}{30}$
$\Rightarrow v_2=-15\text{cm}$
Thus image distance from pole is $-15\text{cm}$
$|P{I}_2|=15\text{cm}$
$\Rightarrow$ length of image will be,
$l_i=|P{I}_1|-|P{I}_2|=20-15=5\text{cm}$