Question

A short linear object of length 15 cm lies along the axis of a concave mirror of focal length 10 cm at a distance 20 cm from the pole. What is the size of image in cm?

Answer 1

According to mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$..................... (i)

For a given mirror, focal length $f=constant$. Therefore, on differentiating, Eq. (i) gives

$-\frac{1}{v^2}dv-\frac{1}{u^2}du=0\Rightarrow dv=-{\left(\frac{v}{u}\right)}^{2}du$................... (ii)

Multiplying Eq. (i) throughout by u, we get

$\frac{u}{v}+1=\frac{u}{f}\Rightarrow \frac{u}{v}=\frac{u}{f}-1=\frac{u-f}{f}$

$\therefore \frac{v}{u}=\frac{f}{u-f}$.......... (iii)

Differential length of the object on the axis, $du=b$ (given).

Therefore, Eq. (ii) gives $dv=-{\left(\frac{f}{u-f}\right)}^{2}b$.

The negative sign shows that the image is longitudinally inverted.

$\therefore$ Size of image $|dv|=b{\left(\frac{f}{u-f}\right)}^2$.