Question

A small linear object of length $b$ is placed at a distance $u$ from the pole of the concave mirror along the axis away from the mirror. The focal length of the mirror is $f$. The length of image will then be?

• A. $b\left(\frac{f}{u-f}{)}^2\right(\frac{f}{u-f+b}{)}^2$
• B. $b\left(\frac{f}{u-f}{)}^2\right(\frac{f}{u-f+b})$
• C. $b\left(\frac{f}{u-f}\right)(\frac{f}{u-f+b}{)}^2$
• D. $b\left(\frac{f}{u-f}\right)\left(\frac{f}{u-f+b}\right)$

Answer 1

Answer: (D)

$b\left(\frac{f}{u-f}\right)\left(\frac{f}{u-f+b}\right)$

Here,
$u_A=-u$
focal length = $-f$
using mirror formula
$\frac{1}{u}+\frac{1}{\nu }=\frac{1}{f}$

$\frac{1}{-u}+\frac{1}{{\nu }_A}=\frac{-1}{f}$

${\nu }_A=-\left(\frac{uf}{u-f}\right)$

Again for end B
$u_B=-(u+b)$

$\frac{1}{u}+\frac{1}{\nu }=\frac{1}{f}$

$\frac{1}{-(u+b)}+\frac{1}{{\nu }_B}=\frac{1}{-f}$
Length of image$={\nu }_A-{\nu }_B$
$=\frac{-uf}{u-f}+\frac{(u+b)f}{(u+b)-f}$
$=\frac{f}{u-f}(\frac{u+b}{1+\frac{b}{u-f}}-u)$
$=\frac{f}{u-f}\left(\frac{u+b-u-\frac{bu}{u-f}}{1+\frac{b}{u-f}}\right)$
$=b\left(\frac{f}{u-f}\right)\left(\frac{1-\frac{u}{u-f}}{1+\frac{b}{u-f}}\right)$
$=b\left(\frac{f}{u-f}\right)\left(\frac{\frac{-f}{u-f}}{\frac{u-f+b}{u-f}}\right)$
$=b\left(\frac{f}{u-f}\right)\left(\frac{-f}{u-f+b}\right)$

We take its absolute value for this for the length of the image.