Question

A short linear object of length l lies along the axis of a concave mirror at a distance u from it. If v is the distance of image from the mirror then size of the image is

• A. $l\times \frac{v}{u}$
• B. $l\times \frac{u}{v}$
• C. $l\times (\frac{v}{u}{)}^2$
• D. $l\times (\frac{u}{v}{)}^2$

Answer 1

Answer: (C)

$l\times (\frac{v}{u}{)}^2$

We know, that

$\frac{1}{v}+\frac{1}{4}=\frac{1}{f}$

diff wrto u both seside

$\begin{array}{cc}\frac{-1}{v^2}\frac{dv}{du}-\frac{1}{u^2}& =0\\ \frac{dv}{du}& =-\frac{v^2}{u^2}\\ \Rightarrow & |dv|=\left|\frac{v^2}{u^2}\right|du\mid \end{array}$

$\begin{array}{cc}\Rightarrow |dv|& =\text{size of image,}|du|=\text{object sise}\\ I& =\frac{v^2}{u^2}\times l\text{{object}\sin e=l)\end{array}$

$\text{Hence, option (c) is correct.}$