Question

A short linear object of length $l$. lies along the axis of a concave mirror, of focal length $f$, at a distance $d$ from the pole of the mirror. The size of the image is then (nearly)

• A. $\frac{lf}{d+f}$
• B. $\frac{d+f}{lf}$
• C. $\frac{l{f}^2}{(d-f{)}^2}$
• D. $\frac{l(d+f{)}^2}{f^2}$

Answer 1

The correct option is C $\frac{l{f}^2}{(d-f{)}^2}$

Given,

The object distance is $u=d$

The focal length is $f$

The short linear distance of object $d\left(d\right)=l$

The mirror formula is given as

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

Differentiating the equation,

$0=-\frac{dv}{v^2}-\frac{d\left(d\right)}{d^2}$

$\frac{dv}{d\left(d\right)}=-\frac{v^2}{d^2}$ …… (1)

Now multiplying $d$ on both side in mirror formula

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

$\frac{d}{v}=\frac{d}{f}-1$ …… (2)

Substitute the value of equation (2) in equation (1)

$\frac{dv}{dd}=|-{\left(\frac{f}{d-f}\right)}^2|$

Given the size of the object $d\left(d\right)=l$, hence

$\frac{dv}{l}=|-{\left(\frac{f}{d-f}\right)}^2|$

$dv=|-l{\left(\frac{f}{d-f}\right)}^2|$

Therefore the size of the image is given as $\frac{l{f}^2}{{\left(d-f\right)}^2}$.