Question

Establish the relation amongst $u$, $v$ and $f$ for concave lens.

Answer 1

Let $MPN$ is a concave lens. In which $F$ is focus, $C$ is centre and $P$ is pole. $AB$ is an object placed up right on the principle axis. Real image of object is $A^{\prime }{B}^{\prime }$. Join $AP$ and $A^{\prime }P$. Then for $AP$ is incident ray and $P{A}^{\prime }$ is reflected ray for $P$.
$\because$ Angle of incidence $=$ Angle of reflection
$\therefore \angle APB=\angle A^{\prime }PB$
Draw a perpendicular from $O$ on the principle axis.
Let it be $OQ$
$\Delta A^{\prime }{B}^{\prime }F$ and $\Delta OQF$ are similar
$\therefore \frac{A^{\prime }{B}^{\prime }}{OQ}=\frac{F{B}^{\prime }}{QF}$
$\because OQ=AB$
$\therefore \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{F{B}^{\prime }}{QF}$ .......(i)
$\Delta A^{\prime }{B}^{\prime }P$ and $\Delta ABP$ is similar
$\therefore \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{P{B}^{\prime }}{PB}$ ......(ii)
From equation (i) and (ii)
$\frac{F{B}^{\prime }}{QF}=\frac{P{B}^{\prime }}{PB}$ ......(iii)
If the aperture of the mirror is small, then the points $Q$ and $P$ will be very near to each other and so, can be replaced by $P$, $QF=PF$ (Approx)
So equation (iii) become
$\frac{F{B}^{\prime }}{PF}=\frac{P{B}^{\prime }}{PB}$
$\frac{P{B}^{\prime }-PF}{PF}=\frac{P{B}^{\prime }}{PB}$ ......(iv)
By cartesian sign convention
Object distance $PB=-u$, Image distance $P{B}^{\prime }=-v$, Focal length $PF=-f$
Hence equation (iv) becomes
$\frac{-v-(-f)}{-f}=\frac{-v}{-u}$
Or $\frac{-v+f}{-f}=\frac{v}{u}$
Or $-vf=-uf+uf$
Or $uv=-uf+uf$
Dividing by $uvf$ on both sides, we get
$\frac{uv}{uvf}=\frac{vf}{uvf}-\frac{uf}{uvf}$
Or $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
Hence, this equation is the relation between $u$, $v$, $f$.