Question

Derive an expression:
$\frac{\mu }{v}-\frac{1}{u}=\frac{\mu -1}{R}$
for refraction of light at spherical surface.

Answer 1

Let $APB$ is a cross-section of any spherical refracting surface. $P$ is its pole and $C$ is its centre of curvature. Left of the surface has air as a medium and in right medium has refractive index $m$.
Any point object is placed at point $O$ on the principle axis of which a virtual image is formed at $I$ due to refracting surface $APB$. According to the figure angle of incidence,
$\angle OMC=i$
Angle of refraction,
$\angle LMN=\angle IMC=r$
Let $\angle MOC=\alpha$, $\angle MIC=\beta$ and $\angle MCP=\theta$
$\therefore$ From Snell's law,
$\mu =\frac{\sin i}{\sin r}$
or $\mu =\frac{i}{r}$ (as $i$ and $r$ are very small)
or $i=\mu r$ ......(i)
From $\Delta OMC$, $\theta =i+\alpha$
(exterior angle $=$ sum of opposite interior angles)
or $i=\theta -\alpha$ ......(ii)
and in $\Delta IMC$, $\theta =r+\beta$
or $r=\theta -\beta$ .....(iii)
From equation (ii) and (iii) putting the values of $i$ and $r$ in equation (i)
$\theta -\alpha =\mu (\theta -\beta )$
or $\mu \beta -\alpha =(\mu -1)\theta$ .....(iv)
Again, $\alpha =\frac{PM}{PO}$, $\beta =\frac{PM}{IP}$ and $\theta =\frac{PM}{PC}$
Substituting above relations in equation (iv),
$\mu \frac{PM}{PI}-\frac{PM}{PO}=(\mu -1)\frac{PM}{PC}$
or $\frac{\mu }{PI}-\frac{I}{PO}=\frac{(\mu -1)}{PC}$ .....(v)
But, $PI=-v$, $PO=-u$ and $PC=-R$
$\therefore \frac{\mu }{-v}-\frac{I}{-u}=\frac{(\mu -1)}{-R}$
or $\frac{\mu }{v}-\frac{1}{u}=\frac{(\mu -1)}{R}$
or $\frac{(\mu -1)}{R}=\frac{\mu }{v}-\frac{1}{u}$