Question

For spherical refracting surface establish the refraction formula $\frac{\mu }{v}-\frac{1}{u}=\frac{\mu -1}{R}$ where symbols have their usual meanings.

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 principal 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,
$\mu =\frac{sini}{\sin r}$
or
$\mu =\frac{i}{r}$(as i and r are very small)
or $i=\mu r$ ...........(i)
From $\Delta$ OMC,
(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 eqns. (ii) and (iii) putting the values of i and r in eqn. (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 eqn. (iv),
$\mu \frac{PM}{PI}-\frac{PM}{PO}=(\mu -1)\frac{PM}{PC}$
or $\frac{\mu }{PI}-\frac{1}{PO}=\frac{(\mu -1)}{PC}$ ..........(v)
But, PI$=-v$, PO$=-u$ and PC$=-R$
$\therefore \frac{\mu }{-v}-\frac{1}{-\mu }=\frac{(\mu -1)}{-R}$
or $\frac{\mu }{v}-\frac{1}{u}=\frac{(\mu -1)}{R}$
or $\frac{(\mu -1)}{R}=\frac{\mu }{v}-\frac{1}{u}$.