Question

Establish the expression for finding the refractive index of material of a prism.

Answer 1

Let $ABC$ be a prism $AB$ and $AC$ are refracting surface. $A$ is the angle of prism, $PQ$ is incident ray. $QR$ is refracted ray, $RS$ is emergent ray, $MO$ and $NO$ are normals, $i_1$ is angle of incidence $r_1$ is angle of refraction, $e$ is angle of emergence, $r_2$ is angle of incidence of $AC$.
Now, from figure,
Angle of deviation, $\delta =\angle UTS$
or $\delta =\angle TQR+\angle TRQ$
(in a triangle exterior angle is sum of interior opposite angles)
$=(i_1-r_1)+(i_2-r_2)$
In the position of minimum deviation
$i_1=i_2=i$ (say)
and $r_1=r_2=r$ (say)
$\therefore {\delta }_m=(i-r)+(i-r)=2i-2r$ ......(i)
In quadrilateral $AQOR$,
$\angle AQO+\angle ARO={180}^o$ (each angle is equal to ${90}^o$)
$\therefore \angle A+\angle QOR={180}^o$ ......(ii)
In $\Delta QOR$,
$\angle QOR+r_1+r_2={180}^o$ ......(iii)
From equations (ii) and (iii),
$\angle A+\angle QOR=\angle QOR+r_1+r_2$
or $\angle A=r_1+r_2$
By condition of minimum deviation,
$\angle A=r+r=2r$ or $r=\frac{A}{2}$
Putting it in equation (i),
${\delta }_m=2i-A$
or $2i=A+{\delta }_m$ or $i=\frac{A+{\delta }_m}{2}$
By Snell's law, $\mu =\frac{\sin i}{\sin r}$
$\mu =\frac{\left(\sin \frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
This is the required formula.