Question

With the help of a neat diagram arrive at an expression for finding the refractive index of a prism.

Answer 1

We know the relation between angle of deviation $\delta$, Angle of incidence $i$, angle of emergence $i^{\prime }$ and prism angle $A$ to be

$\delta$ = $i$ +$i^{\prime }$ -$A$

Also for minimum deviation ${\delta }_m$, $i$ = $i^{\prime }$ and $r$ = $r^{\prime }$

Hence ${\delta }_m$ = $i$+$i^{\prime }$ -$A$ = 2$i$- $A$

or, $i=\frac{A+{\delta }_m}{2}$.....................(i)

Also, since r+r' =A

or, r = A/2.................................(ii)

The refractive index is

$\mu =\frac{sini}{sinr}$

Using (i) and (ii)

$\mu =\frac{sin\frac{A+{\delta }_m}{2}}{sin\frac{A}{2}}$

Also if the angle of prism A is small, ${\delta }_m$ is also small.

Hence the above equation then becomes

$\mu =\frac{\frac{A+{\delta }_m}{2}}{\frac{A}{2}}$