Question

For any prism, show that refractive index of its material is given by:
n or $\mu =\frac{sin\left(\frac{A+\delta m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
where the terms have their usual meaning.

Answer 1

In $\Delta$ QDR
$\delta =i-r+(i^{\prime }+r^{\prime })$
$=(i+i^{\prime })-(r+r^{\prime })$ .....$\left(1\right)$
In Quadrilateral AQER $\angle A+\angle E={180}^o$ .....$\left(2\right)$
In $\Delta$ QER $r+r^{\prime }+\angle E={180}^o$ ............$\left(3\right)$
$r+r^{\prime }=\angle A$ [from equations $\left(2\right)$ and $\left(3\right)$]
Putting value of $r+r^{\prime }$ in equation $\left(1\right)$
$\delta =1+i^{\prime }-\angle A$ ........$\left(4\right)$
In the position of minimum deviation condition.
$i=i^{\prime },r=r^{\prime },\delta ={\delta }_m$,
So, $r+r^{\prime }=\angle A$
$2r=\angle A$
or $r=\angle A/2$ .............$\left(5\right)$
Equations $\left(4\right)$ and $\left(5\right)$ becomes
${\delta }_m=2i\angle A$
or $i=\frac{\delta A+{\delta }_m}{2}$ ........$\left(6\right)$
Putting value of i and r from $\left(5\right),\left(6\right)$, in Snell's law
$n=\frac{sini}{sinr}$

$n=\frac{sin\frac{(A+{\delta }_m)}{2}}{sinA/2}$

A=Angle of deviation
${\delta }_m$=Minimum angle.