Question

For any prism, prove that '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

Let the refractive index of the prism be $\mu$.
Applying Snell's law, $n_{air}\sin i=\mu sinr$
where $n_{air}=1$
Or $sini=\mu \sin r_1$
$⟹\mu =\frac{\sin i}{\sin r_1}$ ........(1)
Deviation by prism $\delta =i+e-A$
where $A=r_1+r_2$
For minimum deviation $i=e$ and $r_1=r_2$
$\therefore A=2r_1$
$⟹r_1=\frac{A}{2}$ .....(b)
Angle of minimum deviation is given by ${\delta }_m=2i-A$
$⟹$ $i=\frac{A+{\delta }_m}{2}$ ......(c)
Putting (b) and (c) in (1), we get
$\mu =\frac{\sin (\frac{A+{\delta }_m}{2})}{sin\frac{A}{2}}$