Question

Find the minimum and maximum deflection angles for a light ray passing through a glass prism with refracting angle $\theta ={60}^{\circ }$

Answer 1

In the case of minimum deviation

$\sin \frac{\alpha +\theta }{2}=n\sin \frac{\theta }{2}$

So, $\alpha =2{\sin }^{-1}\left\{n\sin \frac{\theta }{2}\right\}-\theta ={37}^{\circ }$, for $n=1.5$

Passage of ray for grazing incidence and grazing imergence is the condition for maximum deviation. From fig.

$a=x-\theta =\pi -2{\theta }_{cr}$

(where ${\theta }_{cr}$ is the critical angle)

So, $\alpha =\pi -2\sin \left(1/n\right)={58}^{\circ }$,

for $n=1.5=R.I.$ of glass.