Question

The cross-section of a prism $(\mu =1.5)$ in an equilateral triangle. A ray of light is incident perpendicular on one of the faces. Find the angle of deviation (in degree) of the ray.

Answer 1

Answer: 60.00

Here $I=90^{\circ }$
So $S=i+e-A$
$=90^{\circ }+e-A$
However when $I=90^{\circ }$
for "$AB$"
$1\times \sin 90^{\circ }=\mu \sin r_1$
$\sin r_1=\frac{1}{\mu }=\frac{1}{1.5}=\frac{2}{3}$
$r_1={\sin }^{-1}\left(\frac{2}{3}\right)$
$r_1={\theta }_c$
So ${\theta }_c={\sin }^{-1}\left(\frac{2}{3}\right)$
Now for '$AC$'
$\mu \sin r_2=1\sin e$
$\sin e=\mu \sin \left(A-{\theta }_c\right)$
$e={\sin }^{-1}\left[\mu \sin \left(A-{\theta }_c\right)\right]$
$={\sin }^{-1}\left[1.5\sin \left(90^{\circ }-{\sin }^{-1}2/3\right)\right]$
$e=29^{\circ }$
so ${\delta }_{\max }=90^{\circ }+29-60^{\circ }$
$\left(\text{ Here }A=60^{\circ }\right)$
${\delta }_{\max }=59$