The refracting angle of a glass prism is $30^{\circ}$. A ray is incident on one of the faces and is…
The refracting angle of a glass prism is $30^{\circ}$. A ray is incident on one of the faces and is perpendicular to it. The angle of deviation $\delta$ between the incident ray and that leaves the prism is
(Refractive index of glass $=1.5)\left(\sin \left(30^{\circ}\right)=0.5, \sin (48.6)=0.75\right)$
$17^{\circ}$
$12.6^{\circ}$
$16^{\circ}$
$18.6^{\circ}$
Solution
Given, $A=30^{\circ}, \mu=1.5$ and $i_1=0^{\circ}$
Since, $i_0$, therefore, $r_1$ is also equal to $0^{\circ}$
Further, since, $r_1+r_2=A$
$\therefore r_2=A=30^{\circ}$
Using Snell's Law, $\mu=\frac{\sin i_2}{\sin r_2}$
$\Rightarrow 1.5=\frac{\sin i_2}{\sin 30^{\circ}}$
or $i_2=1.5 \sin 30^{\circ}=1.5 \times \frac{1}{2}=0.75$
or $i_2=1.5 \sin 30^{\circ}=1.5 \times \frac{1}{2}=0.75$
Now, the deviation,
$\delta=\left(i_1+i_2\right)-A=(0+48.6)-30=18.6^{\circ}$