Question

A ray of light from a liquid $(\mu =\sqrt{3})$ is incident on a system of two right angled prisms of refractive indices $\sqrt{3}$ and $\sqrt{2}$ as shown. The ray suffers zero deviation when it emerges into air from face $CD$. The angle of incidence $i$ is $P\times 9^{\circ }$. Then, find the value of $P$.

Answer 1

There is no refraction at surface $AB$ [since R.I of liquid and prism are the same].
First, we apply the law of refraction for the interface $AC$. The incidence angle will be $90–i$.
$\sqrt{3}\sin (90-i)=\sqrt{2}\sin r$
$\sqrt{3}\cos i=\sqrt{2}\sin r$
$\Rightarrow \sin r=\frac{\sqrt{3}}{\sqrt{2}}\cos i$ ... (1)

Now, for the interface $CD$,
$\sqrt{2}\sin (90-r)=\sin i$
[since angle of deviation is zero]
$\Rightarrow \sqrt{2}\cos r=\sin i$
$\Rightarrow \cos r=\frac{1}{\sqrt{2}}\sin i$ ... (2)

We have two equations in $i$ and $r$. Squaring and adding these two,we get
${\sin }^{2}r+{\cos }^{2}r=\frac{3}{2}{\cos }^{2}i+\frac{1}{2}{\sin }^{2}i$
$\Rightarrow 1=1+{\cos }^{2}i$
$\Rightarrow {\cos }^{2}i=\frac{1}{2}$
$\Rightarrow \cos i=\pm \frac{1}{\sqrt{2}}$
or $i={45}^{\circ }=P\times 9^{\circ }$ (given)
$\therefore P=5$