Question 307: Ray Optics

Question

In the given figure, the face $$A C$$ of the equilateral prism is immersed in a liquid of refractive index '$$n$$'. For incident angle $$60^{\circ}$$ at the side $$A C$$, the refractive light beam just grazes along face $$A C$$. The refractive index of the liquid $$n=\frac{\sqrt{x}}{4}$$. The value of $$x$$ is ____________.
(Given refractive index of glass $$=1.5$$ )

Accepted Answer

Given prism is equilateral so, angel of prism, A $=60^{\circ}$
On first surface light is passing straight without any deviation, hence angle of refraction $\mathrm{r}_1=0^{\circ}$.
$$
\begin{aligned}
& r_1+i_2=\mathrm{A} \\
& 0+i_2=60^{\circ} \\
& \Rightarrow i_2=60^{\circ}
\end{aligned}
$$
On $2^{	ext {nd }}$ surface ray, refracting ray is becoming parallel to the junction of surfaces, so $i_2$ will act as critical angle.
From snell's law $\mu_1 \sin 	heta_1=\mu_2 \sin 	heta_2$
$$
\begin{aligned}
& \Rightarrow \frac{3}{2} \sin i_2=\mu \sin 	heta_2 \\
& \Rightarrow \mu=\frac{3}{2} 	imes \frac{\sin 60^{\circ}}{\sin 90^{\circ}}=\frac{3}{2} 	imes \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} 	ext { or } \frac{\sqrt{27}}{4}
\end{aligned}
$$
$\Rightarrow$ Comparing above equation with
given $n=\frac{\sqrt{x}}{4}$, we get $x=27$

Ray Optics

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