Question 33: Ray Optics

Question

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h=\frac{1}{3} \mathrm{~cm}$, and the second and third layers are of equal thickness $d=$ $\frac{\sqrt{3}-1}{2} \mathrm{~cm}$, and refractive indices $\mu_{1}=\sqrt{\frac{3}{2}}$ and $\mu_{2}=\sqrt{3}$, respectively. A light source 0 is placed on the top of the first unit, as shown in the figure. A ray of light from 0 is incident on the second layer of the first unit at an angle of $	heta=60^{\circ}$ to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a distance $l=\frac{8}{\sqrt{3}} \mathrm{~cm}$, as shown in the figure. The value of $n$ is ________.

Accepted Answer

$$
\begin{aligned}
&x_1 =\frac{1}{3} 	imes 	an 60 \\
& =\frac{1}{\sqrt{3}} \mathrm{~cm}
\end{aligned}
$$
By Snell's law
$$
\begin{aligned}
&1 	imes \sin 60 =\frac{\sqrt{3}}{\sqrt{2}} \sin 	heta_2 \\
& \Rightarrow 	heta_2 =45^ \mathrm{o} \\
&	herefore x_2 =d
\end{aligned}
$$
Again, by applying Snell's law
$$
\begin{aligned}
&\sqrt{\frac{3}{2} 	imes \frac{1}{\sqrt{2}}} =\sqrt{3} 	imes \sin 	heta_3 \\
&\Rightarrow 	heta_2 =30^{\circ} \\
& 	herefore x_3 =\frac{d}{\sqrt{3}}
\end{aligned}
$$
Also
$$
\begin{aligned}
& x_1+x_2+x_3 =\frac{1}{\sqrt{3}}+d+\frac{d}{\sqrt{3}} \\
& =\frac{1}{\sqrt{3}}+\left(\frac{\sqrt{3}-1}{2}ight)+\frac{\sqrt{3}-1}{2 \sqrt{3}} \\
& =\frac{2+(3-\sqrt{3})+\sqrt{3}-1}{2 \sqrt{3}} \\
& =\frac{4}{2 \sqrt{3}} \\
& x_1+x_2+x_3  =\frac{2}{\sqrt{3}} \mathrm{~cm} \\
& 	ext {And,}~~ \eta  =\frac{l}{x_1+x_2+x_3} \\
&\eta  =\frac{8 / \sqrt{3}}{2 / \sqrt{3}}=4 \\
&\eta  =4
\end{aligned}
$$

Ray Optics

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