Question

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h=\frac{1}{3}\text{cm}$, and the second and third layers are of equal thickness $d=\frac{\sqrt{3}-1}{2}\text{cm}$, and refractive indices ${\mu }_1=\sqrt{\frac{3}{2}}\text{and}{\mu }_2=\sqrt{3}$, , respectively. A light source $O$ is placed on the top of the first unit, as shown in the figure. A ray of light from $O$ is incident on the second layer of the first unit at an
angle of $\theta ={60}^{\circ }$
to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a distance $l=\frac{8}{\sqrt{3}}\text{cm}$, as shown in the figure. The value of $n$ is

• A. 4
• B. 4.0
• C. 4.00

Answer 1

Answer: (A), (B), (C)

\( 8 / \sqrt{3} \)
For the first layer,
$\frac{X_1}{h}=\tan {60}^{\circ }$
$X_1=h\tan {60}^{\circ }=\frac{1}{3}\sqrt{3}=\frac{1}{\sqrt{3}}$
Now for the total unit consisting of three layers$x=x_1+x_2+x_3\Rightarrow x=\frac{1}{\sqrt{3}}+\frac{\sqrt{3}-1}{2}\tan {45}^{\circ }+\frac{\sqrt{3}-1}{2}\tan {30}^{\circ }$
$\Rightarrow x=\frac{1}{\sqrt{3}}+\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}-1}{2}\frac{1}{\sqrt{3}}\Rightarrow x=\frac{2+(\sqrt{3}-1)\sqrt{3}+\sqrt{3}-1}{2\sqrt{3}}\Rightarrow x=\frac{2+3-\sqrt{3}+\sqrt{3}-1}{2\sqrt{3}}\Rightarrow x=\frac{4}{2\sqrt{3}}\Rightarrow x=\frac{2}{\sqrt{3}}$
Given $n\left(x\right)=\frac{8}{\sqrt{3}};n\left(\frac{2}{\sqrt{3}}\right)=\frac{8}{\sqrt{3}};n=4$