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 ${\mu }_1=\sqrt{\frac{3}{2}}$ 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 angle of $\theta =60°$ 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 _____.

The correct answer is 4.

Step-by-Step Explanation:

From the given image, we observe a configuration of $n$ identical units stacked vertically. Each unit consists of three layers:

1. An air layer of height $h=\frac{1}{3}\text{ cm}$ and refractive index ${\mu }_0=1$.

2. A medium of thickness $d=\frac{\sqrt{3}-1}{2}\text{ cm}$ and refractive index ${\mu }_1=\sqrt{\frac{3}{2}}$.

3. A medium of thickness $d=\frac{\sqrt{3}-1}{2}\text{ cm}$ and refractive index ${\mu }_2=\sqrt{3}$.

1. Angle of refraction in each layer:

Let $\theta =60°$ be the incident angle in air (${\mu }_0=1$).

Using Snell's Law across the interfaces:

${\mu }_0\sin \theta ={\mu }_1\sin r_1={\mu }_2\sin r_2$

For the first layer (air):

The ray travels in air at angle $\theta =60°$ to the normal across a height $h$. The horizontal displacement in air per unit, $x_0$, is:

$x_0=h\tan 60°=\frac{1}{3}\times \sqrt{3}=\frac{1}{\sqrt{3}}\text{ cm}$

For the second layer (refractive index ${\mu }_1$):

$1\times \sin 60°=\sqrt{\frac{3}{2}}\sin r_1$

$\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{2}}\sin r_1⟹\sin r_1=\frac{1}{\sqrt{2}}⟹r_1=45°$

The horizontal displacement in this layer, $x_1$, is:

$x_1=d\tan 45°=d\times 1=\frac{\sqrt{3}-1}{2}\text{ cm}$

For the third layer (refractive index ${\mu }_2$):

$1\times \sin 60°=\sqrt{3}\sin r_2$

$\frac{\sqrt{3}}{2}=\sqrt{3}\sin r_2⟹\sin r_2=\frac{1}{2}⟹r_2=30°$

The horizontal displacement in this layer, $x_2$, is:

$x_2=d\tan 30°=\frac{\sqrt{3}-1}{2}\times \frac{1}{\sqrt{3}}\text{ cm}$

2. Total horizontal shift per unit ($\Delta x$):

The total horizontal shift introduced by one single unit consisting of air, medium 1, and medium 2 is:

$\Delta x=x_0+x_1+x_2$

$\Delta x=\frac{1}{\sqrt{3}}+\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}-1}{2\sqrt{3}}$

Taking $2\sqrt{3}$ as the common denominator:

$\Delta x=\frac{2+\sqrt{3}(\sqrt{3}-1)+(\sqrt{3}-1)}{2\sqrt{3}}$

$\Delta x=\frac{2+3-\sqrt{3}+\sqrt{3}-1}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\text{ cm}$

3. Finding the number of units ($n$):

As shown in the image, the ray enters the top of the first unit at a horizontal position directly below point $\text{O}$ and emerges at the bottom of the configuration after passing through $n$ identical units at a total horizontal distance $L=\frac{8}{\sqrt{3}}\text{ cm}$.

$L=n\times \Delta x$

$\frac{8}{\sqrt{3}}=n\times \frac{2}{\sqrt{3}}$

$n=\frac{8}{2}=4$

Thus, the value of $n$ is 4.