Question

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h=\frac{1}{3} cm$, and the second and third layers are of equal thickness $d=\frac{\sqrt{3}-1}{2} cm$, 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 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 $l=\frac{8}{\sqrt{3}} cm$, as shown in the figure. The value of $n$ is ______

Answer 1

$1 \sin 60^{\circ}=\sqrt{\frac{3}{2}} \sin \theta$
$ \Rightarrow \theta_1=45^{\circ}$
$ \sqrt{\frac{3}{2}} \sin 45^{\circ}=\sqrt{3} \sin \theta_2$
$ =\sqrt{\frac{3}{2}} \frac{1}{\sqrt{2}}=\sqrt{3} \sin \theta_2 $
$ =\theta_2=30^{\circ}$
$h \tan 60^{\circ}+d \tan 45^{\circ}+d \tan 30^{\circ}$
$ \frac{1}{3} \sqrt{3}+\left(\frac{\sqrt{3}-1}{2}\right)+\left(\frac{\sqrt{3}-1}{2}\right) \frac{1}{\sqrt{3}} $
$ \frac{2 \sqrt{3}+3 \sqrt{3}-3+3-\sqrt{3}}{6}$
$ \frac{4 \sqrt{3}}{6} $
$\therefore n \frac{4 \sqrt{3}}{6}=\frac{8}{\sqrt{3}}$
$n = 4$