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

Answer: 4

$1\sin 60^{\circ }=\sqrt{\frac{3}{2}}\sin \theta$
$\Rightarrow {\theta }_1=45^{\circ }$
$\sqrt{\frac{3}{2}}\sin 45^{\circ }=\sqrt{3}\sin {\theta }_2$
$=\sqrt{\frac{3}{2}}\frac{1}{\sqrt{2}}=\sqrt{3}\sin {\theta }_2$
$={\theta }_2=30^{\circ }$
$h\tan 60^{\circ }+d\tan 45^{\circ }+d\tan 30^{\circ }$
$\frac{1}{3}\sqrt{3}+\left(\frac{\sqrt{3}-1}{2}\right)+\left(\frac{\sqrt{3}-1}{2}\right)\frac{1}{\sqrt{3}}$
$\frac{2\sqrt{3}+3\sqrt{3}-3+3-\sqrt{3}}{6}$
$\frac{4\sqrt{3}}{6}$
$\therefore n\frac{4\sqrt{3}}{6}=\frac{8}{\sqrt{3}}$
$n=4$