Consider a configuration of n identical units, each consisting of...

Considering a single unit with three layers. The figure below shows the refraction of light as it moves from one medium to others.

For the first interface,

${x}_{1} = \dfrac{1}{3} \times \tan 60 ° = \dfrac{1}{\sqrt{3}}cm$

Using Snell's law,

${\mu}_{1}sin{\theta}_{1} = {\mu}_{2}sin{\theta}_{2}$

$⇒ 1 \times \dfrac{\sqrt{3}}{2} = \sqrt{\dfrac{3}{2}} \times sin{\theta}_{2}$

$⇒ {\theta}_{2} = 45 °$

For the second interface,

$⇒ {x}_{2} = d \times \tan 45 ° = d$

And again using Snell's law,

$\sqrt{\dfrac{3}{2}} \times \sin 45 = \sqrt{3} \times sin{\theta}_{3}$

$⇒ \sqrt{\dfrac{3}{2}} \times \dfrac{1}{\sqrt{2}} = \sqrt{3}sin{\theta}_{3}$

$⇒ {\theta}_{3} = 30 °$

$⇒ {x}_{3} = d \times \tan 30 ° = \dfrac{d}{\sqrt{3}}$

$∴ {x}_{1} + {x}_{2} + {x}_{3} = \dfrac{1}{\sqrt{3}} + \dfrac{\left(\sqrt{3} - 1\right)}{2}\left(1 + \dfrac{1}{\sqrt{3}}\right)$

$= \dfrac{2}{\sqrt{3}}cm$

$∴ n = \dfrac{1}{{x}_{1} + {x}_{2} + {x}_{3}} = \dfrac{\dfrac{8}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}} = 4$

Thus, the value of $n = 4$.