Consider a configuration of identical units, each consisting of three layers. The first layer is a column of air of height and the second and third layers are of equal thickness and refractive indices and respectively. A light source is placed on th

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 $μ_{1} = \sqrt{\frac{3}{2}}$ and $μ_{2} = \sqrt{3},$ 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 $θ = 60 °$ to the normal. For a specific value of $n$ , the ray of light emerges from the bottom of the configuration at a distance $l = \frac{8}{\sqrt{3}} cm,$ as shown in the figure. The value of $n$ is _____. [2022]

Smart Achievers · Accepted Answer

(4)

$x_{1} = \frac{1}{3} \times \tan 60 ° = \frac{1}{\sqrt{3}} cm$

and, at interface-1,

$1 \sin 60 ° = \sqrt{\frac{3}{2}} \sin θ_{2}$ $\Rightarrow θ_{2} = 45 °$

So, $x_{2} = d \tan 45 ° = \frac{\sqrt{3} - 1}{2}$

at interface-2,

Again, $\sqrt{\frac{3}{2}} \sin 45 ° = \sqrt{3} \sin θ_{3}$

$\Rightarrow θ_{3} = 30 °$

So, $x_{3} = d \tan 30 ° = \frac{\sqrt{3} - 1}{2 \sqrt{3}}$

So, $x_{1} + x_{2} + x_{3} = \frac{1}{\sqrt{3}} + \frac{\sqrt{3} - 1}{2} (1 + \frac{1}{\sqrt{3}})$

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

So, 1 unit shifts the light ray by $\frac{2}{\sqrt{3}}$

Therefore, to shift the light ray by $ℓ = \frac{8}{\sqrt{3}},$ the number of units needed is $n = \frac{\frac{8}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = 4$

Solution

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