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(C) The lateral displacement $x$ for one unit is given by the sum of horizontal shifts in each layer:$x = h 	an 60^{\circ} + d 	an 	heta_1 + d 	an

Vedclass · Accepted Answer

$4$

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(C) The lateral displacement $x$ for one unit is given by the sum of horizontal shifts in each layer:$x = h 	an 60^{\circ} + d 	an 	heta_1 + d 	an 	heta_2$Using Snell's Law at each interface:$1$. At the air-first layer interface: $1 \cdot \sin 60^{\circ} = \sqrt{\frac{3}{2}} \cdot \sin 	heta_1 \Rightarrow \sin 	heta_1 = \frac{\sqrt{3}/2}{\sqrt{3/2}} = \frac{1}{\sqrt{2}} \Rightarrow 	heta_1 = 45^{\circ}$$2$. At the first layer-second layer interface: $\sqrt{\frac{3}{2}} \cdot \sin 45^{\circ} = \sqrt{3} \cdot \sin 	heta_2 \Rightarrow \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \sqrt{3} \cdot \sin 	heta_2 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin 	heta_2 \Rightarrow \sin 	heta_2 = \frac{1}{2} \Rightarrow 	heta_2 = 30^{\circ}$Now,calculate the total horizontal shift $x$ for one unit:$x = \frac{1}{3} 	an 60^{\circ} + \left(\frac{\sqrt{3}-1}{2}ight) 	an 45^{\circ} + \left(\frac{\sqrt{3}-1}{2}ight) 	an 30^{\circ}$$x = \frac{1}{3} \cdot \sqrt{3} + \frac{\sqrt{3}-1}{2} \cdot 1 + \frac{\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3}}$$x = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}-1}{2} + \frac{3-\sqrt{3}}{6} = \frac{2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3}}{6} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}} 	ext{ cm}$For $n$ units,the total distance $l = n \cdot x = n \cdot \frac{2}{\sqrt{3}} = \frac{8}{\sqrt{3}} 	ext{ cm}$$n = 4$

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