Two light beams fall on a transparent materia | vedclass

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(C) Let the angle of incidence with the normal be $i$. Since $	heta_1$ is the angle with the surface,$i = 90^\circ - 	heta_1$. According to Snell's

Vedclass · Accepted Answer

$6$

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(C) Let the angle of incidence with the normal be $i$. Since $	heta_1$ is the angle with the surface,$i = 90^\circ - 	heta_1$. According to Snell's Law: $n_1 \sin(i) = n_2 \sin(r)$,where $r$ is the angle of refraction.$n_1 \sin(90^\circ - 	heta_1) = n_2 \sin(r) \implies n_1 \cos(	heta_1) = n_2 \sin(r)$.Given $	heta_1 = \cos^{-1}\left(\frac{n_2}{2n_1}ight)$,we have $\cos(	heta_1) = \frac{n_2}{2n_1}$.Substituting this into the Snell's Law equation: $n_1 \left(\frac{n_2}{2n_1}ight) = n_2 \sin(r) \implies \frac{n_2}{2} = n_2 \sin(r) \implies \sin(r) = \frac{1}{2}$.Thus,$r = 30^\circ$.From the geometry of the block,the horizontal distance from the point of incidence to the vertical line passing through point $3$ is $d/2$. Let $t$ be the thickness of the block.Then,$	an(r) = \frac{d/2}{t} \implies t = \frac{d}{2 	an(r)}$.Substituting $d = 4\sqrt{3} 	ext{ cm}$ and $r = 30^\circ$: $t = \frac{4\sqrt{3}}{2 	an(30^\circ)} = \frac{2\sqrt{3}}{1/\sqrt{3}} = 2 	imes 3 = 6 	ext{ cm}$.

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