A ray of light passes from vacuum into a medi | vedclass

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(B) According to Snell's law,$n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (vacuum),$n_2 = n$,$i$ is the angle of incidence,and $r$ is the angle of refract

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$2 \operatorname{Cos}^{-1}\left(\frac{n}{2}\right)$

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(B) According to Snell's law,$n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (vacuum),$n_2 = n$,$i$ is the angle of incidence,and $r$ is the angle of refraction.Given that $i = 2r$,we have $r = \frac{i}{2}$.Substituting these into Snell's law:$1 	imes \sin i = n \sin \left(\frac{i}{2}ight)$Using the trigonometric identity $\sin i = 2 \sin \left(\frac{i}{2}ight) \cos \left(\frac{i}{2}ight)$:$2 \sin \left(\frac{i}{2}ight) \cos \left(\frac{i}{2}ight) = n \sin \left(\frac{i}{2}ight)$Dividing both sides by $\sin \left(\frac{i}{2}ight)$ (assuming $i 
eq 0$):$2 \cos \left(\frac{i}{2}ight) = n$$\cos \left(\frac{i}{2}ight) = \frac{n}{2}$$\frac{i}{2} = \cos^{-1} \left(\frac{n}{2}ight)$$i = 2 \cos^{-1} \left(\frac{n}{2}ight)$

$A$ ray of light passes from vacuum into a medium of refractive index $n$. If the angle of incidence is twice the angle of refraction,then the angle of incidence in terms of refractive index is

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