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(D) Let the angle of incidence be $i$ and the angle of refraction be $r$.Given that $i = 2r$,which implies $r = \frac{i}{2}$.According to Snell's Law,

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$2 \cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$

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(D) Let the angle of incidence be $i$ and the angle of refraction be $r$.Given that $i = 2r$,which implies $r = \frac{i}{2}$.According to Snell's Law,$n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (for air) and $n_2 = \sqrt{2n}$.Substituting the values: $1 \cdot \sin i = \sqrt{2n} \cdot \sin\left(\frac{i}{2}ight)$.Using the trigonometric identity $\sin i = 2 \sin\left(\frac{i}{2}ight) \cos\left(\frac{i}{2}ight)$,we get:$2 \sin\left(\frac{i}{2}ight) \cos\left(\frac{i}{2}ight) = \sqrt{2n} \sin\left(\frac{i}{2}ight)$.Dividing both sides by $\sin\left(\frac{i}{2}ight)$ (assuming $i 
eq 0$):$2 \cos\left(\frac{i}{2}ight) = \sqrt{2n}$.$\cos\left(\frac{i}{2}ight) = \frac{\sqrt{2n}}{2} = \sqrt{\frac{2n}{4}} = \sqrt{\frac{n}{2}}$.Therefore,$\frac{i}{2} = \cos^{-1}\left(\sqrt{\frac{n}{2}}ight)$.$i = 2 \cos^{-1}\left(\sqrt{\frac{n}{2}}ight)$.

Consider a light ray travelling in air incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of the refracting angle. Then,the angle of incidence will be

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