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(B) According to Snell's law,the refractive index $\mu$ is given by $\mu = \frac{\sin i}{\sin r}$.Given that the angle of incidence $i$ is twice the a

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

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(B) According to Snell's law,the refractive index $\mu$ is given by $\mu = \frac{\sin i}{\sin r}$.Given that the angle of incidence $i$ is twice the angle of refraction $r$,so $i = 2r$ or $r = \frac{i}{2}$.Substituting this into Snell's law:$\mu = \frac{\sin i}{\sin(i/2)}$.Using the trigonometric identity $\sin i = 2 \sin(i/2) \cos(i/2)$,we get:$\mu = \frac{2 \sin(i/2) \cos(i/2)}{\sin(i/2)}$.$\mu = 2 \cos(i/2)$.Rearranging for $i$:$\frac{\mu}{2} = \cos(i/2)$.$i/2 = \cos^{-1}\left(\frac{\mu}{2}\right)$.Therefore,$i = 2 \cos^{-1}\left(\frac{\mu}{2}\right)$.

The angle of incidence is found to be twice the angle of refraction when a ray of light passes from vacuum into a medium of refractive index $\mu$. The angle of incidence will be

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