The speed of light in media M_{1} and M_{2} a | vedclass

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(A) For total internal reflection to occur, the light must travel from a denser medium to a rarer medium, and the angle of incidence $	heta$ must be

Vedclass · Accepted Answer

$ > \sin^{-1}\left(\frac{3}{4}\right)$

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(A) For total internal reflection to occur, the light must travel from a denser medium to a rarer medium, and the angle of incidence $	heta$ must be greater than the critical angle $C$.Given speeds of light in media $M_{1}$ and $M_{2}$ are $v_{1} = 1.5 	imes 10^{8} 	ext{ m/s}$ and $v_{2} = 2 	imes 10^{8} 	ext{ m/s}$.Since $v_{1} The critical angle $C$ is given by $\sin C = \frac{v_{1}}{v_{2}}$.Substituting the values: $\sin C = \frac{1.5 	imes 10^{8}}{2 	imes 10^{8}} = \frac{1.5}{2} = \frac{3}{4}$.Therefore, $C = \sin^{-1}\left(\frac{3}{4}ight)$.For total internal reflection, the angle of incidence $	heta$ must satisfy $	heta > C$.Hence, $	heta > \sin^{-1}\left(\frac{3}{4}ight)$.

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