Light travels from an optically denser medium | vedclass

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Question

(A) The refractive index of a medium is inversely proportional to the speed of light in that medium,given by $\mu = \frac{c}{v}$.Given speeds are $v_{

Vedclass · Accepted Answer

$\sin ^{-1}\left(\frac{2}{3}\right)$

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(A) The refractive index of a medium is inversely proportional to the speed of light in that medium,given by $\mu = \frac{c}{v}$.Given speeds are $v_{A} = 1.8 	imes 10^{8} \ m/s$ and $v_{B} = 2.7 	imes 10^{8} \ m/s$.The relative refractive index of medium $A$ with respect to medium $B$ is ${}_{B}\mu_{A} = \frac{\mu_{A}}{\mu_{B}} = \frac{v_{B}}{v_{A}}$.Substituting the values: ${}_{B}\mu_{A} = \frac{2.7 	imes 10^{8}}{1.8 	imes 10^{8}} = \frac{27}{18} = \frac{3}{2}$.The formula for the critical angle $C$ is $\sin C = \frac{1}{{}_{B}\mu_{A}}$.Therefore,$\sin C = \frac{1}{3/2} = \frac{2}{3}$.Thus,the critical angle is $C = \sin^{-1}\left(\frac{2}{3}ight)$.

Light travels from an optically denser medium $A$ into the optically rarer medium $B$ with speeds $1.8 \times 10^{8} \ m/s$ and $2.7 \times 10^{8} \ m/s$ respectively. The critical angle between them is ($\mu_{A}$ and $\mu_{B}$ are the refractive indices of media $A$ and $B$ respectively).

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