The speed of light in media M_1 and M_2 are 1 | vedclass

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(C) The refractive index of medium $M_1$ is given by $\mu_1 = c/v_1 = (3 	imes 10^8) / (1.5 	imes 10^8) = 2$.The refractive index of medium $M_2$ is

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$> \sin^{-1}(3/4)$

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(C) The refractive index of medium $M_1$ is given by $\mu_1 = c/v_1 = (3 	imes 10^8) / (1.5 	imes 10^8) = 2$.The refractive index of medium $M_2$ is given by $\mu_2 = c/v_2 = (3 	imes 10^8) / (2.0 	imes 10^8) = 1.5 = 3/2$.For total internal reflection to occur,the light must travel from a denser medium to a rarer medium,and the angle of incidence $i$ must be greater than or equal to the critical angle $C$.The critical angle $C$ is defined by $\sin C = \mu_2 / \mu_1$.Substituting the values,$\sin C = (3/2) / 2 = 3/4$.Therefore,for total internal reflection,$i \geq C$,which implies $\sin i \geq \sin C$.Thus,$i \geq \sin^{-1}(3/4)$.

The speed of light in media $M_1$ and $M_2$ are $1.5 \times 10^8 \ m/s$ and $2.0 \times 10^8 \ m/s$ respectively. $A$ ray of light enters from medium $M_1$ to $M_2$ at an incidence angle $i$. If the ray suffers total internal reflection,the value of $i$ is

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