A light wave enters from medium 1 to medium 2 | vedclass

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(A) Given that the velocity in the $2^{	ext{nd}}$ medium $(v_2)$ is double the velocity in the $1^{	ext{st}}$ medium $(v_1)$,we have $v_2 = 2v_1$.Si

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(A) Given that the velocity in the $2^{	ext{nd}}$ medium $(v_2)$ is double the velocity in the $1^{	ext{st}}$ medium $(v_1)$,we have $v_2 = 2v_1$.Since the refractive index $\mu$ is inversely proportional to velocity $(\mu \propto \frac{1}{v})$,we have $\frac{\mu_1}{\mu_2} = \frac{v_2}{v_1} = \frac{2v_1}{v_1} = 2$.For total internal reflection to occur,light must travel from a denser medium to a rarer medium,which is satisfied here as $\mu_1 > \mu_2$.The critical angle $C$ is given by $\sin C = \frac{\mu_2}{\mu_1} = \frac{1}{2}$.Thus,$C = \arcsin(0.5) = 30^{\circ}$.For total internal reflection,the angle of incidence $i$ must be greater than the critical angle,so $i > 30^{\circ}$.

$A$ light wave enters from medium $1$ to medium $2$. Its velocity in the $2^{\text{nd}}$ medium is double that in the $1^{\text{st}}$ medium. For total internal reflection,the angle of incidence must be greater than ....... $^{\circ}$.

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