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

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

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$30$

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(A) Given that the velocity in the $2^{nd}$ medium $(v_2)$ is double the velocity in the $1^{st}$ medium $(v_1)$,we have $v_2 = 2v_1$.Since the refractive index $\mu$ is inversely proportional to velocity $(\mu = c/v)$,the ratio of refractive indices is $\frac{\mu_1}{\mu_2} = \frac{v_2}{v_1} = 2$.Total internal reflection occurs when light travels from a denser medium to a rarer medium. Here,$\mu_1 > \mu_2$.The critical angle $C$ is given by $\sin C = \frac{\mu_2}{\mu_1}$.Substituting the values,$\sin C = \frac{1}{2}$.Therefore,$C = \arcsin(0.5) = 30^o$.For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$. Thus,$i > 30^o$.

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

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