The critical angle of light going from medium | vedclass

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Question

(A) The condition for the critical angle $	heta$ when light travels from a denser medium $A$ to a rarer medium $B$ is given by Snell's Law:$n_A \sin

Vedclass · Accepted Answer

$\frac{v}{\sin 	heta}$

Vedclass · Answer

(A) The condition for the critical angle $	heta$ when light travels from a denser medium $A$ to a rarer medium $B$ is given by Snell's Law:$n_A \sin 	heta = n_B \sin 90^\circ$Since $\sin 90^\circ = 1$,we have:$n_A \sin 	heta = n_B$$\frac{n_A}{n_B} = \frac{1}{\sin 	heta}$We know that the refractive index $n$ is inversely proportional to the speed of light $v$ in that medium $(n = \frac{c}{v})$,so:$\frac{n_A}{n_B} = \frac{v_B}{v_A}$Equating the two expressions:$\frac{v_B}{v_A} = \frac{1}{\sin 	heta}$Given $v_A = v$,we find:$v_B = \frac{v_A}{\sin 	heta} = \frac{v}{\sin 	heta}$

The critical angle of light going from medium $A$ to medium $B$ is $\theta$. The speed of light in medium $A$ is $v$. The speed of light in medium $B$ is:

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