The critical angle of light going from medium | vedclass

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Question

(B) The condition for the critical angle $	heta$ when light travels from medium $x$ to medium $y$ is given by Snell's Law: $\mu_x \sin 	heta = \mu_y

Vedclass · Accepted Answer

$v / \sin 	heta$

Vedclass · Answer

(B) The condition for the critical angle $	heta$ when light travels from medium $x$ to medium $y$ is given by Snell's Law: $\mu_x \sin 	heta = \mu_y \sin 90^\circ$,where $\mu_x$ and $\mu_y$ are the refractive indices of the media.Since $\sin 90^\circ = 1$,we have $\frac{\mu_y}{\mu_x} = \sin 	heta$.The speed of light in a medium is inversely proportional to its refractive index,$v = \frac{c}{\mu}$.Therefore,the speed of light in medium $y$ $(v_y)$ is related to the speed of light in medium $x$ $(v_x = v)$ by the ratio of refractive indices: $v_y = v_x 	imes \frac{\mu_x}{\mu_y}$.Substituting the ratio $\frac{\mu_x}{\mu_y} = \frac{1}{\sin 	heta}$,we get $v_y = \frac{v}{\sin 	heta}$.

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

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