It is found that electromagnetic signals sent | vedclass

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(B) This is a case of total internal reflection at point $B$.From the geometry of the circle,the angle of incidence $	heta$ at point $B$ is $45^{\cir

Vedclass · Accepted Answer

$2.4 	imes 10^8 	ext{ m/s}$

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(B) This is a case of total internal reflection at point $B$.From the geometry of the circle,the angle of incidence $	heta$ at point $B$ is $45^{\circ}$.For the signal to reach point $C$ after reflection at $B$,total internal reflection must occur at $B$. Thus,the angle of incidence $	heta$ must be greater than the critical angle $	heta_C$.$	heta > 	heta_C \Rightarrow 45^{\circ} > 	heta_C \Rightarrow \sin 45^{\circ} > \sin 	heta_C$Since $\sin 	heta_C = \frac{1}{\mu}$,we have $\frac{1}{\sqrt{2}} > \frac{1}{\mu} \Rightarrow \mu > \sqrt{2}$.Given the speed of light in glass $v = \frac{c}{\mu}$,where $c = 3 	imes 10^8 	ext{ m/s}$,we have $\mu = \frac{c}{v}$.Substituting this into the inequality: $\frac{c}{v} > \sqrt{2} \Rightarrow v $v Among the given options,only $2.4 	imes 10^8 	ext{ m/s}$ is greater than $2.12 	imes 10^8 	ext{ m/s}$,so it is not possible.

It is found that electromagnetic signals sent inside a glass sphere from $A$ towards $B$ reach point $C$. The speed of electromagnetic signals in glass cannot be:

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