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(B) The problem states that signals sent from $A$ towards $B$ reach point $C$. This implies that the signals undergo total internal reflection at the

Vedclass · Accepted Answer

$2.4 	imes 10^8 \,m/s$

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(B) The problem states that signals sent from $A$ towards $B$ reach point $C$. This implies that the signals undergo total internal reflection at the glass-vacuum interface at point $B$.For total internal reflection to occur,the angle of incidence $	heta$ must be greater than the critical angle $	heta_C$.From the geometry of the circle,the angle of incidence at $B$ is $	heta = 45^{\circ}$.Thus,$	heta > 	heta_C \implies 45^{\circ} > \sin^{-1}(1/\mu)$.$\sin 45^{\circ} > 1/\mu \implies 1/\sqrt{2} > 1/\mu \implies \mu > \sqrt{2}$.Since the speed of light in glass is $v = c/\mu$,where $c = 3 	imes 10^8 \,m/s$,we have:$v Therefore,the speed $v$ must be less than $2.12 	imes 10^8 \,m/s$. Among the given options,$2.4 	imes 10^8 \,m/s$ is greater than this limit and thus cannot be the speed of the signals in glass.

It is found that all electromagnetic signals sent from $A$ towards $B$ reach point $C$. The speed of electromagnetic signals in glass can not be

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