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(A) The situation is shown in the figure. Light will not emerge from the vertical side $BC$ if the angle of incidence $i$ at the side $BC$ is greater

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) The situation is shown in the figure. Light will not emerge from the vertical side $BC$ if the angle of incidence $i$ at the side $BC$ is greater than the critical angle $	heta_c$.Condition for total internal reflection: $i > 	heta_c$ or $\sin i > \sin 	heta_c$.Since $\sin 	heta_c = \frac{1}{\mu}$,we have $\sin i > \frac{1}{\mu} \quad ......(i)$Applying Snell's law at point $O$ on the bottom surface: $1 	imes \sin 	heta = \mu \sin r$.In $	riangle OPR$,$r + 90^\circ + i = 180^\circ$,so $r + i = 90^\circ$,which means $r = 90^\circ - i$.Therefore,$\sin 	heta = \mu \sin(90^\circ - i) = \mu \cos i$,which gives $\cos i = \frac{\sin 	heta}{\mu}$.Using $\sin i = \sqrt{1 - \cos^2 i}$,we get $\sin i = \sqrt{1 - \left(\frac{\sin 	heta}{\mu}ight)^2} \quad ......(ii)$.Substituting $(ii)$ into $(i)$:$\sqrt{1 - \frac{\sin^2 	heta}{\mu^2}} > \frac{1}{\mu} \implies 1 - \frac{\sin^2 	heta}{\mu^2} > \frac{1}{\mu^2} \implies 1 > \frac{1 + \sin^2 	heta}{\mu^2}$.$\mu^2 > 1 + \sin^2 	heta$.For the letters to be invisible for all angles $	heta$,we consider the maximum value of $\sin^2 	heta$,which is $1$.$\mu^2 > 1 + 1 = 2 \implies \mu > \sqrt{2}$.Thus,the minimum refractive index is $\mu_{\min} = \sqrt{2}$.

$A$ rectangular glass block is placed on a printed paper lying on a horizontal surface. Find the minimum refractive index of the glass for which the letters on the paper are not visible from any of the vertical sides of the block.

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