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(B) For total internal reflection $(TIR)$ to occur at point $B$,the angle of incidence $	heta^{\prime \prime}$ must be greater than or equal to the c

Vedclass · Accepted Answer

$\sin ^{-1} \frac{\sqrt{7}}{3}$

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(B) For total internal reflection $(TIR)$ to occur at point $B$,the angle of incidence $	heta^{\prime \prime}$ must be greater than or equal to the critical angle $C$. The condition for grazing emergence at point $B$ is $\sin 	heta^{\prime \prime} = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.From the geometry of the triangle,we have $	heta^{\prime} = 90^{\circ} - 	heta^{\prime \prime}$.Applying Snell's Law at point $A$ on the top surface:$1 	imes \sin 	heta = \mu 	imes \sin 	heta^{\prime}$$\sin 	heta = \frac{4}{3} 	imes \sin(90^{\circ} - 	heta^{\prime \prime})$$\sin 	heta = \frac{4}{3} 	imes \cos 	heta^{\prime \prime}$Since $\sin 	heta^{\prime \prime} = \frac{3}{4}$,we find $\cos 	heta^{\prime \prime} = \sqrt{1 - \sin^2 	heta^{\prime \prime}} = \sqrt{1 - (3/4)^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$.Substituting this into the equation for $\sin 	heta$:$\sin 	heta = \frac{4}{3} 	imes \frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3}$$	heta = \sin ^{-1} \left( \frac{\sqrt{7}}{3} ight)$.

$A$ ray of light enters from air into a denser medium of refractive index $\mu = \frac{4}{3}$,as shown in the figure. The light ray undergoes total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to -

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