A planar structure of length L and width W is | vedclass

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(B) For the maximum time,the ray of light must undergo $TIR$ at all surfaces at the minimum possible angle,which is the critical angle $	heta_c$.For

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) For the maximum time,the ray of light must undergo $TIR$ at all surfaces at the minimum possible angle,which is the critical angle $	heta_c$.For $TIR$,the condition is $n_1 \sin 	heta_c = n_2$.$\sin 	heta_c = \frac{n_2}{n_1} = \frac{1.44}{1.5} = 0.96$.The speed of light in the medium $n_1$ is $v = \frac{c}{n_1} = \frac{3 	imes 10^8}{1.5} = 2 	imes 10^8 \ m/s$.Let the total path length of the ray inside the structure be $D$. The horizontal length is $L$. The angle the ray makes with the normal to the surface is $	heta_c$. Therefore,the horizontal component of the path is $D \sin 	heta_c = L$.Thus,$D = \frac{L}{\sin 	heta_c}$.The time taken is $t_{total} = \frac{D}{v} = \frac{L}{v \sin 	heta_c} = \frac{9.6}{(2 	imes 10^8) 	imes 0.96} = \frac{9.6}{1.92 	imes 10^8} = 5 	imes 10^{-8} \ s = 50 	imes 10^{-9} \ s$.Therefore,$t = 50$.

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Reason: Because critical angle $\theta_c = \sin^{-1} \left( \frac{1}{\mu} \right)$ and $\mu \propto \frac{1}{\lambda}$.

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