The figure shows a ray incident at an angle i | vedclass

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(A) According to Snell's Law,$\mu_2 \sin i = \mu_1 \sin r,$ which implies $\sin r = \frac{\mu_2}{\mu_1} \sin i = \frac{1}{k} \sin i.$$1$. For $|r - i|

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the value of $k_1$ is $\frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) According to Snell's Law,$\mu_2 \sin i = \mu_1 \sin r,$ which implies $\sin r = \frac{\mu_2}{\mu_1} \sin i = \frac{1}{k} \sin i.$$1$. For $|r - i| = 0,$ we have $r = i,$ which implies $\sin r = \sin i,$ so $k = \frac{\mu_1}{\mu_2} = 1.$ Thus,$k_o = 1.$$2$. At $k = k_1,$ the ray undergoes Total Internal Reflection $(TIR)$ or reaches the critical angle,meaning $r = \pi / 2.$ From Snell's Law: $\sin(\pi / 2) = \frac{1}{k_1} \sin(\pi / 3) \Rightarrow 1 = \frac{1}{k_1} \cdot \frac{\sqrt{3}}{2} \Rightarrow k_1 = \frac{\sqrt{3}}{2}.$$3$. At $k = k_1,$ the value of $|r - i| = |\pi / 2 - \pi / 3| = \pi / 6.$ Thus,$	heta_1 = \pi / 6.$$4$. As $k ightarrow \infty,$ $\sin r = \frac{1}{k} \sin i ightarrow 0,$ so $r ightarrow 0.$ Then $|r - i| ightarrow |0 - \pi / 3| = \pi / 3.$ Thus,$	heta_2 = \pi / 3.$Comparing these results with the options,option $A$ states $k_1 = 2/\sqrt{3},$ which is incorrect as $k_1 = \sqrt{3}/2.$ Therefore,the wrong alternative is $A$.

The figure shows a ray incident at an angle $i = \pi /3.$ If the plot drawn shows the variation of $|r - i|$ versus $\frac{\mu_1}{\mu_2} = k,$ ($r =$ angle of refraction) choose the wrong alternative.

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