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(C) The critical angle $	heta_C$ at an interface between two media with refractive indices $n_{dense}$ and $n_{rare}$ is given by $\sin 	heta_C = \f

Vedclass · Accepted Answer

$\sin^{-1}(\frac{5}{6})$

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(C) The critical angle $	heta_C$ at an interface between two media with refractive indices $n_{dense}$ and $n_{rare}$ is given by $\sin 	heta_C = \frac{n_{rare}}{n_{dense}}$.Given $n_2 > n_1$,$\sin 	heta_{1C} = \frac{n_1}{n_2}$.Given $n_3 > n_1$,$\sin 	heta_{2C} = \frac{n_1}{n_3}$.We are given $\sin 	heta_{2C} - \sin 	heta_{1C} = \frac{1}{2}$.Substituting the expressions: $\frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2}$.We can write $\frac{n_1}{n_3} = \frac{n_1}{n_2} \cdot \frac{n_2}{n_3}$.Since $\frac{n_2}{n_3} = \frac{2}{5}$,we have $\frac{n_1}{n_2} \cdot \frac{2}{5} - \frac{n_1}{n_2} = \frac{1}{2}$.Let $x = \frac{n_1}{n_2} = \sin 	heta_{1C}$.$x(\frac{2}{5} - 1) = \frac{1}{2} \implies x(-\frac{3}{5}) = \frac{1}{2} \implies x = -\frac{5}{6}$.Since $\sin 	heta_{1C}$ must be positive and less than $1$,there is a contradiction in the problem statement's provided values ($n_3 > n_2 > n_1$ implies $\sin 	heta_{2C} < \sin 	heta_{1C}$,so $\sin 	heta_{2C} - \sin 	heta_{1C}$ should be negative). Assuming the magnitude is intended: $\sin 	heta_{1C} = \frac{5}{6}$,so $	heta_{1C} = \sin^{-1}(\frac{5}{6})$.

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