At the interface between two materials having refractive indices $ n_1 $ and $ n_2 $, the critical angle for reflection of an EM wave is $ \theta_c $. The $ n_1 $ material is replaced by another material having refractive index $ n_3 $, such that the critical angle at the interface between $ n_1 $ and $ n_3 $ materials is $ \theta_{c3} $. If $ n_1 n_2 n_3 $, $ \frac{n_2}{n_3} = \frac{2}{5} $, and $ \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} $, then $ \theta_{c1} $ is:

Question

At the interface between two materials having refractive indices $ n_1 $ and $ n_2 $, the critical angle for reflection of an EM wave is $ \theta_c $. The $ n_1 $ material is replaced by another material having refractive index $ n_3 $, such that the critical angle at the interface between $ n_1 $ and $ n_3 $ materials is $ \theta_{c3} $. If $ n_1 > n_2 > n_3 $, $ \frac{n_2}{n_3} = \frac{2}{5} $, and $ \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} $, then $ \theta_{c1} $ is:

cdquestions Admin · Accepted Answer

Using the given relations for the critical angle and refractive indices, we can derive the expression for $ \theta_{c1} $. The critical angle is related to the refractive index by the equation $ \sin \theta_c = \frac{n_2}{n_1} $, and the difference between the sines of the critical angles for the two interfaces gives the desired result.

Answer

$ \sin^{-1} \left( \frac{5}{6n_1} \right) $

Answer

$ \sin^{-1} \left( \frac{2}{3n_1} \right) $

Answer

$ \sin^{-1} \left( \frac{1}{3n_1} \right) $

Answer

$ \sin^{-1} \left( \frac{1}{6n_1} \right) $

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