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_1n_2n_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 relationship between critical angles and refractive indices, and applying Snell's law at the interface, we can derive the necessary expression for $ \theta_{c1} $. The given condition $ \sin\theta_{c2} - \sin\theta_{c1} = \frac{1}{2} $ helps in finding the value of $ \theta_{c1} $. Thus, the correct expression for $ \theta_{c1} $ is $ \sin^{-1} \left( \frac{5}{6n_1} \right) $.

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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The Correct Option is A

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