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Chapter 18: Problem 123

The relation between \(\theta\) (if slope of the trajectory of the path of ray at a point \(B(x, y)\) in the medium is \(\tan \theta\) ) and refractive index \(n\) at point \(B\) is (A) \(\frac{\sin \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\) (B) \(\frac{\cos \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\) (C) \(\frac{\tan \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\) (D) \(\frac{\cot \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\)

Short Answer

Expert verified
The correct relationship between the slope of the ray's trajectory at a point in the medium and the refractive index at that point is given by option (A) \(\frac{\sin \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\).

Step by step solution

01

Apply Snell's Law

According to Snell's law, which governs the behavior of light when it crosses from one medium to another with a different refractive index, we have that \(n = \sin \theta / \sin r\), where \(r\) is the angle of refraction.
02

Express \(\sin r\) in terms of \(n\)

From Snell's law, it can be derived that \(\sin r = 1 / n\).
03

Apply the arcsine function to both sides

This gives us \(\sin ^{-1}(1 / n) = r\).
04

Make the comparison

Considering the given options, the correct one is (A): \(\frac{\sin \theta}{\theta}=\frac{\sqrt{n^{2}-1}}{\sin ^{-1}(1 / n)}\), because the relationship between \(n\), \( \theta \) and \( r \) is correctly applied through Snell's Law and its derived expressions.

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