Question

The critical angle for a monochromatic light going from medium A to medium B is \(\theta\). If the speed of light in medium A is V, then the speed of light in medium B is

• A. V (1 - \(\cos\theta\))
• B. \(\frac{V}{\cos\theta}\)
• C. \(\frac{V}{\sin\theta}\)
• D. V (1 - \(\sin\theta\))

Hint:

Remember that light travels faster in a rarer medium. Since \(\sin \theta\) is always \(\le 1\), dividing by it will give a value larger than \(V\), which makes physical sense as medium B is rarer.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to find the speed of light in the second medium (B) using the given critical angle and the speed of light in the first medium (A).
Step 2: Key Formula or Approach:
1. Relationship between critical angle (\(\theta\)) and refractive indices: \(\sin \theta = \frac{n_2}{n_1} = n_{21}\)
2. Relationship between refractive index and speed: \(n = \frac{c}{v} \Rightarrow \frac{n_2}{n_1} = \frac{v_1}{v_2}\)
Step 3: Detailed Explanation:
For total internal reflection to occur, light must go from a denser medium (A) to a rarer medium (B). Thus \(n_A>n_B\).
From the definition of critical angle:
\[ \sin \theta = \frac{n_B}{n_A} \]
Since refractive index is inversely proportional to the speed of light in the medium (\(n \propto 1/v\)):
\[ \frac{n_B}{n_A} = \frac{v_A}{v_B} \]
Substituting the given speed \(v_A = V\):
\[ \sin \theta = \frac{V}{v_B} \]
Solving for \(v_B\):
\[ v_B = \frac{V}{\sin \theta} \]
Step 4: Final Answer:
The speed of light in medium B is \(\frac{V}{\sin\theta}\).