Question

The critical angle for light going from medium A into medium B is $\theta$. The speed of light in medium A is $V_A$. What is the speed of light in medium B?

• A. $V_A\sin\theta$
• B. $V_A\tan\theta$
• C. $\frac{V_A}{\tan\theta}$
• D. $\frac{V_A}{\sin\theta}$

Hint:

Total internal reflection only happens when light goes from a denser medium to a rarer medium. This means light travels faster in medium B than in medium A ($V_B > V_A$). Since $\sin\theta \le 1$, dividing $V_A$ by $\sin\theta$ results in a value larger than $V_A$. This immediately rules out option (A), as multiplying by $\sin\theta$ would make the speed smaller!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A light ray travels from an optically denser medium A into an optically rarer medium B, with a critical angle $\theta$. Given the velocity of light in medium A ($V_A$), we need to find the speed of light in medium B ($V_B$).

Step 2: Key Formula or Approach:
1. The relationship defining the critical angle $\theta$ for two media is given by: $$\sin\theta = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \frac{\mu_B}{\mu_A}$$ 2. The refractive index $\mu$ of any medium is inversely proportional to the speed of light in that medium ($\mu = \frac{c}{v}$). Therefore: $$\frac{\mu_B}{\mu_A} = \frac{V_A}{V_B}$$ 3. Equating both relationships gives: $$\sin\theta = \frac{V_A}{V_B}$$

Step 3: Detailed Explanation:
Let's set up our critical angle equation using the velocities: $$\sin\theta = \frac{V_A}{V_B}$$ We want to solve for the speed of light in medium B ($V_B$). Rearrange the equation to isolate $V_B$: $$V_B \cdot \sin\theta = V_A$$ $$V_B = \frac{V_A}{\sin\theta}$$

Step 4: Final Answer:
The speed of light in medium B is $\frac{V_A}{\sin\theta}$, which corresponds to option (D).