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:
Apply the conditions for total internal reflection to deduce the relationship between light speeds in two media.

Step 2: Key Formula or Approach:
Total internal reflection occurs only when light travels from an optically denser medium to a rarer one. Thus V_B>V_A. The critical angle relation is sin θ_c = V_A/V_B, so V_B = V_A/sin θ_c.

Step 3: Detailed Explanation:
Since sin θ_c is always less than or equal to 1, dividing V_A by sin θ_c yields a value larger than V_A. This immediately rules out any option where V_B is obtained by multiplying V_A by sin θ_c, as that would produce a smaller speed—contradicting the denser-to-rarer requirement. The division correctly reflects that light must travel faster in the rarer medium B.

Step 4: Final Answer:
V_B = V_A/sin θ_c, ruling out any multiplicative option.