Question

A light ray enters from medium 1 to medium 2. Its velocity in medium 1 is \( 2 \times 10^{8} \) m/s and in medium 2 is \( 1.5 \times 10^{8} \) m/s. The critical angle for the pair of media is:

• A. \(\sin^{-1}(0.75)\)
• B. \(\sin^{-1}(0.5)\)
• C. \(\sin^{-1}(0.66)\)
• D. \(\sin^{-1}(0.8)\)
• E.

Hint:

To avoid confusion about which velocity goes on top, remember that \(\sin \theta\) can never be greater than 1. Therefore, in the ratio of velocities, the smaller velocity must always be in the numerator.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The critical angle is defined when light travels from a denser medium (lower velocity) to a rarer medium (higher velocity). It is the angle of incidence for which the angle of refraction is $90^\circ$.
Step 2: Key Formula or Approach:
Using Snell's Law at the critical angle $C$:
\[ n_{\text{denser}} \sin C = n_{\text{rarer}} \sin 90^\circ \]
Since refractive index $n = \frac{c}{v}$, we can write:
\[ \sin C = \frac{n_{\text{rarer}}}{n_{\text{denser}}} = \frac{v_{\text{denser}}}{v_{\text{rarer}}} \]
Step 3: Detailed Explanation:
Given velocities:
Velocity in medium 1, $v_1 = 2 \times 10^8 \text{ m/s}$ (higher speed, so it is the rarer medium).
Velocity in medium 2, $v_2 = 1.5 \times 10^8 \text{ m/s}$ (lower speed, so it is the denser medium).
Substituting the values into the formula:
\[ \sin C = \frac{1.5 \times 10^8}{2 \times 10^8} = \frac{1.5}{2} = 0.75 \]
Therefore, the critical angle is:
\[ C = \sin^{-1}(0.75) \]
Step 4: Final Answer:
The critical angle for the pair of media is $\sin^{-1}(0.75)$.