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 (\(\theta_c\)) is the angle of incidence in a denser medium for which the angle of refraction in the rarer medium is 90°. For total internal reflection or critical angle to exist, light must travel from a denser medium (lower velocity) to a rarer medium (higher velocity).

Step 2: Key Formula or Approach
1. Refractive index \(n = \frac{c}{v}\).
2. \(\sin \theta_c = \frac{n_{rarer}}{n_{denser}} = \frac{v_{denser}}{v_{rarer}}\).

Step 3: Detailed Explanation
1. Identify Media: - Velocity in medium 1 (\(v_1\)) = \(2 \times 10^8\) m/s (Rarer medium). - Velocity in medium 2 (\(v_2\)) = \(1.5 \times 10^8\) m/s (Denser medium).
2. Apply Formula: To find the critical angle, the ray must originate in medium 2. \[ \sin \theta_c = \frac{v_{denser}}{v_{rarer}} = \frac{v_2}{v_1} \] \[ \sin \theta_c = \frac{1.5 \times 10^8}{2 \times 10^8} = \frac{1.5}{2} = 0.75 \]
3. Final Calculation: \[ \theta_c = \sin^{-1}(0.75) \]

Step 4: Final Answer
The critical angle is \(\sin^{-1}(0.75)\).