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Difficulty: MediumRefraction of Light, Total Internal Reflection, and Prisms

A light wave travels through medium A at a speed of 2.25×108 m s−12.25 \times 10^8\text{ m s}^{-1} and enters medium B, where its speed decreases to 1.50×108 m s−11.50 \times 10^8\text{ m s}^{-1}. What is the value of the sine of the critical angle for total internal reflection between these two media, and in which medium must the light ray originate?

  1. 23\frac{2}{3}, originating in medium BAnswer
  2. B
    23\frac{2}{3}, originating in medium A
  3. C
    32\frac{3}{2}, originating in medium B
  4. D
    32\frac{3}{2}, originating in medium A

Answer

The sine of the critical angle is 23\frac{2}{3}, and the light ray must originate in medium B.
Total internal reflection occurs when light travels from an optically denser medium to an optically rarer medium. Since the speed of light is lower in medium B (1.50×108 m s−11.50 \times 10^8\text{ m s}^{-1}) than in medium A (2.25×108 m s−12.25 \times 10^8\text{ m s}^{-1}), medium B is the denser medium. The critical angle CC satisfies sin⁡C=vdensevrare=1.50×1082.25×108=23\sin C = \frac{v_{\text{dense}}}{v_{\text{rare}}} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}. Therefore, the light must originate in medium B and the sine of the critical angle is 23\frac{2}{3}.

Step-by-Step Solution

1
Determine the relative optical densities of medium A and medium B from wave speed
Medium B has a lower light speed (1.50×108 m s−11.50 \times 10^8\text{ m s}^{-1}) than medium A (2.25×108 m s−12.25 \times 10^8\text{ m s}^{-1}), so medium B is optically denser than medium A.
Refractive index is inversely proportional to wave speed (n∝1vn \propto \frac{1}{v}).
2
Identify the required direction of light propagation for total internal reflection
The light ray must originate in medium B and travel toward medium A.
Total internal reflection occurs only when light attempts to pass from a medium of higher refractive index (denser) to a medium of lower refractive index (rarer).
3
Calculate the sine of the critical angle
sin⁡C=vBvA=1.50×1082.25×108=23\sin C = \frac{v_B}{v_A} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}.
By Snell's law at the critical angle, sin⁡C=nAnB=vBvA\sin C = \frac{n_A}{n_B} = \frac{v_B}{v_A}.

Key Concept

Conditions for Total Internal Reflection and Critical Angle calculation from wave speeds
Estimated Time:1m 30s
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Topics

Refraction of Light, Total Internal Reflection, and Prisms

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