Question

Difficulty: Very hardWave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

A beam of light traveling in air is incident on a transparent liquid at the polarizing angle (Brewster's angle) of 53.153.1^\circ, where tan53.1=1.33\tan 53.1^\circ = 1.33. What is the critical angle for total internal reflection when light travels from this liquid into air?

  1. sin1(0.75)\sin^{-1}(0.75)Answer
  2. B
    sin1(0.60)\sin^{-1}(0.60)
  3. C
    sin1(1.33)\sin^{-1}(1.33)
  4. D
    53.153.1^\circ

Answer

The critical angle for total internal reflection at the liquid-air boundary is sin1(0.75)\sin^{-1}(0.75).
According to Brewster's law, the refractive index of the liquid is given by n=tan(53.1)=1.33=43n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}. When light travels from the denser liquid medium to the rarer air medium, the critical angle θc\theta_c for total internal reflection satisfies sinθc=1n\sin\theta_c = \frac{1}{n}. Substituting n=43n = \frac{4}{3} gives sinθc=34=0.75\sin\theta_c = \frac{3}{4} = 0.75, so θc=sin1(0.75)\theta_c = \sin^{-1}(0.75).

Step-by-Step Solution

1
Determine the refractive index of the liquid using Brewster's law.
n=tan(53.1)=1.33=43n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}.
Brewster's law states that when light in air (n1=1n_1 = 1) is incident at the polarizing angle θB\theta_B on a medium of index nn, tanθB=n\tan\theta_B = n.
2
Apply the total internal reflection condition for light passing from liquid to air.
sinθc=1n=14/3=34=0.75\sin\theta_c = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4} = 0.75.
Total internal reflection occurs at an interface when light travels from a denser medium (nn) to a less dense medium (11) at an angle greater than θc\theta_c, where sinθc=1n\sin\theta_c = \frac{1}{n}.
3
Solve for the critical angle θc\theta_c.
θc=sin1(0.75)\theta_c = \sin^{-1}(0.75).
Taking the inverse sine of 0.750.75 yields the critical angle.

Key Concept

Synthesizing Brewster's law of polarization with total internal reflection critical angle
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