Polarization by Total Internal Reflection

Task number: 2329

The angle of total internal reflection for transparent enamel is 58°. What is the index of refraction of enamel?

Determine the polarization angle of diamond with index of refraction 2,42.

  • Hint 1

    How does the polarization by reflection work and which relation holds for Brewster’s angle of incidence?

  • Solution

    The light must come under a particular angle of incidence, so called Brewster’s angle αB, to be reflected with complete polarization. At this angle reflected and refracted rays form an angle of 90°. We can get the following relation:

    \[\alpha_\mathrm{B}+ 90° + \beta = 180°.\]

    For detailed derivation see the Hint 1 solution.

    Simplifying we get:

    \[\beta = 90° − \alpha_\mathrm{B}. \]

    Modifying the Snell’s law, we get:

    \[\frac{\sin\, \alpha_\mathrm{B}}{\sin\, \beta}=\frac{\sin\, \alpha_\mathrm{B}}{\sin\,(90° − \alpha_\mathrm{B})}=\frac{\sin\, \alpha_\mathrm{B}}{\cos\,\alpha_\mathrm{B}}=\mathrm{tg}\,\, \alpha_\mathrm{B}= \frac{n_\mathrm{2}}{n_\mathrm{1}}.\]

    As the task assignment did not mention differently, we assume that the problem is happening in vacuum or in air, so we can substitute for n1 = 1.

    \[\mathrm{tg}\,\, \alpha_\mathrm{B}= \frac{n_\mathrm{2}}{1},\]


    \[\mathrm{tg}\,\, \alpha_\mathrm{B}= n_\mathrm{2}.\]

    We can substitute for the Brewster’s angle αB = 58° from the assignment and calculate the index of refraction of enamel:


    We can also calculate Brewster’s angle for diamond with the index of refraction n2 = 2,42:

    \[\mathrm{tg}\,\, \alpha_\mathrm{B}= n_\mathrm{2} = 2{,}42,\] \[\alpha_\mathrm{B} \dot=\,68°.\]
  • Answer

    The index of refraction of enamel is n = 1,6 and the polarization angle of diamond is αB = 68°.

Difficulty level: Level 2 – Upper secondary level
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