Critical angle

The critical angle is the angle of incidence at which light passing into a lower index medium refracts to exactly 90 degrees and grazes along the boundary. Past that angle no light is transmitted at all.

θcritical=sin⁡−1 ⁣(n2n1)\theta_{\text{critical}} = \sin^{-1}\!\left(\frac{n_2}{n_1}\right)

The CED labels this a derived equation in EK 13.3.A.5.i, which is its way of saying it is not printed on the exam sheet. Rebuild it from Snell's law by setting the refracted angle to 90 degrees: n1sin⁡θc=n2sin⁡90∘=n2n_1 \sin\theta_c = n_2 \sin 90^\circ = n_2.

The formula contains its own existence test. An arcsine needs an argument of at most 1, so a critical angle exists only when n2<n1n_2 < n_1, light travelling from the higher index medium toward the lower index one. Try it the other way and the algebra fails before the physics does. EK 13.3.A.5 states the same condition in words.

Three regimes, split at θc\theta_c.

  • Below it: some light refracts out, some reflects back.
  • At it: EK 13.3.A.5.ii, the ray refracts at 90 degrees and travels along the surface of the material.
  • Above it: EK 13.3.A.5.iii, all light is reflected and none is transmitted into the other medium.

It depends on the ratio and nothing else. Not on the shape of the boundary, not on the brightness of the beam, only on n2/n1n_2/n_1. Push the two indices together and θc\theta_c climbs toward 90 degrees, so almost nothing is trapped. Pull them apart and θc\theta_c collapses, so almost everything is. With water taken as n=1.33n = 1.33 against air at 1, θc=sin⁡−1(1/1.33)=48.8\theta_c = \sin^{-1}(1/1.33) = 48.8 degrees.

The phenomenon this angle gates is total internal reflection.

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