If the surface of prism is coated with an anti-reflective coating, (specifically an index matching, refractive index gradient, or moth-eye structure), and light impinges at a greater than critical angle (i.e. would otherwise be reflected by total internal reflection), at what angle does the light exit the prism?


1 Answer 1


Answer: you can ignore the coating (assuming monotonic index of refraction); light does not exit if incident ray is beyond critical angle

Reasoning: Snell's Law states $$n_1 \sin \theta_1 = n_2 \sin \theta_2,$$ where $n_1$ and $n_2$ are indexes of reflection within media $M_1$ and $M_2$, and $\theta_1$ and $\theta_2$ are the angle between the normal and the light ray within the respective media.

I presume by "prism" you mean an interface $M_1$ to $M_2$, with $n_1 > n_2$.

Total internal reflections (TIR) occur when $\sin \theta_2 = \frac{n_1}{n_2} \sin\theta_1 > 1$, i.e., where transmission would be absurd since there is no real solution for $\theta_2$.

Now consider the case with $3$ media $M_1$, $M_2$, $M_3$ with parallel interfaces. Snell's law becomes $$n_1 \sin \theta_1 = n_2 \sin \theta_2 = n_3 \sin \theta_3 = \text{constant}.$$ So it doesn't mater what $n_2$ is. As long as $\sin\theta_3 = \frac{\text{constant}}{n_3} > 1$, then $\theta_3$ has no real solution, and light will not enter $M_3$ (in fact, TIR may occur from the $M_1$-$M_2$ interface and never reach $M_3$ in the first place).

We can generalize the above to multiple layers. In the limit we would have a (assumed monotonic) gradient of index of fractions from $n_1$ to $n_2$, for each plane parallel to a common plane (from which the normal is defined).

Now consider a light ray coming from $M_1$ that would normally undergo TIR against $M_2$. For the gradient case, the light ray would bend and become increasing parallel to the plane, until it encounters a layer $M_i$ with $\frac{n_1}{n_i} \sin \theta_1 = 1$. Thereafter the ray bends back toward $M_1$, effectively undergoing TIR.

(Edit: updated "Answer" section part to match what the question is asking)

  • $\begingroup$ Thanks for the detailed response, arccosh. What about if the coating instead utilised 'frustrated total internal reflection'? I had assumed anti-reflective coating and frustrated total internal reflection were closely related. $\endgroup$
    – Xavier
    Jul 9, 2014 at 6:33
  • $\begingroup$ At the opposite side of the TIR the electric field of the light wave is non-zero; is has a spatially decaying solution. This is know as an evanescent wave. If you have another surface very close to the TIR surface then some of the evanescence waves amplitude will enter the other medium. This is frustrated TIR. $\endgroup$
    – boyfarrell
    Jul 9, 2014 at 11:24
  • $\begingroup$ No argument with your answer; just pointing out that your statement "monotonic index of refraction" is important. A cleverly-designed multilayer stack could act almost like a lens and allow the effective TIR angle to be increased. $\endgroup$ Jul 9, 2014 at 12:01
  • $\begingroup$ Thanks boyfarrel for the addendum re. evanescent wave. I used to think that it was "ugly", but realized that it's necessary; otherwise a $M_1$-$M_2$-$M_1$ sandwich with infinitely think $M_2$ would have huge influence. $\endgroup$
    – arccosh
    Jul 9, 2014 at 14:13
  • $\begingroup$ @ Carl Witthoft: I emphasized on the "monotonic index of refraction" to constrain the problem, and prevent the case where the index got smaller than $M_1$, and then the layer would matter as it would block light at even more angles. Would the multi layer stack you mentioned have periodic structures at the resolution of wavelength, so interference becomes important? I didn't consider that in my answer.If that were the case, monotonicity would not suffice, so I'd fall back to linearly increasing. :) $\endgroup$
    – arccosh
    Jul 9, 2014 at 14:18

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