# Does thin film interference (anti-reflective coating) let more light through?

The theory of an anti-reflective coating is that the reflected light off the coating and the reflected light off the substrate is 180 degrees out of phase, causing destructive interference and subsequently no light is 'reflected'. But how does this process allow more light to pass through the substrate. Because although the reflected light is causing destructive interference, before the light leaves the coating, it is still reflected, meaning that light has still been lost.

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I agree that this idea can seem weird. There are two was you can resolve the "paradox" you raise.

1. You can think of the coating as a multiple reflexion: this is done both in Born and Wolf "Principles of Optics" i section 7.6 in the Sixth Edition and in Hecht "Optics" (which I don't have before me). The first wave bounces off the coating's outer interface as you say, but it is soon joined and mostly cancelled by the coating's inner, out-of-phase bounce. One gets an infinite sequence of bounces at different phases that sum up as an infinite geometric sequence. So, when the light wavefront first reaches the antireflexion coating, the loss you talk about indeed does happen, but it is only for a fleetingly short time. All the bounces progressively add up over time so that the steady state behaviour is reached and the initially big losses are cancelled.

2. We can analyse the behaviour of light by using a Fourier transform to split it up into its perfectly time-harmonic components. So we get system transmission $t(\omega)$ and reflexion $r(\omega)$ co-efficients as functions of frequency $\omega$ and use these to analyse the behaviours of the the frequency components separately and then use the inverse Fourier transform to build up the total, transient system response. Now think for a moment about a single frequency component of light. It has no beginning, no end and no cause and effect behaviour: it is simply "there", delocalised over all space and you solve Maxwell's equations for its propagation subject to all boundary conditions everywhere at once. So the seemingly non-causal (or acausal) behaviour is not paradoxical for an entity like a lone frequency wave. The causal behaviour of the time-limited light pulse is a phenomenon that emerges when we gather all the acausal, lone frequency behaviours and build them up into the pulse that fulfills the initial time conditions;

The first approach gives you what you need more directly, but the second I think is more interesting. Maxwell's equations are invariant with respect to a time flip i.e. $t\mapsto-t$. Every retarded wave solution to Maxwell's equations, for example, the spherical wave electric potential pulse $V(x,y,z) = f(\sqrt{x^2 + y^2 + z^2} - c\,t)/\sqrt{x^2 + y^2 + z^2}$ can be replaced by the corresponding advanced wave solution wherein we replace $t$ by $-t$ and still have a valid solution. Causality must be "put into our solutions by hand"; we do this by choosing retarded waves over advanced waves when we add up time-harmonic solutions to fulfill our initial time conditions. In the first explation above, we do this by assuming that the electromagnetic field has an abrupt front which travels towards the substrate from the surrounding air and reaches the coating first. There is no requirement for this order of process to get solutions of Maxwell's equations and this order does not arise from Maxwell's equations, we are imposing this order grounded on our causal intuition.

Indeed, Wheeler Feynman Absorber Theory was an intriguing early attempt by Feynman and Wheeler to get rid of the self interaction of the electron (which causes divergences in quantum electrodynamics) by making all solutions to Maxwell's equations symmetric in time as well as Maxwell's equations themselves. Both the anitcausal advanced and causal retarded solutions are present with equal weight in such solutions, but Wheeler and Feynman showed how causality can be preserved when all the electric charge of the universe is considered as a whole.

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Oops, sorry for the overly hasty downvote. If you make a trivial edit, such as inserting whitespace, I can undo the downvote. – Ben Crowell Sep 6 '13 at 14:27
What led me to initially misread your answer was the part about "when the light pulse first reaches..." I visualized this as a pulse shorter than twice the thickness of the coating, in which case the remainder of your discussion would be wrong. You might want to clarify that you're considering a pulse that is many times longer than this. – Ben Crowell Sep 6 '13 at 14:48
@BenCrowell Done. I changed "pulse" to wavefront - pulse is a bad word anyway. – WetSavannaAnimal aka Rod Vance Sep 6 '13 at 21:53
@ChrisRyder I added some comments on general causality, which you may or may not find interesting. The points 1 and 2 are unchanged. – WetSavannaAnimal aka Rod Vance Sep 9 '13 at 4:49

The usual equation for the reflected amplitude, $R_{12}=(v_2-v_1)/(v_2+v_1)$, is always valid, modulo an over-all sign that depends on your choice of how to define amplitude (e.g., for a light wave, defining it as $\textbf{E}$ or $\textbf{B}$). It's always valid because it depends only on the requirement that the wave and its derivative be continuous at the boundary, and this requirement is universal wave behavior. For an optical coating, we have two of these reflections. Let the labels 1, 2, and 3 refer to the air, coating, and glass.

When the incoming wave train has a length in the film that is shorter than twice the thickness of the film, there is no interference effect, and optimal transmission is achieved simply by maximizing the product $(1-R_{12}^2)(1-R_{23}^2)$. This is maximized by the geometric mean $v_2=\sqrt{v_1 v_3}$.

The condition for perfect transmission through an optical coating is a relation between the thickness and the wavelength. Since most waves don't have a well-defined wavelength, this clearly can't be anything general that applies to all waves, and it definitely doesn't apply to pulses shorter than the length described in the preceding paragraph. Suppose that the condition for complete transmission is satisfied for wavelength $\lambda$. If you send in a sinusoidal wave train of finite length, with this wavelength, it will not be completely transmitted. The leading portion of it undergoes some reflection at interface 12, and this reflection of the leading part suffers no destructive interference, since the later reflections can never catch up with it. Some energy is reflected. Only near the middle of a sufficiently long pulse do we approach perfect transmission.

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