# How does light interact with the whole surface without the angles of reflection and incidence being equal?

Suppose we have a light source shining all over the surface , and there is a photon detector separated by a block Q . I would expect it to have only one line at the center the center , where the angle of incidence is equal to angle of reflection getting reflected and reach the photon detector but that it is not true , then what it is true then ?

• A proper reference should be given: from pg. 43 of Feynman, QED: The Strange Theory of Light and Matter. Commented Mar 5, 2022 at 15:23

The picture in the question seems to be from a text discussing the way the electromagnetic contributions of individual paths of light sum up, producing the classical effect since most paths interfere with each other, leaving the path with the minimum time as the main contribution (Huygen's principle).

Why doesn't these paths obey angle of incidence = angle of reflection? This is because the model is based on Huygen's idea that each particle in the surface scatters light equally in all directions: there will be light all over the place. The cool thing is that this microscopic description of light-matter interaction then produces an emergent macroscopic behavior which is similar to what we expect.

In practice "incidence = reflection" is only partially true for real surfaces. A perfectly sharp light beam will produce a distribution with more or less intensity for different angles. A rough surface gives a broad distribution, a polished one a sharp one, centred on the "right" angle.

The diagram that you have cited is an attempt to explain the answer why.

# A lay explanation

In plain words: interference. You can assume that light of a single frequency takes all possible paths from a given source to a given destination. Like a tiny little bit of it goes in every direction. You can just postulate that at the beginning and get the same behavior that you would always get. How? Interference.

The basic mechanism is that if we make this assumption, then we don't really have to worry about the paths where light curves around the obstacles, say, because any path that curves around obstacles is next to other paths which take a slightly longer or slightly shorter time, and therefore have a slightly different phase. Actually, as this diagram shows, they can be quite wildly different because the frequency of light is so high. Because all of these nearby paths interfere destructively with each other, there is no appreciable loss in intensity if you place a large obstacle in that space, you are removing just as many bits of the light that would have destructively interfered with the final wave, as the amount of bits that would have interfered constructively. So curvy paths are allowed but they interfere with their nearby paths.

Once you see this logic you worry that maybe every path interferes destructively with nearby paths, so maybe this postulate doesn't really work. What the diagram argues is, that objection is not correct. In fact, very often a path will be stationary: meaning nearby pants take exactly as long because we're at a minimum or a maximum or an inflection point or something like that. Usually a minimum.

So the reason that light takes a straight path is because the straight path is legitimately the fastest way between two points. All of the nearby paths take about the same time because the time is minimized. And so all of the waves can constructively interfere, at the minimum time path.

This is called Fermat's principle, after mathematician Pierre de Fermat, who came up with it trying to rewrite Snell's law in a more “perfect” way to spite his rival intellectual Renee Descartes: Fermat realized that light always takes a minimum time path, whether in reflection or refraction or straight line motion.

# A more technical explanation

In the much more terse language of modern mathematical physics, the free wave equation $$\frac{\partial ^2 f}{\partial t^2}=\nabla f$$ is a linear equation and therefore obeys the principle of superposition, making it amenable to Green’s function methods. Usually a Green’s function approach would be used to handle some driving force, but the free equation of course doesn't have that. However the equation, like all PDEs, depends on a complicated boundary condition, say for example $$f(\mathbf r,t_0)=f_0(\mathbf r)$$, and we can use superposition on that to write $$f(x,y,z,t)=\int_{\mathbb R^3} \mathrm d^3r'~f_0(\mathbf r')~G(\mathbf r,t,\mathbf r'),$$where G is a sort of Green’s function solving the problem with a 3D Dirac $$\delta$$-function as its $$f_0.$$ but what does that function do? It propagates out a spherical bubble of light centered at $$\mathbf r=\mathbf r'$$.

This doesn't immediately look like it includes the curvy paths but on further reflection it has to: allow a Green’s function to propagate over a short time $$\delta t$$ and it will create a bubble of size $$c~\delta t$$, now express that bubble in terms of Green’s functions! Those given chunks of light also travel in all directions the same way, and 100% of those paths are not on a straight line (everything except for a set of measure zero), describing instead a curving light path from 0 to $$\delta t$$ to $$2\delta t$$ to $$3\delta t$$... .

• Wonderful explanation , yet just one question , Is this assumption correct , with the assumption being " That light of a single frequency takes all possible paths from a given source to a given destination" , if this correct , then why is it correct ? Commented Mar 6, 2022 at 5:48
• It's mathematically equivalent to the standard wave equation approach, so no experiment even in principle can tell the difference... once they are mathematically equivalent they are either both correct or both incorrect. In this case the description has been very good first classically, then in quantum electrodynamics. Feynman offers a “proof” in his New Zealand lectures that photons take all paths: erase parts of the mirror periodically and light will no longer reflect with equal angles. This can be seen on the bottom of CDs and DVDs, as the angles are different for different frequencies. Commented Mar 6, 2022 at 15:36