Concerning the dark bands in the light "wave" interference pattern

Question

I'll begin with a with a brief and familiar example to frame the question:

|_>EXAMPLE

    When water waves pass through a double slit experiment everyone knows that an 
    interference pattern is created. The interference pattern is simply a combination 
    of crests and troughs, but the "dark bands" here represent flat water (no up/down 
    motion). This means that water is still reaching the observed wall in these dark 
    band regions. The interference pattern is thus defined with crests, troughs, and 
    flats.

|_> QUESTION

    When light passes through a double slit experiment, an interference pattern is
    created (with no recording instruments). Following the example above, the dark 
    bands created should instead be horizontal 'flat light' (light which no longer 
    exhibits wave properties, only the particle of light itself should be here). 
    Thus, light should still be reaching the observed wall in these dark band regions 
    if analogous. Why then is there no light reaching these "dark band" regions 
    instead of a flat horizontal line of light or other expected outcome based on 
    standard wave/particle motion?

I have many other questions and of course, more to read. But I think this is the most important start.

The question has been slightly addressed here, but I welcome more complicated answers: Are double-slit patterns really due to wave-like interference?

Answer 1

For water waves, the quantity that is subject to interference is the amplitude of the surface disturbance (i.e. the water's height above its undisturbed value). So as you correctly stated at regions of destructive interference the water's height remains at its undisturbed value for all times.

For light waves, the quantity that is subject to interference is the total amplitude of the light, so at regions of destructive interference the amplitude of the light is zero for all times, and so we observe no light emitted from these regions.

It is not true that two water waves shot at each other will produce a flat surface for all times. In fact, if you superimpose two traveling waves moving in opposite directions, you will get what is referred to as a standing wave. This is the case no matter what type of wave we consider (water or light or sound, etc.) The most familiar example of a standing wave, for visualization purposes, is a vibrating string. Just like a simple oscillator, energy in a standing wave is continuously converted from kinetic energy to potential energy as the wave's amplitude grows, and vice versa as its amplitude diminishes. There are, then, instants in time during the standing wave's evolution where its amplitude is zero at all points along the wave. These instants are what you refer to when you say the surface of the water is flat. But these are just instants, and moments before or after these instants the wave will be shrinking and growing, respectively. So if you shoot two lasers beams at each other you will get a standing wave. This is in fact exactly what you get in a laser cavity, where light traveling in one direction reflects off the cavity mirrors and superimposes its leading portion on top of its trailing portion, producing a standing wave.

Answer 2

There's an important difference between water waves and light waves. Water waves require a medium to travel through (i.e. water). Light waves require no medium. In fact, even postulating they do contradicts special relativity.

So when the waves in the water cancel out, the water is left behind, but when the light waves cancel out, nothing is left behind.

Answer 3

Since no one else has contributed, and the most satisfactory answer was the pdf I found, I will state that that is the answer to my question:

"The Illusion of Light as Transverse Electromagnetic Waves" (bit.ly/1etSb0n)