I am trying to understand why the intensity of a light wave spreads out along a "back wall" which is near a light source (assume of a narrow bandwidth, ie. red LED). From our experience the answer is obvious, and intuitive, as the "bright spot" will be found at the shortest distance between the light and the back wall, and the light intensity will drop off as you go in any direction away from this bright spot along the wall.

I'm using a 2D water wave as an analogy, and here is how I am devising the setup: at (0, 0) a source is making oscillations in the water with a fixed frequency. Assuming a steady-state system (where the water waves have propagated a sufficient distance away from the source), if we place some measurement wall at x = x', we would notice that the amplitude of the wave is strongest at (x', 0) and would drop off as we go in either direction on the y-axis (sub question- what is the exact mathematical form of this distribution? is it Gaussian?).

The intuition here is that as we go further away from this point on the wall, (x', 0), the wave will impart more energy to the water itself (the medium) as compared with the maximum point path. In turn, this reduces the amplitude that will be seen at the wall, at say, (x', y').

What then, is the reasoning for light intensity dropping off as you go away from the bright spot? This puzzles me as light does not need a medium to propagate through.

I realize this might be a simple question, but let's just say I am a student of physics and sometimes the simplest things really stump me.

Note: I am assuming this has nothing to do with wave interference, as there is only one wave source in the question.

up vote 1 down vote accepted

The intuition here is that as we go further away from this point on the wall, (x', 0), the wave will impart more energy to the water itself (the medium) as compared with the maximum point path. In turn, this reduces the amplitude that will be seen at the wall, at say, (x', y').

When a wave propagates through a uniform medium with no obstacles, it loses its intensity with distance for two reasons: first, because some of its energy may be absorbed by the medium, second, because the wave front is expanding.

The rate of intensity loss with distance, $r$, due to the expansion of the wave front is different for $3$D and $2$D waves. For $3$D waves, the intensity or power flow per unit area of the wave front, will be decreasing as $\frac 1 {r^2}$. For $2$D waves, the intensity or power flow per unit length of the wave front, will be decreasing as $\frac 1 r$.

If we apply the above to your examples, we can say that the intensity of the water wave will be decreasing not only due to the absorption by the medium, but also due to the linear expansion of the wave front. For the LED light, the percentage of the losses due to the absorption would be lower, but, assuming that it is a $3$D wave, the intensity loss due to the wave front expansion will be greater.

  • Thanks! Expanding wave front is what I was missing. That is perfect. Of course that makes sense, as the wave front expands and propagates through space, the localized intensity at a further point (where the wave is less "dense" so to speak) will be lower. – Michael Burt Sep 28 at 4:05

You could also use the particle view, many particles each their own a small wave. An led is a good source to consider, the light distribution is said to be "Lambertian" after famous scientist Lambert. This pattern is a distribution where the photons are leaving in all directions but the profile is proportional to cosine (lots of math to explain it). Most diffuse surfaces produce this kind of pattern as well. Obviously for a laser the spread would be small, the geometry of a filament can affect the distribution at close range as well. The cosine pattern disappears at greater distances and the 1 over r squared relationship is observed for light.

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