Waves

Look over edge

diffraction at edge

Another simple case. Consider the border of light and shadow from plane screen with a straight edge (black rectangle in the figure). Surprises await us here again (or simply the “specifics” of optics, which should not be forgotten).

The screen covers half of a wide beam of parallel rays (of plane wave). As its predicted by geometric optics and by everyday experience, light and dark areas will appear behind the screen, separated by a clear, sharp border.

However, if the border of light and shadow is considered in more details (or to observe from far distance), the picture becomes ... not obvious. The border of shadow and light will blur, but how exactly?

Let's reason using the Fresnel zones. We take the intensity of light in a completely open area as 1.

If the observation point is far away from the edge of the screen (from the projection of the edge onto the observation plane), then only its far Fresnel zones will be screened, which do not make a significant contribution to the light intensity at the observation point. Thus, the intensity will be almost the same as in the absence of screen.

As the observation point approaches the edge projection, the closer Fresnel zones are partially overlapped by the screen. And, depending on the sign of the nearest screened zone, the intensity at the observation point can be either less or more than 1 (!).

The intensity will reach its greatest value near the edge, when most of the second Fresnel zone is closed, and the first zone is almost completely open.

The light intensity exactly under the edge will be less than the intensity in the open area, but not zero. If the intensity of light without screen is 1, then what value should be under the edge? Half of the beam/wave is closed... This means that half of the wave does not participate in the formation of the wave front behind the screen...

As a result of "diffusion" of the light to shadow area, the intensity exactly under the edge of the screen will be equal to ... 0.25 (a quarter!) of the intensity in a completely open area. It would seem that it should be 0.5? The fact is that half of the wave is closed, but we observe intensity. Square of half is a quarter.

So, result of screening of half of plane wave is not just a smooth transition from shadow to light. The intensity in the "bright region" fluctuates higher and lower than 1; the light intensity at the screen edge is only a quarter (of full). These are examples of the diffraction effects.

This is also an illustration of several features of wave optics and confirmation of the Huygens and Fresnel models.

An E-M wave can “go around” obstacles (like water waves go behind rocks sticking out of the water). This results in partial "highlighting" in the area of the geometric shadow. (Confirmation of the Huygens-Fresnel principle)
Under the boundary of the geometric shadow, half of the E-M wave is screened. And the observed intensity (energy) is proportional to the square of the modulus of the wave amplitude. Thus, under the boundary of the shadow we have the intensity of half of the E-M field, or (1/2)² = (¼) of the total intensity.
The intensity oscillations in the bright region are explained by the absence of some terms (sources of secondary waves hidden by the screen). The intensity becomes greater or smaller than the norm depending on these missing terms - if they were negative or positive. (Remember the Fresnel zones)

In the case considered above, one half of the plane wave was blocked by the obstacle. As a result, an asymmetric (with respect to the boundary) light intensity distribution was obtained.

In the case of screening the other half of the same plane wave, the observed intensity (its distribution) will be a mirror copy of the first distribution.

Now, based on these data, let's try to answer: what will be the result if we remove the screen and open both halves of the plane wave at the same time?

It is obvious that the addition of two intensity distributions will give a meaningless result. Since it is incorrect to add the intensities of one wave. Once again: waves (E-M fields) interact with each other, they need to be added up (taking into account phases and amplitudes). And the result of this addition is raised to the second power to get the observed intensity. This is not the same as the sum of the intensities.
Mathematically, this can be expressed simply:

If everything is done correctly, we get the correct result: a constant value for the intensity of a plane wave in the absence of obstacles (blue line in the figure).

Note that the "strange" distribution - the sum of two (in red) intensities - can still be obtained. For example, as the result of two "independent" (successive) illuminations of the same photo plate. First exposure with one half of the wave blocked. Second exposure with another half of the wave blocked. In this case, two separate waves that do not interfere with each other illuminate the photographic plate. This is the fundamental difference from the case of exposure of one initial wave.

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