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I am currently studying Optics, fifth edition, by Hecht. In chapter 2.9 Spherical Waves, the author says the following:

The outgoing spherical wave emanating from a point source and the incoming wave converging to a point are idealizations. In actuality, light can only approximate spherical waves, as it can only approximate plane waves.
As a spherical wavefront propagates out, its radius increases. Far enough away from the source, a small area of the wavefront will closely resemble a portion of a plane wave (Fig. 2.29). enter image description here enter image description here enter image description here

It is this part that I found interesting:

The outgoing spherical wave emanating from a point source and the incoming wave converging to a point are idealizations. In actuality, light can only approximate spherical waves, as it can only approximate plane waves.

I've asked questions in the past about idealized conceptions of light, such as "collimated light" and "monochromatic light", so I'm comfortable with the fact that these are just idealizations. But, in those other cases, people were able to explain what the actual manifestations of those phenomena were in reality, in contrast to their idealizations. So in this case, if "spherical waves" and "plane waves" (and other waves) are just idealizations, then what is the actual manifestations of these phenomena in reality? Clearly, by virtue of the fact that we know that these are idealizations, we must also know what their actual form is, right?

In thinking about this myself, I suspect that it has to do with the concept conveyed by this image:

enter image description here

Our theories in physics are approximations (of varying degrees of accuracy) of reality. Over time, our goal is to produce increasingly accurate theories that are closer and closer to the "true" reality. And, for the context of the field of optics, that is what is conveyed by the above image.

Spherical waves and plane waves are an approximation, since "wave optics" itself is just an approximation. And to get a more accurate idea of what is actually going on, we would have to use the theory of electromagnetic optics (and then the theory of quantum optics for an even more accurate idea). Am I thinking about this correctly?

Thank you.

Related: The pulse has the same extent in space at any point along any radius $r$

EDIT

I'm assuming that electromagnetic optics doesn't describe light in terms of waves (plane or otherwise), but, rather, something more accurate. And I then assume that this knowledge is why we know that "planes" are just an approximation (that is, since we know that light is actually more accurately described in some other form, as theorised by electromagnetic/quantum optics). I haven't finished studying optics yet, so I don't actually know if this is correct (which is why I'm asking); all I'm saying is that this is the best I could do when trying to think about this myself (with my current understanding of optics).

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  • $\begingroup$ What distinction are you making between "wave optics" and "electromagnetic optics"? $\endgroup$
    – The Photon
    Jun 27, 2020 at 5:40
  • $\begingroup$ @ThePhoton I'm assuming that electromagnetic optics doesn't describe light in terms of waves (plane or otherwise), but, rather, something more accurate. And I then assume that this knowledge is why we know that "planes" are just an approximation (that is, since we know that light is actually more accurately described in some other form, as theorised by electromagnetic/quantum optics). I haven't finished studying optics yet, so I don't actually know if this is correct (which is why I'm asking); all I'm saying is that this is the best I could do when trying to think about this myself. $\endgroup$ Jun 27, 2020 at 5:44
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    $\begingroup$ @ThePhoton Light is a wave in electromagnetism. Usually the distinction is that in "wave optics" light is taken to be a scalar wave that obeys a scalar wave equation. This is in contrast with "electromagnetic optics", where E- and H-fields both obey a vector wave equation. Wave optics can describe some phenomena like interference and diffraction, but not phenomena related to polarization. $\endgroup$
    – Puk
    Jun 27, 2020 at 5:55
  • $\begingroup$ @Puk thanks for the clarification. Can you please elaborate by explaining what you meany by “scalar wave”? Also, when you say “E- and H-fields”, does this mean electric and magnetic fields? $\endgroup$ Jun 27, 2020 at 6:03
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    $\begingroup$ A scalar wave is a scalar field obeying some wave equation, basically a function that depends on position and time. This is how wave optics describes light. Yes, $\vec{E}$ and $\vec{H}$ are the electric and magnetic field, respectively. They are vector fields, meaning unlike a scalar fields, they also have a direction at every position and time. $\endgroup$
    – Puk
    Jun 27, 2020 at 6:08

1 Answer 1

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As discussed in the comments an electromagnetic wave is (in general) a vector field. Hence, its true that the scalar field model discussed by Hecht in chapter 2.9 is only applicable if certain conditions are met. However, taking the context of chapter 2.9 into account, I believe that this is not the argument of the author. Instead, I believe that the author intended to point out that the assumption of a point-like source is an approximation in itself. As stated above the text reads (emphasised added):

The outgoing spherical wave emanating from a point source and the incoming wave converging to a point are idealizations. In actuality, light can only approximate spherical waves, as it can only approximate plane waves.

Light sources usually have a finite dimension (height, length, width). Therefore, if we consider the light field at a "small" distance from the source, the field will not follow the $e^{-ikr}/r$ law, but reflects the shape of the source. E.g. if we assume that a square with diameter $d=10\mu m$ acts as a light source, the wave at a distance $s=10m$ will be indistinguishable from a spherical wave (true for almost every detector). However, if we reduce the distance to $s=20\mu m$ the light field still contains the information of the square nature of the source. The latter must be true, because by looking at the source with a microscope we are able to distinguish between the actual squared source and a theoretical point-source. Therefore, if the light source is not a point source, but possesses a finite extension, the fact that the source contains "many points" becomes evident at "small" distances.

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  • $\begingroup$ Thanks for the answer. Why does light sources having finite dimension imply that we should expect that the spherical wave is only an approximation? I didn't get this point. $\endgroup$ Jun 27, 2020 at 9:16
  • $\begingroup$ I still don't understand why any of this necessitates expecting a spherical wave. Also, wasn't your example 3D (height, length, width)? $\endgroup$ Jun 27, 2020 at 9:45
  • $\begingroup$ Are you able to narrow your problem down? Please be more specific. Yes, my original text considers 3D dimensions, but the physics is true also in 2D. Personally, I have less trouble to think about a wave in 2D, that's why I choose to consider the square instead of a cube. Finally, I feel like we started to talk about about "why is the wave not spherical in the near field" and now we talk about "why is the wave spherical in the far field". $\endgroup$
    – Semoi
    Jun 27, 2020 at 10:18

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