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I am currently studying geometrical optics, and it is stated that the ray equation is invalid at the edge of a plane wave traveling through a pinhole. I know that, when the amplitude of the wave is changing slowly, the ray equation is a valid solution to Maxwell's equations, and can therefore be used to evaluate optical systems. However, I don't understand how this implies that the ray equation is invalid for the aforementioned case.

One reason I find this fact perplexing is that the ray equation is valid for the case of a plane wave refracting through the centre of a lens, so I also wonder what the difference between this case and the pinhole one is? Why is it true for this case and not the pinhole?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ I am not familiar with ray optics, but I can see the difference between a pinhole and a lens. The lens distorts the space for the plane wave, all the energy is there, the pinhole allows only a tiny dx to go through, the rest hitting the barrier, $\endgroup$
    – anna v
    Nov 19, 2019 at 6:02
  • $\begingroup$ @annav thanks for the comment. I thought this was the case, but I was more-so wondering what the difference is that makes it so that the ray equation is valid for the case without the lens, whilst for the case where, basically, nothing has changed, since a pinhole is just some space and not a different medium such as a lens, the ray equation is invalid. It seems odd, because, for the pinhole case, it seems like nothing has really changed, and the wave is still travelling through some space consisting of the same medium, [...] $\endgroup$ Nov 19, 2019 at 6:09
  • $\begingroup$ [...] and so, it might naively seem like the case where the wave must now pass through a new medium (the lens) is where some change occurs to make the ray equation invalid. $\endgroup$ Nov 19, 2019 at 6:14
  • $\begingroup$ Well, in the pinhole problem, it cannot pass through a medium, only through the tiney aperture of the pin. Most of the wave will scatter off the barrier. $\endgroup$
    – anna v
    Nov 19, 2019 at 7:00

1 Answer 1

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The OP has given the answer himself.

when the amplitude of the wave is changing slowly, the ray equation is a valid solution to Maxwell's equations, and can therefore be used to evaluate optical systems.

In case of the lens, the wave is changing slowly.

The wall of a pinhole is "hard", the wave goes abruptly to zero, this is not a slow change, and thus the ray equation does not apply, diffraction sets in.

You don't need a pinhole for this, a straight hard wall (half an open plane, half an opaque plane) is enough to cause refraction.

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  • $\begingroup$ Thanks for the answer. But the ray equation is valid for a plane wave refracting at an infinitely flat surface, right? How is this different from the examples you’ve provided? $\endgroup$ Nov 19, 2019 at 7:06
  • $\begingroup$ I see. So the ray equation is invalid at any points at which we have refraction? $\endgroup$ Nov 19, 2019 at 7:12
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    $\begingroup$ Infinitely flat surface causes slow change. Half infinite transparent, half infinite opaque has as abrupt change on the limit. Hence diffraction. $\endgroup$
    – Alfred
    Nov 19, 2019 at 7:16
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    $\begingroup$ Abrupt change along the direction of propagation is not an issue, like hitting a medium with different refraction index. Only abrupt change perpendicular to that direction causes diffraction. If you send a wide beam on a small lens, the effect of most of the lens will be described by the wave equation but there will be diffraction from the side of the lens if it is within the beam. $\endgroup$
    – Alfred
    Nov 19, 2019 at 7:16
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    $\begingroup$ @The Pointer Sorry, crossed messages. No, refraction is where the wave equation is valid. Sorry for not mentioning earlier : abrupt change along the direction Is OK. It is abrupt change perpendicular to the direction that causes diffraction. $\endgroup$
    – Alfred
    Nov 19, 2019 at 7:19

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