2
  1. Does light always take the shortest path?

  2. And is it possible to change the probability of a photon travelling to a point by only disturbing the paths that are far away from the shortest path?

Qmechanic
  • 201,751
Azzinoth
  • 2,375

2 Answers2

1

@Azzinoth My comment is too long and contains a sketch, so I'm posting it this way.

Your sketch shows only one point Z. For all other points on a radius from Q the same sketch must be drawn. I have done this in a sketch.
The large yellow circle on the left is a light source with a certain intensity represented by the diameter of the source. The yellow circles on the right represent the intensity of light, which in reality is of course a continuum. Except the area around the obstacle.

On obstacles the light gets deflected in a way that fringes of light occur. Not in the best way drawn, but perhaps you see the different intensities represented by (slightly) different diameters of the last three dots on the top. And in between there is no continuum but there are not exposured areas. Because portions of light on obstacles get deflected this way.

enter image description here

The shortest path model works excellently under the assumption of deflection of light at obstacles.

HolgerFiedler
  • 10,334
  • I only consider photons that hit my detector at Z, everything else is discarded. I don't need to consider all possible target points to disprove that light always takes the shortest path. Of course the placement of my obstacles is tailored for point Z and will not work for all other points. That does not change any of my conclusions.
  • The shortest path model does not work. The shortest path from Q to Z is through M, right? So please explain how a photon traveling along that path is influenced by obstacles on the red paths in your model.
  • – Azzinoth Dec 26 '20 at 18:54
  • You can create the same result without diffraction at objects if you replace the window with a mirror if you like that more. Light travels from the source Q to a point on the mirror and is then reflected to Z. The direct path from Q to Z is blocked, such that the light has to use the mirror. I can now scratch out some parts of the mirror such that the amplitude at Z changes, even tough these parts are not on the shortest path (which is the path that has incident angle=reflected angle).
  • – Azzinoth Dec 26 '20 at 18:54
  • If you want to discuss this further I suggest we move to this chat. https://chat.stackexchange.com/rooms/117709/does-light-always-take-the-shortest-path Long discussions in comments are afaik not allowed. – Azzinoth Dec 26 '20 at 19:01