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Quote from: hamdani yusuf on 30/12/2022 07:36:25Quote from: Bored chemist on 29/12/2022 09:49:01Go and look up what the words mean.Does a silver mirror transparent, as per your definition?Does it what?Or did you use the wrong word?
Quote from: Bored chemist on 29/12/2022 09:49:01Go and look up what the words mean.Does a silver mirror transparent, as per your definition?
Go and look up what the words mean.
Quote from: Bored chemist on 30/12/2022 09:43:59Quote from: hamdani yusuf on 30/12/2022 07:36:25Quote from: Bored chemist on 29/12/2022 09:49:01Go and look up what the words mean.Does a silver mirror transparent, as per your definition?Does it what?Or did you use the wrong word?Is the silver mirror transparent?
Quote from: hamdani yusuf on 11/06/2022 14:04:23I just got an even stronger evidence that diffracted light is produced by the edges of the obstacle, instead of the space between those edges. The experiment involves linear polarization. I've finally uploaded the video. //www.youtube.com/watch?v=oMj4l0rM2qw
I just got an even stronger evidence that diffracted light is produced by the edges of the obstacle, instead of the space between those edges. The experiment involves linear polarization.
I've come to this discussion a bit late, but isn't that obvious? Something that isn't there can't have an effect on anything that is, and the simplest definition of a space is surely "something that isn't there"?
Something that isn't there can't have an effect on anything that is
Not "every point in space" but every point on a wavefront. Huygens describes the geometry of its propagation. There's nothing to "agree with" - it just works.A problem arises when you talk about slit interference.The slit is mostly irrelevant: what you see according to Huygens is the interference between two edge-truncated sources and the primary geometric beam, and as the slit gets narrower, so the primary beam component becomes less relevant.
What would happen if those edges are removed altogether?What if those edges are made of polarizing material?
Quote from: hamdani yusuf on 08/01/2023 08:31:53What would happen if those edges are removed altogether?What if those edges are made of polarizing material?In principle, nothing.
The diffraction and interference pattern disappears. The bright spot on the screen returns to show the shape of original light beam.
In the wave model, every point in space becomes a wavefront at different point in time.What would happen if those edges are removed altogether?
Quote from: hamdani yusuf on 08/01/2023 15:28:31The diffraction and interference pattern disappears. The bright spot on the screen returns to show the shape of original light beam.Which part of this do you not yet understand?I have tried to point it out several times.The " shape of original light beam" is a "diffraction and interference pattern".Some part of the laser acts as an aperture. That aperture has edges (or one roughly circular edge).Those edges cause diffraction.Also, how did you address the cylindrical lens effect?
A good laser pointer produces a single circular spot on the screen.
When you say that nothing changes, you are saying that removing or not removing the edges produce the same result, whether or not you are intentionally making that implication.
But the incoming light only has two possible fates.It hits the obstacle and is absorbed, or it misses it and carries on.
There are other possibilities. It can be reflected, diffracted, or refracted.
whose size equals the first dark ring
Quote from: hamdani yusuf on 12/01/2023 06:50:40There are other possibilities. It can be reflected, diffracted, or refracted.Only if it hits the obstacle.And, in the classic experiment, the obstacle is black so any light that hits it is absorbed,So, as I said, it either gets absorbed, or it passes straight on.
Quote from: hamdani yusuf on 12/01/2023 06:46:57whose size equals the first dark ringWould that be the inner or the outer edge of the dark ring? You might enjoy doing the experiment with microwaves!