[hamdani yusuf (OP)]

Very neat experiment!

What do you think is happening at 1:24?

What wavelength are you using?

Thanks.
I think it's caused by reflection. It's not counted as diffraction because the microwave doesn't go behind the obstacle in the line of sight from the transmitter.

The transmitter is 10.5 GHz, so the wavelength is around 28.6 mm

[hamdani yusuf (OP)]

This is my earlier video showing that diffraction can occur on the edge of transparent materials.
It even occurs when the refractive index is very similar to the medium.

[hamdani yusuf (OP)]

My conclusion so far is that diffraction can occur if the penetration depth of the light through the obstacle is significantly more than the wavelength. In case of microwave on aluminum plate, the penetration depth is much lower than the wavelength. That's why diffraction effect can't be detected.

I don't know any material which has penetration depth in visible spectrum much less than 500 nm. So I used another method to produce impenetrable barrier for visible light, using total internal reflection in glass.

As shown in the video, it doesn't produce diffraction effect.

[hamdani yusuf (OP)]

With two total internal reflection edges, we can effectively produce a single slit aperture of impenetrable barrier. It produces no interference pattern.

[hamdani yusuf (OP)]

I'm preparing a new video investigating diffraction of light by producing one sided interference pattern. One edge of the single slit aperture is made of total internal reflection barrier, while the other edge is made of folded aluminium foil as a normal barrier.

[hamdani yusuf (OP)]

If my experiments posted in this thread don't convince people that diffraction and interference pattern formed in single slit experiment come from the edges instead of the space between them,  I don't know what will.

[alancalverd]

I think you have misinterpreted your excellent microwave experiments.

Single-edge diffraction produces a single bright peak outside the shadow of the edge - see  https://sciencedemonstrations.fas.harvard.edu/presentations/edge-diffraction - and I think this is what you are seeing at 1:24. There is a secondary, much smaller peak around 1:25.

Things are a bit different with the partial absorber. The diffracted wave interferes with scattered radiation through the plate, and produces a bright peak inside the shadow area

Likewise the  paraffin wax block acts as a single edge and a scatter source. Bringing up the second block creates a second edge so you now have the beginning of a diffraction grating, which will produce bigger maxima and minima according to the grating equation mλ = d sinθ where d is the grating spacing, m is an integer, and in your case θ ≈ 30°. Note how you get a maximum when the edges area about λ/2 apart, and a minimum when they overlap by about λ. 

[evan_au]

If my experiments posted in this thread don't convince people that diffraction and interference pattern formed in single slit experiment come from the edges instead of the space between them,  I don't know what will.

The distance between the edges is defined by the space between them.
See: https://en.wikipedia.org/wiki/Negative_space

[hamdani yusuf (OP)]

Things are a bit different with the partial absorber. The diffracted wave interferes with scattered radiation through the plate, and produces a bright peak inside the shadow area

The reflector and partial reflector came with the microwave transceiver kit as accessories. They don't absorb much of the microwave. Most of it is either reflected or transmitted.

[hamdani yusuf (OP)]

Likewise the  paraffin wax block acts as a single edge and a scatter source. Bringing up the second block creates a second edge so you now have the beginning of a diffraction grating, which will produce bigger maxima and minima according to the grating equation mλ = d sinθ where d is the grating spacing, m is an integer, and in your case θ ≈ 30°. Note how you get a maximum when the edges area about λ/2 apart, and a minimum when they overlap by about λ.

I interpret the second block as additional source of diffracted wave. The optical distance to the receiver is almost equal with the first edge, thus produced constructive interference.

[hamdani yusuf (OP)]

If my experiments posted in this thread don't convince people that diffraction and interference pattern formed in single slit experiment come from the edges instead of the space between them,  I don't know what will.

The distance between the edges is defined by the space between them.
See: https://en.wikipedia.org/wiki/Negative_space

I have three types of single slit experiments.
1. Penetrable barriers are used to create the edges. Interference pattern shows on both sides. This is the type usually used in physics demonstrations as teaching materials.
2. Impenetrable barriers are used to create the edges. No interference pattern shows on either side.
3. One edge is penetrable barrier, while the other is impenetrable. Only one side shows interference pattern. The side behind the impenetrable barrier doesn't show interference effect.

The width of the slit is almost the  same in all of those cases. But the results are different.

[alancalverd]

And I forgot to mention refraction! Paraffin wax doesn't just absorb and diffract a bit of the radiation, it also bends it.

https://news.mit.edu/2012/new-metamaterial-lens-focuses-radio-waves-1114 describes an interesting microwave lens that exploits the negative refractive index of a conductive metamaterial. This explains why the peaks of your metal-edge diffraction pattern appear on the "wrong" side of the beam central plane.

https://phyweb.physics.nus.edu.sg/~L3000/Level3manuals/Microwave.pdf  discusses a number of microwave optical experiments. Definitely worth remembering that your source is strongly polarised so the effect of an asymmetric slit, lens or whatever will depend on its orientation.

[hamdani yusuf (OP)]

And I forgot to mention refraction! Paraffin wax doesn't just absorb and diffract a bit of the radiation, it also bends it.

Where does it bend to on a flat surface?

[hamdani yusuf (OP)]

https://news.mit.edu/2012/new-metamaterial-lens-focuses-radio-waves-1114 describes an interesting microwave lens that exploits the negative refractive index of a conductive metamaterial. This explains why the peaks of your metal-edge diffraction pattern appear on the "wrong" side of the beam central plane.

I used a plain aluminium plate. Noone would call it a metamaterial.

[alancalverd]

You can see the bending of light passing through a flat surface with your glass blocks. Obviously there is no displacement if the beam is perpendicular to the surface, but in the case of microwaves, your source was delivering a cone of radiation so the first encounter with the  paraffin wax was at an angle of around 10 - 15 degrees to perpendicular.

Aluminium isn't a metamaterial but it is an excellent conductor, so one edge can display the properties of a metamaterial in a limited geometry.

[hamdani yusuf (OP)]

You can see the bending of light passing through a flat surface with your glass blocks. Obviously there is no displacement if the beam is perpendicular to the surface, but in the case of microwaves, your source was delivering a cone of radiation so the first encounter with the  paraffin wax was at an angle of around 10 - 15 degrees to perpendicular.

Refraction by a dielectric plate with parallel surface doesn't change the direction of light at the end of the process. It's only translated, which depends on the thickness of the plate.

In my experiment, the plates aren't that thick to produce significant effect of refraction.

[hamdani yusuf (OP)]

Aluminium isn't a metamaterial but it is an excellent conductor, so one edge can display the properties of a metamaterial in a limited geometry.

Or it's just a simple reflection by a convex surface located at the edge of the aluminum plate.

[alancalverd]

Aluminium isn't a metamaterial but it is an excellent conductor, so one edge can display the properties of a metamaterial in a limited geometry.

Or it's just a simple reflection by a convex surface located at the edge of the aluminum plate.

Unlikely to be significant - the wavelength is around 50 times the radius of any possible convexity! But the phenomenon you demonstrated is exactly what is predicted by a wavelet model of diffraction.

[hamdani yusuf (OP)]

Unlikely to be significant - the wavelength is around 50 times the radius of any possible convexity! But the phenomenon you demonstrated is exactly what is predicted by a wavelet model of diffraction.

Why not? A radio transmitter antenna can transmit radio wave with wavelength a thousand times its diameter.
The aluminum plate doesn't produce diffracted microwave, unlike paraffin wax plate.

[alancalverd]

I must repeat what I said in #46 above, with a bit of emphasis

Single-edge diffraction produces a single bright peak outside the shadow of the edge - see  https://sciencedemonstrations.fas.harvard.edu/presentations/edge-diffraction - and I think this is what you are seeing at 1:24. There is a secondary, much smaller peak around 1:25.

Aluminum is a conductor, paraffin wax is a dielectric, so you wouldn't expect them to behave the same.