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if you have just 1 vertical slit, you will only get light which is travelling at that angle, turning a 3-dimensional beam of light into a (more or less) 2-dimensional ribbon of light ...

You must be aware that behind edges light gets deflected and instead of a ribbon one get a wavelike intensity distribution (source) behind edges and slits.

[![enter image description here][1]][1]enter image description here

... while one-coloured light will more or less be just the oscillating sine-like curve, if there is only one light source

A common light source isn’t spatial coherent for the outgoing light. Instead of a beam with periodically increasing and decreasing intensity one get a continuous stream of light with a quantum noise.

You get an oscillating beam with a wave generator for radio waves:

Radio frequency (RF) is the oscillation rate of an alternating electric current or voltage or of a magnetic, electric or electromagnetic field or mechanical system

Even behind a slit (if it is oriented in the direction of the radio wave polarization) you will receive the radio signal and could listen to the broadcast program.

So, if I put 2 slits behind each other, so looking at them from the front they will be a +

But now again the electromagnetic radiation is deflected behind the edges and the intensity behind the second slit is diffracted into several areas. The main fringe will be less intense than the main fringe in front of the second slit.

Now, if the first and the second one are exactly a multiple of the wavelength of any colour apart from each other, shouldn't exactly that colour pass through that polarizing filter (of course only those that are 90° to the vertical allignment of the filter, happen to be precisely centered and happen to enter the 1st polarizing filter exactly at the beginning of a wave in the exact middle)?

Any wavelength below some limit (depends from the size of the slit) will go through the pinhole. Note that dichroic filters, which only pass a certain wavelength, consist of alternating layers of optical layers with different refractive indices. This is very different from what happens with a slit or a pinhole.

The slit will do another thing. It polarizes the incoming light. Roughly spoken, 50% of the incoming light gets deflected or absorbed and 50% gets rotated into the direction of the slit. That will happens with every photon in the light beam, or deflection/absorption, or rotation of its electric field component (together with the same amount of rotation of the magnetic field component) in relation to the direction of the slit.

Your question is a very interesting. Indeed at some point the electric field component of the photon is zero and if one place the second slit at quarter wavelength behind the first slit, the photon should be go through without any influence from the second slit.
In reality the thickness of the slit will spoil the idea. No way that the setup is realizable.

That would probably be a very small amount of light, but wouldn't it mean that it's wrong to say that no light passes through them?

No light passe through the two slits. [1]: https://i.sstatic.net/d0kdbm.jpg

if you have just 1 vertical slit, you will only get light which is travelling at that angle, turning a 3-dimensional beam of light into a (more or less) 2-dimensional ribbon of light ...

You must be aware that behind edges light gets deflected and instead of a ribbon one get a wavelike intensity distribution (source) behind edges and slits.

[![enter image description here][1]][1]

... while one-coloured light will more or less be just the oscillating sine-like curve, if there is only one light source

A common light source isn’t spatial coherent for the outgoing light. Instead of a beam with periodically increasing and decreasing intensity one get a continuous stream of light with a quantum noise.

You get an oscillating beam with a wave generator for radio waves:

Radio frequency (RF) is the oscillation rate of an alternating electric current or voltage or of a magnetic, electric or electromagnetic field or mechanical system

Even behind a slit (if it is oriented in the direction of the radio wave polarization) you will receive the radio signal and could listen to the broadcast program.

So, if I put 2 slits behind each other, so looking at them from the front they will be a +

But now again the electromagnetic radiation is deflected behind the edges and the intensity behind the second slit is diffracted into several areas. The main fringe will be less intense than the main fringe in front of the second slit.

Now, if the first and the second one are exactly a multiple of the wavelength of any colour apart from each other, shouldn't exactly that colour pass through that polarizing filter (of course only those that are 90° to the vertical allignment of the filter, happen to be precisely centered and happen to enter the 1st polarizing filter exactly at the beginning of a wave in the exact middle)?

Any wavelength below some limit (depends from the size of the slit) will go through the pinhole. Note that dichroic filters, which only pass a certain wavelength, consist of alternating layers of optical layers with different refractive indices. This is very different from what happens with a slit or a pinhole.

The slit will do another thing. It polarizes the incoming light. Roughly spoken, 50% of the incoming light gets deflected or absorbed and 50% gets rotated into the direction of the slit. That will happens with every photon in the light beam, or deflection/absorption, or rotation of its electric field component (together with the same amount of rotation of the magnetic field component) in relation to the direction of the slit.

Your question is a very interesting. Indeed at some point the electric field component of the photon is zero and if one place the second slit at quarter wavelength behind the first slit, the photon should be go through without any influence from the second slit.
In reality the thickness of the slit will spoil the idea. No way that the setup is realizable.

That would probably be a very small amount of light, but wouldn't it mean that it's wrong to say that no light passes through them?

No light passe through the two slits. [1]: https://i.sstatic.net/d0kdbm.jpg

if you have just 1 vertical slit, you will only get light which is travelling at that angle, turning a 3-dimensional beam of light into a (more or less) 2-dimensional ribbon of light ...

You must be aware that behind edges light gets deflected and instead of a ribbon one get a wavelike intensity distribution (source) behind edges and slits.

enter image description here

... while one-coloured light will more or less be just the oscillating sine-like curve, if there is only one light source

A common light source isn’t spatial coherent for the outgoing light. Instead of a beam with periodically increasing and decreasing intensity one get a continuous stream of light with a quantum noise.

You get an oscillating beam with a wave generator for radio waves:

Radio frequency (RF) is the oscillation rate of an alternating electric current or voltage or of a magnetic, electric or electromagnetic field or mechanical system

Even behind a slit (if it is oriented in the direction of the radio wave polarization) you will receive the radio signal and could listen to the broadcast program.

So, if I put 2 slits behind each other, so looking at them from the front they will be a +

But now again the electromagnetic radiation is deflected behind the edges and the intensity behind the second slit is diffracted into several areas. The main fringe will be less intense than the main fringe in front of the second slit.

Now, if the first and the second one are exactly a multiple of the wavelength of any colour apart from each other, shouldn't exactly that colour pass through that polarizing filter (of course only those that are 90° to the vertical allignment of the filter, happen to be precisely centered and happen to enter the 1st polarizing filter exactly at the beginning of a wave in the exact middle)?

Any wavelength below some limit (depends from the size of the slit) will go through the pinhole. Note that dichroic filters, which only pass a certain wavelength, consist of alternating layers of optical layers with different refractive indices. This is very different from what happens with a slit or a pinhole.

The slit will do another thing. It polarizes the incoming light. Roughly spoken, 50% of the incoming light gets deflected or absorbed and 50% gets rotated into the direction of the slit. That will happens with every photon in the light beam, or deflection/absorption, or rotation of its electric field component (together with the same amount of rotation of the magnetic field component) in relation to the direction of the slit.

Your question is a very interesting. Indeed at some point the electric field component of the photon is zero and if one place the second slit at quarter wavelength behind the first slit, the photon should be go through without any influence from the second slit.
In reality the thickness of the slit will spoil the idea. No way that the setup is realizable.

That would probably be a very small amount of light, but wouldn't it mean that it's wrong to say that no light passes through them?

No light passe through the two slits.

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HolgerFiedler
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if you have just 1 vertical slit, you will only get light which is travelling at that angle, turning a 3-dimensional beam of light into a (more or less) 2-dimensional ribbon of light ...

You must be aware that behind edges light gets deflected and instead of a ribbon one get a wavelike intensity distribution (source) behind edges and slits.

[![enter image description here][1]][1]

... while one-coloured light will more or less be just the oscillating sine-like curve, if there is only one light source

A common light source isn’t spatial coherent for the outgoing light. Instead of a beam with periodically increasing and decreasing intensity one get a continuous stream of light with a quantum noise.

You get an oscillating beam with a wave generator for radio waves:

Radio frequency (RF) is the oscillation rate of an alternating electric current or voltage or of a magnetic, electric or electromagnetic field or mechanical system

Even behind a slit (if it is oriented in the direction of the radio wave polarization) you will receive the radio signal and could listen to the broadcast program.

So, if I put 2 slits behind each other, so looking at them from the front they will be a +

But now again the electromagnetic radiation is deflected behind the edges and the intensity behind the second slit is diffracted into several areas. The main fringe will be less intense than the main fringe in front of the second slit.

Now, if the first and the second one are exactly a multiple of the wavelength of any colour apart from each other, shouldn't exactly that colour pass through that polarizing filter (of course only those that are 90° to the vertical allignment of the filter, happen to be precisely centered and happen to enter the 1st polarizing filter exactly at the beginning of a wave in the exact middle)?

Any wavelength below some limit (depends from the size of the slit) will go through the pinhole. Note that dichroic filters, which only pass a certain wavelength, consist of alternating layers of optical layers with different refractive indices. This is very different from what happens with a slit or a pinhole.

The slit will do another thing. It polarizes the incoming light. Roughly spoken, 50% of the incoming light gets deflected or absorbed and 50% gets rotated into the direction of the slit. That will happens with every photon in the light beam, or deflection/absorption, or rotation of its electric field component (together with the same amount of rotation of the magnetic field component) in relation to the direction of the slit.

Your question is a very interesting. Indeed at some point the electric field component of the photon is zero and if one place the second slit at quarter wavelength behind the first slit, the photon should be go through without any influence from the second slit.
In reality the thickness of the slit will spoil the idea. No way that the setup is realizable.

That would probably be a very small amount of light, but wouldn't it mean that it's wrong to say that no light passes through them?

No light passe through the two slits. [1]: https://i.sstatic.net/d0kdbm.jpg