How do interference and AR coatings work?

[Karagoz]

How does AR coating works on cameras?

How do they decrease reflection?

And how does it improve the quality?

 

[sophiecentaur]

How does AR coating works on cameras?

How do they decrease reflection?

And how does it improve the quality?

It's a clever system and this is the simplest form of it. There is a thin coating of a substance with a refractive index that is less than the glass. The thickness is a quarter wavelength and the light that is reflected at the first surface is canceled buy the light reflected by the second surface, which is a half wavelength delayed. So, at one wavelength, at least and for light at normal incidence, there is no reflection and all the light goes through.
In real life, they use several layers with different thicknesses and materials to spread the 'blooming' effect over most of the visible bandwidth.
Google "Anti reflection coating lenses". This wiki article will start you off in the right direction.
Quality is improved because more light gets through and, at the same time as letting more light in, there are fewer multiple reflections and less flare. Early cameras had no such coatings and the excellent pictures that were taken with some cameras needed very well controlled lighting conditions and angles to avoid these reflection problems.

 

[davenn]

How does AR coating works on cameras?

honestly ... did you even bother to google this ?

there's dozens of hits

go have a read and then come back with ant specific questions on anything you didnt understand

 

[Karagoz]

It's a clever system and this is the simplest form of it. There is a thin coating of a substance with a refractive index that is less than the glass. The thickness is a quarter wavelength and the light that is reflected at the first surface is canceled buy the light reflected by the second surface, which is a half wavelength delayed. So, at one wavelength, at least and for light at normal incidence, there is no reflection and all the light goes through.
In real life, they use several layers with different thicknesses and materials to spread the 'blooming' effect over most of the visible bandwidth.
Google "Anti reflection coating lenses". This wiki article will start you off in the right direction.
Quality is improved because more light gets through and, at the same time as letting more light in, there are fewer multiple reflections and less flare. Early cameras had no such coatings and the excellent pictures that were taken with some cameras needed very well controlled lighting conditions and angles to avoid these reflection problems.

The articles show how the reflected light waves interfere destructively with each other (canceling each other) and thus "disappear" or become "invisible". But at the end, the amount of light waves that are reflected are the same, but with AR coating the reflected waves interfere destructively with each other and become "invisible"?

But how does this increase the transmission of light waves through the lens?

From what I read, flare is caused by light waves reflecting and refracting inside the lens causing a wash out. How does such "AR coatings" reduce flares?

When you add one more matter on the lens, won't the light waves have to travel more through different solid matters, and hence refract and reflect more?

 

[Khashishi]

When you have destructive interference in the reflection, you have constructive interference in the transmission.

 

[sophiecentaur]

When you have destructive interference in the reflection, you have constructive interference in the transmission.

The energy has to go somewhere (conservation) so it goes through.
Flare is reduced because any (already low level) internally reflected goes out at the next surface it meets.

 

[Karagoz]

The energy has to go somewhere (conservation) so it goes through.
Flare is reduced because any (already low level) internally reflected goes out at the next surface it meets.

How does the energy go through the lenses?

Why and how the internally reflected goes out at the next surface it meets?

 

[sophiecentaur]

How does the energy go through the lenses?

Why and how the internally reflected goes out at the next surface it meets?

I think you should show evidence that you have actually researched this yourself. You are simply repeating your initial question. PF doesn't work like that.

 

[Drakkith]

How does the energy go through the lenses?

In the form of EM waves. The waves destructively interfere in one direction and constructively interfere in the other. Wave interference is the crux here, and you can't understand this without getting into the details of interference.

Why and how the internally reflected goes out at the next surface it meets?

The AR coating prevents most of the reflection that would otherwise happen with the initial wave passes from one medium to another (air-to-lens or lens-to-air). So right off the bat the initial internal reflection is greatly reduced. Then, when this internally reflected wave hits the medium boundary again, most of it passes through with very little reflected. The part that passes through simply leaves the system (since it just passed back out of the first lens of the system). The reflected portion undergoes numerous reflections, but at a greatly reduced intensity compared to uncoated lenses. This reduction in the intensity of the internally reflected waves means that much less stray light makes it through the lenses to hit the sensor or your eye.

The increase in the intensity of the light that makes it through the camera lenses, along with the reduction in the stray light, increases the signal-to-noise ratio of the resulting image. More light = more signal, less stray light = less noise.

 

[sophiecentaur]

The increase in the intensity of the light that makes it through the camera lenses, along with the reduction in the stray light, increases the signal-to-noise ratio of the resulting image. More light = more signal, less stray light = less noise.

I know that is the usual answer but the significance of SNR is 'logarithmic' and AR coating will only improve SNR by perhaps 1dB or so, which represents a very small difference in exposure required. I think the overwhelming advantage is in the improvement of picture 'sharpness' due to increased contrast as flare etc. is reduced. Replacing the good quality Kit Lens that came with your camera, with a super dooper equivalent lens will improve the contrast and, possibly, sharpness of your pictures.

 

[Drakkith]

I know that is the usual answer but the significance of SNR is 'logarithmic' and AR coating will only improve SNR by perhaps 1dB or so, which represents a very small difference in exposure required. I think the overwhelming advantage is in the improvement of picture 'sharpness' due to increased contrast as flare etc. is reduced.

True, the increased contrast is the most noticeable improvement (though even this is related to the SNR).

 

[Karagoz]

In the form of EM waves. The waves destructively interfere in one direction and constructively interfere in the other. Wave interference is the crux here, and you can't understand this without getting into the details of interference.

From what I know, to make interference happen, it must be two waves in phase that move into same direction.

In this model, they show one wave that reflect two times, and each is interfering with each other, and cancelling each other (destructive interference).

But only one wave is passing through the glass. How does it interfere with another wave to make a constructive interference?

. . . Then, when this internally reflected wave hits the medium boundary again, most of it passes through with very little reflected. The part that passes through simply leaves the system (since it just passed back out of the first lens of the system).

If I understand you correctly, you mean: most of the waves that hits the medium boundary between the coat and lens do reflect again, and these reflected waves then hits the boundary between coat and air, and then moves out to the air?

 

[sophiecentaur]

But only one wave is passing through the glass. How does it interfere with another wave to make a constructive interference?

Ha ha, you have asked the question that challenges the simple explanation. There are in fact many reflections between the layers and the 'through' ray is enhanced by interference from the other internal bounces, resulting in all (most) the light being transmitted. The layer has the effect of cancelling all reflections from both surfaces. If the angle of incidence is not zero then this process will not work as well and this will result in a colour caste of the surface as some wavelengths are reflected more than others.
My (simpler) original point about the Conservation of energy (this has to happen) justifies the principle in that no reflection from the first face must mean total transmission.

True, the increased contrast is the most noticeable improvement (though even this is related to the SNR).

I think it is more of an effect of modifying the transfer function of the lens than the effect of random noise. There will be a noise-like effect, rather like the quantisation 'noise' in a digitised signal which is really a form of distortion. It takes contributions from other parts of an image and puts them into other parts.

 

[Drakkith]

I think it is more of an effect of modifying the transfer function of the lens than the effect of random noise. There will be a noise-like effect, rather like the quantisation 'noise' in a digitised signal which is really a form of distortion. It takes contributions from other parts of an image and puts them into other parts.

I agree. Perhaps I'm using the term "noise" incorrectly or differently than most.

 

[Drakkith]

From what I know, to make interference happen, it must be two waves in phase that move into same direction.

Not in general, no. Two different waves will interfere with each other when they overlap, regardless of the direction of propagation. The amplitude of each wave is added together and you get either constructive or destructive interference depending on if the sum is larger or smaller than the amplitude of either wave.

In this model, they show one wave that reflect two times, and each is interfering with each other, and cancelling each other (destructive interference).

But only one wave is passing through the glass. How does it interfere with another wave to make a constructive interference?

The reflected portion of the initial wave is itself another wave. So in total you have 3 waves in that picture, not just one. It's just that two of the waves undergo destructive interference and have near-zero amplitude in the backwards direction. As for the constructive interference I mentioned in my post, I confess I don't know much about that. I was mostly seconding Khashishi's explanation.

Also, keep in mind that you're looking at an extremely simplified picture of the process. In real life there will be an entire wavefront, not just a single ray, and there will be multiple reflections and other issues that arise.

If I understand you correctly, you mean: most of the waves that hits the medium boundary between the coat and lens do reflect again, and these reflected waves then hits the boundary between coat and air, and then moves out to the air?

Actually I meant that most of the light reflected from the back side of the lens (at the rearward lens-AR coat boundary) will pass back out the front side instead of bouncing around. That is my understanding of the process at least.