# Optical Forums > Ophthalmic Optics >  Question about AR

## EyeFitWell

:Nerd:  I'm curious if anyone can answer a question I have about AR.
I read in a magazine that AR works by shortening the wavelength of reflected light so that it creates destructive interfearance(sp?) with the light coming in, thus negating the reflection.
My question is this: if it negates the reflection by causing destructive interfearance, wouldn't that also negate some of the light coming in?  I mean, if two waves interact with destructive, both waves are negated.  So does AR _really_ let more light in to the lens?  Or does it just keep reflections off the surface?

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## lensgrinder

Because of destructive interference

"As there is no net reflected energy (and energy must go somewhere), the light that would have been lost by reflcetion is acually added to the energy transmitterd by the lens."  Mo Jalie, Opthalmic Lenses and Dispensing

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## EyeFitWell

That doesn't really make any sense to me.  You mean that the two cancel each other out, and increase the amplitude of the waves coming in?  or soemthing?

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## lensgrinder

What is being said is that since the light was not reflcected back, due to destructive interference, you still have the energy from the light wave which is transmitted through the lens.
Just because you negated the reflection does not mean that the energy from the light stops, it must go somewher and that is transmitted through the lens.
Example if you take a pebble and drop it in still water it creates a wave if you drop another pebble  to negate those waves that you just created the waves are no longer there but the water is still moving.

Hope this helps.

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## HarryChiling

no longer playing in this sand box

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## John Sheridan

> That doesn't really make any sense to me. You mean that the two cancel each other out, and increase the amplitude of the waves coming in? or soemthing?


Wow, I visit the "Ophthalmic Optics" forum for the first time in 6 months and find a question that someone with physics degree (like me) can answer!

First start with an uncoated lens.  A portion of the light that hits the lens surface gets reflected back.

Now you add the AR coat. The coat is half a wavelength thick, so some light gets reflected from the AR coat surface, some gets reflected from the lens surface, and since these are a half-wavelength out of phase, these undergo destructive interference and cancel out.

Now think about the light that gets reflected off the lens surface, then hits the AR coat surface and gets reflected back towards the lens. That light travels through the AR coat twice, so the distance it travels is a full wavelength. This then interferes constructively with the light that is passing right through the lens, thereby increasing its intensity.

The above is essentially a mechanical description of how the principle of the conservation of energy works. Whenever you come up with a mechanism that decreases energy flow in one direction, that same mechanism must increase the flow in some other direction. Or else the energy must be dissipated as heat. It can never just "disappear".

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## harry a saake

john your correct, that is why Ar coated lenses are more efficient, a clear lens loses apx 10 per cent of the light passing through, with ar the lens is now apx 99 percent efficient, thus one can see the advantage of wearing ar at night when there is the least amount of light, especially while you are driving and the average windshield is losing another  apx 20 percent. There was a valid point made on the board one day that in some applications you may want to think about more light coming in

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## David Wilson

John,
Fannin and Gosvenor take the same approach as you. My problem with that theory is that you have the light reflecting at the first surface of the coating AND interfering with the incoming light. I believe that Harry Chiling is on the money here. In fact, the best answer comes fromn Richard Feynman who won a Nobel prize in physics for his work on quantum electrodynamics (QED). Feynman actually addresses the problem in one of his books, titled QED. While we teach wave theory and destructive interference as the cause of 'anti-reflection', only quantum mechanics, in my view, can explain why transmittance is increased.

Regards
David

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## Darryl Meister

I usually describe the effect in terms of constructive interference, as John has. While the conservation of energy principle serves as a nice guideline, telling us that the energy must go _somewhere_, it certainly isn't an explanation as to _how_ the light transmittance is increased. After all, the excess energy could simply be absorbed by the materials and converted into heat.

That said, the fact that light seems to "know" that a portion is being reflected, while the remainder is being transmitted and reinforced by constructive interference, can only be explained by the magic of quantum mechanics. This is similar to the "strange" interaction seen between photons during double slit interference.

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## HarryChiling

no longer playing in this sand box

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## chip anderson

HarryAsakke:
Have things changed, are new figures out.  I was always  taught that lenses reflected light (of course this was glass) at rate of 4% per surface, making reflection (as least of light strikeing the surface @90 degrees, when it get up to 42% angle of incidence reflection becomes "total internal reflection) a total of 8%.   Now you got it up to 5% per surface.

I have always wondered of AR claims that one get 7 or 8% more light as in the latter case one would have  100% transmission which no lens has as such a lens would be totally invisible.

Chip

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## HarryChiling

no longer playing in this sand box

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## Darryl Meister

> It is easier to explain interference in terms of wave theory, I think thats why it is taught that way for opticians.


Certainly. Wave theory was the basis of John's original explanation (i.e., using constructive interference), and the one I prefer.

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## John Sheridan

I agree that QED is the only way to "really" explain what happens with light propagation. I was afraid to say that here, since most people look at me like I have three heads when I say the words "quantum electrodynamics". :D  I'm really surprised to even see it mentioned here.

There are many other "funny" things light can do, like in the double-slit experiement where it acts like a wave when it passes the slits (i.e. showing a diffraction pattern with constructive/destructive interference) then acts like a particle when it hits the screen (i.e. makes a distinct "ping" in a localized place on the screen).

I could talk lots more about QED "funny business", but I think it would be off-topic on a web site devoted to eyecare!

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## Kyle

Can you imagine a world in which all opticians were this interested in quantum theory?

What I really want to know is what PAL the rabbit wears...;)

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## EyeFitWell

> I would venture to say that quantum optics is a bit overkill for most opticians. It is easier to explain interference in terms of wave theory, I think thats why it is taught that way for opticians. I know that it would be a pain to try and teach some of the people I work with quantum theory as I don't totally understand every concept myself. Maybe when opticians become more educated.


For the record, I'm in school for my license in one of the hardest states to get a license. So, although I might not be a quantum physics major, I'd wager that I know as much or more about this stuff than the average optician you've encountered. I debated for two years whether I wanted my optician's license or to get a degree in Phyics or Optical Science. My background is in Chemistry. So, I'm not a dunce. I finally decided that even if I don't stop there, a license is a good step (at least financially!;) ) and decided to go for it. In NC, even the ABO only covers two sections out the 9 sections on the test.

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## EyeFitWell

In fact, the single most frustrating part of my job is that I am a bright science-minded individual.  I find myself working for and with people who don't know squat, and it's annoying.  In fact, while enrolled in the 'apprenticeship' route to be eligable to sit for the exam, I asked my manager a question.  This is a licensed optician.  I asked her (and her boss, the owner-also licensed) if there was an equation to determine the necessary blank size if using a stock lens.  We had wasted several pairs of stock polarized lenses that summer when big was hot.  I had thought about it all night and decided there MUST be a way to catch that ahead of time.  I came up with my own formula and asked my bosses.  They BOTH told me to quit being a little miss smarty pants.  Come to find out, the equation was correct and ON THE TEST!!!
It's annoying when I want to know how stuff REALLY works, even if you go way over my head, I've already heard the dummed down stuff.  You know what I mean? :Rolleyes:

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## HarryChiling

no longer playing in this sand box

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## Dave Nelson

Ouch. People selling eyeglasses and managing optical stores who don't know how to determine blank size? When I started in optical, I knew how to do that before lunch on the first day. I HATE when people want to stifle curiousity and initiative on the job, or anywhere. Eyefitwell, I hope your desire to better yourself and keep learning in the art and science of opticianry is lifelong, and I hope when you become a manager and master optician (if you are'nt already) you foster the same appreciation for learning and initiative in your staff, and never forget the "little people" who rewarded your initiative with rude comments. they were probably afraid of their ignorance showing.

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## Aarlan

> For the record, I'm in school for my license in the hardest state to get a license. .


What are the licensing requirements in NC?

AA

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## lensgrinder

> What are the licensing requirements in NC?
> 
> AA


Follow this link

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## ziggy

> In fact, the single most frustrating part of my job is that I am a bright science-minded individual. I find myself working for and with people who don't know squat, and it's annoying.


If you work in the field very long you will come to understand this is the norm,,not the exception.

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## drk

I have a practical side question:

Since reflectance is directly related to index of refraction, and since AR coatings are not 100&#37; effective...

...what would be the order of difference in effectiveness of a theoretical given AR process between, say, a 1.7 index lens and a 1.5 index lens? Is the difference 1% reflectance or less? Or is it a greater effect?

In other words, if you AR-coated a 1.5 and a 1.7 lens, and used a light meter to measure reflectance, would the 1.5 outdo the 1.7 by very much? 

Thanks.

P.S.  As to the wave-particle duality of light, I was taught that when dealing with light on a "macro" level, the wave theory explains phenomena quite well.  It's only when dealing on very small scales that particle theory becomes important.

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## EyeFitWell

Hmm....DRK, that's a good question.  I know that higher indexes reflect more light, making the use of AR all the more important.

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## Fezz

> In other words, if you AR-coated a 1.5 and a 1.7 lens, and used a light meter to measure reflectance, would the 1.5 outdo the 1.7 by very much?


 
My understanding is that the reflectance would be all but the same. Efficiency of the coating is determined by the correct selection of coating materials for refractive index, the thickness in which they are applied, and adherence to the previous layer. So a "correctly" applied AR coating would level the playing field among the various index lens materials.

Hopefully, someone more up on this will correct me if I am wrong.

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## drk

Fezz, that sounds awfully compelling.  I vote that you're right.

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## Darryl Meister

In some cases, you can have even lower reflectance with AR on high-index lens materials if they more closely satisfy the _amplitude_ condition of the available coating materials. Although this is less of an issue with multi-layer coatings.

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## drk

That confirms it.  Thanks, D-man.

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## Darryl Meister

You guys are good at coming up with these thought-provoking questions.

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## drk

Not to breach etiquitte, but please see the minus aspherics thread, Darryl.  It needs you.

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## Chris Ryser

> *My understanding is that the reflectance would be all but the same*.


Actually if you had a coating that would be  100% reflex absorbing, it would make no difference what kind of lens material you had underneath. You might have a difference in light transmittance but not reflection.

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## Darryl Meister

> Actually if you had a coating that would be 100% reflex absorbing, it would make no difference what kind of lens material you had underneath.


Perhaps you could elaborate on this a bit (I'm assuming you're still referring to a transparent coating).

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## jherman

I thought AR's were 1/4 thick?







> Wow, I visit the "Ophthalmic Optics" forum for the first time in 6 months and find a question that someone with physics degree (like me) can answer!
> 
> First start with an uncoated lens. A portion of the light that hits the lens surface gets reflected back.
> 
> Now you add the AR coat. The coat is half a wavelength thick, so some light gets reflected from the AR coat surface, some gets reflected from the lens surface, and since these are a half-wavelength out of phase, these undergo destructive interference and cancel out.
> 
> Now think about the light that gets reflected off the lens surface, then hits the AR coat surface and gets reflected back towards the lens. That light travels through the AR coat twice, so the distance it travels is a full wavelength. This then interferes constructively with the light that is passing right through the lens, thereby increasing its intensity.
> 
> The above is essentially a mechanical description of how the principle of the conservation of energy works. Whenever you come up with a mechanism that decreases energy flow in one direction, that same mechanism must increase the flow in some other direction. Or else the energy must be dissipated as heat. It can never just "disappear".

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## OPTIDONN

1/4 in and 1/4 out will put the wave a total of 1/2 out of sequence creating destructive interference.

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## jherman

or tato chip it down for me.

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## HarryChiling

Here is a quick mock up.

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## jherman

:d

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## Julian

> I thought AR's were 1/4 thick?


It is correct if we are talking about optical thickness. If light is split into two components by reflection at the top and the bottom surface of a thin film (AR coat), then the beams will recombine in such a way that the total amplitude will be the difference of the amplitudes of the two components. We say that the beams interfere destructively if the relative phase shift is 180 degree. To ensure that the relative phase shift is 180 degree, the optical thickness of the  film should be one quarter of wavelength. What is important here, it works when the reflection will take place in a medium of lower index then adjoining medium.

"1/4 in and 1/4 out will put the wave a total of 1/2 " it is true if we consider optical way, but optical thickness is stiil the same- one quarter of wavelength.

Best regards,
Julian

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## Darryl Meister

We should clarify that "optical thickness," or more specifically, _optical path length_ is the product of the physical thickness and the refractive index of the film.

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## Julian

> We should clarify that "optical thickness," or more specifically, _optical path length_ is the product of the physical thickness and the refractive index of the film.


Right, it is the physical thickness multiply by refractive index (and this is optical path) and divided by reference wavelenght.

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## Aarlan

we discussed a lot of the A/R theory in this thread as well 


http://www.optiboard.com/forums/showthread.php?t=13188

AA

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