View Full Version : lens coating, how does it work?

John D Gerndt
5-Dec-2004, 17:50
I am trying to answer a seemingly simple question my brother asked me: "How do lens coatings work? When I went to answer the question I found my own knowledge foggy, and the net seems only to acknowledge that they DO work. Can someone out there help?


Michael S. Briggs
5-Dec-2004, 18:33
Lens coatings, both single and multiple, use the wave nature of light. The coating thickness is carefully controlled so that reflections from the glass and from the coating have waves that are out of phase and destructively interfere, thereby suppressing the reflection. With a single coating, the coating thickness can only have exactly the correct thickness for one particular wavelength and so at other wavelengths the reflectivity is higher. A coating with multiple layers can have very low reflectivity over a broader wavelength range.

There is a explaination with a figures showing the phases of the waves at http://www.edmundoptics.com/techSupport/DisplayArticle.cfm?articleid=247 (http://www.edmundoptics.com/techSupport/DisplayArticle.cfm?articleid=247). You can find more on the physics using search words "thin film interference".

Steve McKinney
5-Dec-2004, 18:37
Try this:

http://www.geocities.com/nhwade2002/ (http://www.geocities.com/nhwade2002/)



Robert A. Zeichner
5-Dec-2004, 19:45
Here's some historical background as related to me by the late glass chemist Howard Ross. The accidental discovery of improving light transmission through optical glass was a result of a phenomenon called "blooming". As the surface of the glass oxidized, it created what appeared to be a rainbow-like coating that at first, seemed to be a defect, but actually resulted in more light passing through the lens. There are a number of occurences of this sort in nature, the most common being the appearance of soap bubbles. In fact, the first lens coatings were thin layers of soap which could only be applied to inner surfaces, for obvious reasons. A brilliant scientist at General Electric by the name of Catherine Blodgett conducted research into the phenomenon known as optical interference at the surface. She developed a method of measuring thin films by observing the color they reflected. This research formed the basis for the development of lens coatings and Blodgett was awarded a Nobel prize for her work. The modern coatings used today are super thin layers of Magnesium Flouride or other compounds deposited under high vacuum. As a general rule, the index of refraction of the coating layer should be approximately equal to the square root of the index of refraction of the glass element on which it is deposited.

David Karp
6-Dec-2004, 11:25
Here is some more information for you:

http://www.wisner.com/coatings.htm (http://www.wisner.com/coatings.htm)

John D Gerndt
6-Dec-2004, 16:36
Thanks guys, I must be denser than Barium glass though because I still don't get it. .25 wavelength constructive interference makes sense but I can only picture this happening with a wave that passes through the coating to the glass, bounces off the glass in the opposite direction and then back in the original direction off the inside interface of the coating and air. SO how can a coating (in my mind's picture) allow the light to pass through in one direction and reflect it in the other? Is it possible because the transition from air to coating is low index to high and in the other direction it is high to low? This is speculation on my part, I am quite confused, still..


Struan Gray
7-Dec-2004, 14:22
Conservation of energy is your friend: if you suppress reflection you necessarily enhance transmission - the light has to go somewhere.

If you want the gory detail, look up the Kramers-Kronig relation: a universal equation which links reflection, transmission and absorbtion for all materials.

7-Dec-2004, 17:18
Sorry if this is a bit long.

There are simple answers and there are better answers. I can attempt a simple answer, although I warn that I've been struggling with how to put it. It might help to draw the picture.

Draw the advancing light as a sine wave with a wavelength of w. Keep in mind that when the light reflects off of an interface in which it has gone from a lower index of refraction substance to a higher index of refraction substance the wave will flip. This is the same for our sine wave as advancing it by 1/2 a wavelength.

Assume that the coating on the lens has a thickness of a 1/4 the wavelength of light and that its index of refraction is greater than air, but less than that of the glass of the lens.

To a first approximation, some of the light will reflect off of the air-coating interface and it will shift by w/2. Of the light that passes through the interface (most of it) some will reflect off of the coating-glass interface. This will shift the wave by w/2, but the additional two trips through the coating will shift it by another w/2. These two waves destructively interfere because they are exactly out of phase and the reflected light is suppressed.

Now, one can invoke the law of conservation of energy and show that since no light is reflected it must all be transmitted. You can, instead continue the above reasoning to account for this without appealing to the magical law.

Remember that some of the light went through the air-coating interface, bounced off the coating-glass interface, and came back through the coating. Some of this will instead bounce off the air-coating interface again (without a phase shift) and go through the coating-glass interface. If you drew the picture right, this light is in phase with the original light you had going through the lens, and so interferes constructively. Here's some of your light being added back in.

To get the calculation correct, the sum has to be done over an infinite number of reflections, but the corrections keep getting smaller. What you find is that with the correct coating all of the corrections to the reflection keep interfering destructively with the reflection from the front reflection and sum to zero. Likewise, all of the transmitted portions add together to recreate the original wave. Nature keeps track of all possible ways the light could go, and in the end all of the light is accounted for. Pretty cute.

To actually perform this calculation, you have to keep in mind that this is a fundementally quantum mechanical operation. The waves you're summing represent the probability amplitudes of the light, the square of which gives the probability of the light either being reflected or being transmitted.

If you want a better explanation and you're like me---not a physicist, but want to get a feeling for what's really going on---I'd check out the book Q.E.D. by Richard Feynman. He was not only a lot brighter than me, but he was a lot better at explaining these things. Although he does not exactly explain a coated lens, he covers some examples that are pretty darn close.

The book is a slim (and inexpensive) volume in which he explains to a non-physicist how electrons and photons do their thing: quantum electrodynamics. This theory explains most of what we experience on a day-to-day basis, including optics. (The theory does not talk about nuclear forces, which we don't normally notice, and gravity, which we do.)

The book requires very little actual math, although I'd say you might need to be "mathematically minded" to get it (or, perhaps, care). As Feynman warns: "I hope you can accept Nature as She is---absurd."


Emmanuel BIGLER
8-Dec-2004, 05:27
To John D Gerndt. Let me put something clearly. Consider a glass plate which has an anti-reflection coating on one side only. If you illuminate from the coated side or from the un-coated side, the behavior at the coated interface is similar in both directions, the coated side does not act as a 'light diode'. But if the 2-nd side of the plate is un-coated, you'll get as usual a prasitic reflection either on entrance or exit according from which side you illuminate the plate.
If we concentrate on a single coated interface, things are simple : everythig not transmitted is either reflected or absorbed. A thin metal layer of sub-micron thickesss wil act as a semi-reflective mirror but will absorb light as well.
The progress in modern dielectric coatings is that they do not absorb light, either they reflect or transmit, the sum of both energies being close to 100%.
Consider for example a non-photographic coating, the one used for lasers, a highly reflective coating. A fresh silver mirror reflects 95% of light but absorbs 5%. Even this 5% is unacceptable for a laser. Modern reflective coatings can go up to 99.99% with almost zero absorption ; symmetrically engineers can design something than will transmit 99.99% and reflect amost nothing but the two coatings are of course totally different ; in common they have something : almost zero energy is withdrawn from light propagation.

John D Gerndt
9-Dec-2004, 18:53
Thank you Isaac for the only explanation that sifted down to the more basic understanding I have of the universe. Thank you too for the book recommendation. It sounds like my kind of physics.

I don't think I can yet explain it to my brother but I can make the phase drawing you suggested (much like the one I had cobbeled up on my own), read him you words and hand him the QED book...when I am through with it!