Question about AR

**drk** · 01-31-2007, 11:10 AM

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

**Darryl Meister** · 01-31-2007, 12:35 PM

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.

**drk** · 01-31-2007, 01:55 PM

That confirms it. Thanks, D-man.

**Darryl Meister** · 01-31-2007, 02:03 PM

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

**drk** · 01-31-2007, 02:13 PM

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

**Chris Ryser** · 02-11-2007, 03:19 PM

Originally Posted by Fezz

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.

**Darryl Meister** · 02-11-2007, 04:49 PM

Originally Posted by Chris

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).

**jherman** · 02-20-2007, 05:53 PM

I thought AR's were 1/4 thick?

Originally Posted by John Sheridan

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".

**OPTIDONN** · 02-20-2007, 06:57 PM

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

**jherman** · 02-21-2007, 10:26 AM

or tato chip it down for me.

**HarryChiling** · 03-07-2007, 01:20 PM

Here is a quick mock up.

**jherman** · 03-08-2007, 11:43 AM

:d

**Julian** · 03-15-2007, 08:52 AM

Originally Posted by James Herman

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

**Darryl Meister** · 03-15-2007, 04:41 PM

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.

**Julian** · 03-16-2007, 03:05 AM

Originally Posted by 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.

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