Principle of anti-reflective coating is very simple: thin layer between air and glass returns light wave in opposite phase. Or to be more precise: anti-reflective layer shifts light wave about 1/2 of wavelength. So light waves reflected from surfaces 'air/coating' and 'coating/glass' are subtracted (destructive interference). The reason is to increase light transmission, decrease inner lens flares and increase contrast of lenses.

I still remember when I calculated the thickness of layer (with given refraction index) to increased or decreased appropriate wavelength at school...

But: why is anti-reflective coating on front lens surface and are not on inner lenses? Or why is anti reflective coating on filters??

There is no impact to picture quality what filter reflects outside of the lens. And, as I understand, the amount of light entering into lens is still the same. The amount of light (energy) reflected out of lens (or filter) is still the same too. Just the both outgoing waves are subtracted and are not visible for outside observer. How can anti-reflective coatings on outer lens surface influence inner light inside the lens?

Additional comments (added later it because limited text capacity in comment sections below)

I know the AR coating is working. Just don't understand where is mistake in my understanding.... :-(

Let's imagine simple old lenses like Planar, Sonar, Tessar. Tessar has 4 elements in 3 groups. It means 2 lenses are glued together (with canada balsam because its refractive index is same as glass). Consider these two glued lenses are one optical element. There is no coating between them. So to simplify it consider we have 3 elements only. Each of them has 2 surfaces. Let's name surfaces (from outside towards the chip) as A-B, C-D, E-F. A-B is the first glass/lens element, E-F is the last lens/element closest to chip. I don't consider any filter (but if you want, add filter as 1 additional element and just add 2 letters). I don't consider any diffraction nor absorbtion on glass and coating layer (lens elements are perfectly smooth and glass is clear).

1.: No coating on "A": there is approx. 4% reflected light from "A", transmission is 96%.

2.: AR coating on "A": there is approx. 2.2% reflected light from "A", transmission is 97.8%. (But these 2.2% of reflected light are eliminated during destructive interference because of opposite phases of waves reflected from glass and from coating-as explained above.)

Calculation is based on indexes n=1.5 (glass), n=1.3 (coating), total light falling to lens is 100%, formulas for refraction and reflection are below (see Steven Kersting - thank you). Coating is on the "A" surface only.

Difference in transmission between coating and non-coating is 1.8% (=97,8%-96%) Improvement is 1.8% only, NOT 50%!!

Alan Marcus wrote about light reflected from "B" surface backwards to "A" in comment below (thanks Alan). There is no coating on "B" so 4% of light are reflected from "B" backwards to "A". It is exactly 3.9% (=4% x 97.8%). Returned light is transmitted through "A" ( 3.81%=97.8% x 3.9%) and 0.086% (=2.2% x 3.9%) is reflected back from inner side of "A".

0.086% is reflected back from coated "A" only! And causes some glares, flares, less contrast...

But 0.086% is about -10EV difference only (=log ( 0,086% / 100% ; 2) ). Such flares are barely visible... And where is 50% improvement of transmission?

And I didn't consider different behavior of coating on "A" for inner light (light reflected backwards from "B" to "A"). Inner light enters from more dense to environments with less density so light wave has the same (not opposite) phase. There are no coatings on B, C, D...

If there is no coating on "A", the calculation will be similar and result will be slightly worse (about 1,5% light will be reflected from inner surface of "A" back to "B"). Negligible difference..

In this example I considered only the first lens element (A-B). There are much less light from the second and from the third element. Another -10EV difference.. There are other reflections, of course. From C to B and A, from D to C, B and A... I ignored them to simplify this text...

Again-I believe it works. So I had to make a mistake somewhere .... But where??

  • 1
    \$\begingroup\$ You can read all about it in this technical article from Zeiss: About the reduction of reflections for camera lenses: How T*-coating made glass invisible \$\endgroup\$
    – osullic
    Commented Aug 19, 2022 at 9:04
  • 1
    \$\begingroup\$ The only important part isn't the thickness (which allows ideal anti-reflective coating only for a single wavelength of light). It's the reduced refractive index contrast too. If you have an abrupt change from n = 1.0 to n = 2.0, it's worse than n = 1.0, then n = 1.4, then n = 2.0. \$\endgroup\$
    – juhist
    Commented Aug 19, 2022 at 13:04
  • \$\begingroup\$ What makes you think only the outermost lens surface is coated? \$\endgroup\$
    – Eric S
    Commented Aug 20, 2022 at 4:45
  • \$\begingroup\$ @EricS In prehistoric lenses of the mid-20th century that was the case. Not until the rise of digital sensors, which are much more reflective than the emulsions smeared on the surface of films, did most lens makers start coating the back side of lens elements. \$\endgroup\$
    – Michael C
    Commented Aug 22, 2022 at 10:13
  • \$\begingroup\$ "I know the AR coating is working. Just don't understand where is mistake in my understanding...." There is no free lunch in physics. A specific amount of light energy falls on the surface of an optical element. Light that is reflected away from a refractive surface does not also go through the refractive medium. Reflected light is LOST light. Coatings that reduce reflection absorb only an infinitesimally small amount of light and let the rest of the light that would otherwise be reflected pass through the refractive medium (lens element). \$\endgroup\$
    – Michael C
    Commented Aug 22, 2022 at 10:37

3 Answers 3


A front surface coating does a few useful things.

First, contrary your understanding, it does increase transmission. An uncoated glass surface reflects about 4% of the light that passes through it (entering or leaving the glass), which is much of why lenses with more than four elements were uncommon before the advent of coating technology. Multiply 4% loss over 6 glass surfaces even in an 1890s technology Cooke Triplet type lens, and you're already down to only 78% transmission (not even accounting for filtration losses inside the glass elements). A Tessar type adds a cemented interface, which loses less than 4%, but still has those six glass-air surfaces. The f/2 Xenon on my 1941 Weltini, with six elements in four groups, has no more than 72% transmission (probably a little less, since there are cemented interfaces as well). Obviously, this is all air/glass surfaces, but even reducing the first 4% loss is significant for low light applications.

Second, coating reduces scatter at each air/glass interface -- scatter being light that is neither transmitted directly nor reflected. In general, the lower the reflectivity of a surfac, the less it scatters as well. Scatter produces "flare" -- which isn't what you see in a Transformers CG shot, but instead is an overall loss of contrast due to light from bright areas being scattered into what should be dark areas.

Third, reducing reflection at the first surface limits the strength of light source image reflections from whatever filter(s) might be mounted in front of the lens (these are the dazzle spots you see in an image where a bright light source like the Sun is in or barely out of frame). These are the Michael Bay dazzle spots.

In the edit of the original question, we see this: "Just the both outgoing waves are subtracted and are not visible for outside observer." Sorry, you don't understand how destructive interference works. Light that's phase shifted a half wavelength isn't just invisible, it's forbidden by physics. That is to say, the light never reflects at all if doing so would produce a half-wave interference. This isn't what happens with coatings, however; if it were, there would be color fringes like in a soap bubble, as different viewing/reflecting angles interacted with the thickness of the coating. Instead, the coating makes the interface less "abrupt" to the incoming (or exiting) light, like finding a drop curb at a driveway instead of trying to climb the regular curb.

  • \$\begingroup\$ Thanks for comments, but.... It is not the answer to my question. I have no doubts it's working. :-) I just don't understand it. Don't understand why is coating just on the first lens surface... Btw. coating doesn't reduce scatter in glass/air interfaces by itself. Reducing scatter is consequences of less light diffused on all lens surfaces. The inner surfaces especially (and coating help to reduce it). But there is no scattering from the light diffused by the first lens surface (or the filter). This light is emitted out of lenses and have no impact on scattering nor flares... \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 12:19
  • \$\begingroup\$ ...to reduce flares/scatter you must put coating to all surfaces. Not just to the first surface. \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 12:21
  • \$\begingroup\$ I don't know of any lens that's coated only on the first surface. If they're coated at all, they're coated on every glass/air interface. Scattering happens both in reflection and transmission, so you can and do get scattering from the first surface, and coating there reduces it. \$\endgroup\$
    – Zeiss Ikon
    Commented Aug 19, 2022 at 12:47
  • \$\begingroup\$ OK, maybe lenses are coated on all elements. But I am really not sure about that. But why is coating on filters?? It is working as well-less flares, scatters, ghosts... But how does it works?? How the coating and reflections from the first surface influences inner reflections inside lens? \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 13:20
  • \$\begingroup\$ Once again, a coated filter will reduce the strength of re-reflections of light sources. Reflection of lens front element, off filter, produces some of those dazzle spots. Coated filters do less of this, as well as avoiding ghosting from rear and front reflection (like the overlapped double in a pane of window glass). \$\endgroup\$
    – Zeiss Ikon
    Commented Aug 19, 2022 at 14:59

The British Physicist (Nobel Prize) John William Strutt, 3rd Baron Rayleigh, in 1886 discovered that old lenses on the shelf passed more light than new ones of the same design. Carrying on, the English optician Harold Taylor, in 1892 figured out why. Seems old lenses were blemished with soot. This was during the industrial revolution and the air was laden with smoke and soot from the coal fires that powered the steam engines and gave warmth. This coating of atmospheric pollution settled on lenses on the shelf and etched them. He discovered that this thin transparent coat somehow reduced surface reflections allowing more light to transverse the lens.

Taylor experimented and found a way to artificially bloom (age) lenses. This truly was an important discovery because new lenses suffer a 4 to 6% loss in light due to light being reflected from their polished (mirror like) surfaces. Now lenses used in cameras and telescopes are complex systems with many lens elements sandwiched together. Thus, multi-lens element systems can suffer a light loss of 40 – 50%.

This discovery and remedy are important as modern lenses often use many elements and groups. Losing 4 to 6% at each surface (glass to air junction). In a multi-element lens system, this translates to a very high loss, could be 50% or more. Plus, each internal reflection caused light rays to go astray, many misdirected rays will bath the film/chip with light scatter called flare. Flare is devastating; it degrades the image by reducing contrast. Gross reflections cause glare spots.

Many coating methods are used. One method is to place the lens to be coated in a vacuum chamber. The air is evacuated and the mineral that will be the coat is heated causing it to vaporize. This vapor condenses on the glass lens and coats and etches. It is the thickness of the coat plus the material that does the trick. Each coat is optimized for just one color of light. A modern lens has multiple coats applied. Each coat is different in thickness. A high-quality lens can have as many as 7 thru 11 coats.

  • \$\begingroup\$ Thanks for interesting comments. But it doesn't explain my question, unfortunately. Just to add some facts: several layers (MC, multicoating) helps to decrease reflection of several different light waves and waves from different angles. There are not so many ways and methods how to coat lenses. Vacuum coating, as I know, is the only usable way how to put thin coating layers on lens. \$\endgroup\$
    – Ferd
    Commented Aug 21, 2022 at 7:31
  • \$\begingroup\$ Modern nano-coating layers are based on different principles. There is no such interference of entering and outgoing waves. But different behavior of light on the surface of the lens. But.... again.... why just on the first surface??? It doesn't involve light passing through lens. – \$\endgroup\$
    – Ferd
    Commented Aug 21, 2022 at 8:02
  • \$\begingroup\$ No coat, light hits polished surface of lens. About 4% is reflected away and lost. This happens at each element junction be it air, glass, or cement. A complex lens will suffer a loss of 50% or more. Now we coat. Light hits the coat and some is reflected away. Light continues to the glass surface; some is reflected away. This light hits coating air junction and some is reflected backwards, and this ray path is in the same direction as original. Now this wave traverses the lens reducing the light loss due to reflection. \$\endgroup\$ Commented Aug 21, 2022 at 15:21
  • \$\begingroup\$ Thanks Alan, light reflected from inner side of the the first lens surface is too small. I put more details above to my question to explain and calculate it. There is limited text here... \$\endgroup\$
    – Ferd
    Commented Aug 21, 2022 at 18:07
  • \$\begingroup\$ @AlanMarcus Your explanation in comments is just wrong. The effect of lens coatings has nothing to do with interference and everything to do with impedance matching. Going through one or more layers of more gradually changing refractive index reduces reflection losses relative to a single interface with a large change. \$\endgroup\$
    – Zeiss Ikon
    Commented Aug 22, 2022 at 11:05

An AR coating is not intended to decrease light transmission; in fact, it is the opposite. "Anti-reflection" can be equally considered to be "increased transmission/transparency". Incorporating a coating with a lower refractive index on a lens than the refractive index of the optic itself reduces the overall refractive index of the constructed element. In this example a single coating with an RI of 1.3 increases the light transmission to 0.978 (97.8%).

Any single coating that reduces losses (scatter/stray reflection/etc) is beneficial, and it is that first interface that is most critical... but most modern lenses/filters have all/most glass-air interfaces coated.


  • \$\begingroup\$ The first: mentioned equation is for "thick" layers (like air/glass), not for thin layers which are used for anti reflecting coating. For these layers you must consider light interference. The second: it is not true the first interface is most critical. There is difference just 4% between the first and the second surface. 96% of light fall to the second surface. The third: any lights which is reflected (or is not reflected) from the first surface doesn't involve glares on inside elements of lens. \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 14:05
  • \$\begingroup\$ And the forth: there is no involvement of coating on the first surface to transmission, transmission is still 96%. The reflected 4% is just interfere so there are no visible reflection from outside of lense. I know I am sure wrong-but I don't know where... :-( \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 14:11
  • \$\begingroup\$ And sorrrrrrry for typo, I repaired it: "..reason for coating is to INCREASE transmission.." You are right!! \$\endgroup\$
    – Ferd
    Commented Aug 19, 2022 at 14:22
  • \$\begingroup\$ It is not only about cancelling the light that is reflected outwards (visible reflection from outside)... it is about reducing the reflections that cause light to end up in the wrong place in the image. If the first interface (or element/filter) causes a high degree of refraction/scatter it makes it that much harder to correct for with additional elements/coatings. I.e. the first interaction is the most important... that's part of why coated lenses were not common in phone cameras (the uncoated protective glass); and it's why adding an uncoated filter to a lens can ruin an image. \$\endgroup\$ Commented Aug 19, 2022 at 14:26

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