The MTF is formally defined as the magnitude (absolute value) of the complex OTF. The image on the right shows the optical transfer functions for two different ...

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.

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.

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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).

Antireflectioncoatingformula

The calculator below converts between the focal length f and the field of view (FOV) of a rectilinear lens. The formula that it implements is FOV = 2 arctan (x / (2 f)), where x is the diagonal of the film. The FOV is measured across the frame’s diagonal, and is therefore smaller across the horizonal dimension, and even smaller across the vertical dimension.

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

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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.

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

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".

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.

Anti reflective coatingiPad

Until now all Pentax DSLRs use a sensor smaller than the standard 24 x 36 mm film frame. The lens still produces a 24 x 36 mm image, but the sensor only captures the central portion of that image — your wide-angle is not so wide and your tele is even longer. The calculator below takes the 1.5 x crop factor into account.

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.

The Pentax fish-eye zoom lens has a diagonal FOV varying between 180° at the wide end and 90° at the long end. The Birds-Eye lens forms a circular fish-eye image with FOV of 180° across the image circle. All other Fish-Eye lenses form full-frame images with a diagonal FOV equal to 180°.

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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?

Anti Reflective coatingPhysics

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%).

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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...

Anti reflective coatingmaterial

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%.

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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..

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.

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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.

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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...

Antireflectioncoatingprinciple PDF

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.

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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.)

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.

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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?

by TW Cronin · 2011 · Cited by 148 — ... polarized, but polarization of light ... polarization will convert depolarized light to linearly polarized light. ... This is obvious in the example illustrated in ...

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.