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How do anti-reflective coatings let MORE light in through the lens?

  1. Jul 14, 2013 #1
    In the context of anti-reflective coatings put on glasses lenses, I recently read, "No light reflecting means more light in through the lens - energy must be conserved after all."? How does MORE light get in?

    The best I could imagine is either:

    1. That by using destructive interference on reflecting rays, more coherent light can enter the lenses and get to the lenses without colliding with reflecting light which might change their direction.


    2. The anti-reflective coatings keep the critical angle wide so that more light is refracted and less is reflected.

    How do you suspect that MORE light gets into a lens with an anti-reflective coating, than without that coating?

    (I am currently not a student. I am studying for the medical college admissions test on my own and without a course to hopefully get into medical school. Many of my questions are inspired by my studying.)
  2. jcsd
  3. Jul 14, 2013 #2


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    Light is an electromagnetic wave. When a wave enters a medium with a different refractive index, such as when light enters a glass lens, a certain portion of the wave will always be reflected. The coatings use destructive interference to make it so that the reflected portion of the wave contains a much smaller percentage of energy than it would without the coatings. Less energy in the reflected portion of the wave means less light is reflected and more passes through the lens.
  4. Jul 14, 2013 #3
    What causes more light to pass through the lens? Sorry if I'm being dense.
  5. Jul 14, 2013 #4


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    It is because of conservation of energy. If the glass does not absorb, the intensity of the incident ray (incoming energy in unit time) equals to the sum of the reflected intensity and the intensity of the refracted ray. If there is no reflected ray because of the anti-reflective coating, all light is refracted. Reaching the other side of the lens, also covered with anti-reflecting layer, the whole light beam traverses the interface with air, without reflection loss.

  6. Jul 14, 2013 #5
    Really? My understanding or misunderstanding was that with an anti-reflective coating, the same amount of light was initially reflected (and then about half a wavelength later would face destructive interference) like in this illustration: The first illustration here if you scroll down, http://physics.stackexchange.com/qu...flective-coating-makes-glass-more-transparent

    The drawing shows R1 and R2 facing destructive interference after about half a wavelength, essentially cancelling each other out. Unless I read it wrong. I know there's something I'm missing, and am not sure what?
  7. Jul 15, 2013 #6


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    They interfere immediately, not after a half wavelength.
  8. Jul 15, 2013 #7


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    The anti-reflective coating works for light beams much broader than the thickness of the layer. You can imagine that a ray reflected from the interface with glass and stepping out into air "finds" a directly reflected ray out of phase.


    Attached Files:

  9. Jul 15, 2013 #8


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    This is not a case of getting something for nothing and no fundamental law is being tinkered with. It's merely a matter of re-directing some energy from being reflected to being transmitted. It just boils down to the effect of wave interference.
  10. Jul 15, 2013 #9
    Note that when the index of refraction changes gradually over a distance that is large compared to the wavelength of the rays, the rays will refract under exactly the same angle as if there were an interface - but there is no reflection.

    If you look at the side of the refracted ray... what is the ray that can interfere with the refracted ray and reinforce it?
  11. Jul 15, 2013 #10


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    All the rays that aren't reflected or absorbed?
  12. Jul 15, 2013 #11


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    There are multiple reflected rays, and there are multiple transmitted rays. The reflected rays interfere destructively, the transmitted rays interfere constructively.

    Without getting into the math or other details, that is the plain simple answer.

    Getting a little more detailed, the following figure may help. Just imagine that the incident ray is at normal incidence, so all the reflected rays overlap each other, and all the transmitted rays overlap each other as well:


    For a perfect AR coating, when you add all the reflected rays accounting for the phase of each, you get an amplitude of zero. When you do the same for all the transmitted rays, you get an amplitude equal to that of the original incident ray. So 0% is reflected and 100% is transmitted.

    Going further with this, the coating pictured above is AR only for some discrete wavelengths. To get "broad band AR" coatings, you might have 20 to 30 coating layers, alternating between two materials of different refractive indexes:


    The thickness of each layer can be adjusted, and companies that make such coatings can calculate a set of thickness that typically give under 0.25% to 0.5% reflectivity (considered acceptably small in many applications) over a fairly wide range of wavelengths (eg. visible, near IR, etc.)


    The above figures were found at
    Last edited: Jul 15, 2013
  13. Jul 15, 2013 #12
    Anti-relection means more transmission all right.
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