0
$\begingroup$

enter image description here

I've got two questions on anti-glaze coating. This is how I understand it: the phase difference between the light reflected by the front surface and the back surface of the coating is pi/2 so they would cancel each other out

1) Surely the two different path, one reflected off the front and the other from the back, aren't on top of each other. So why would they interfere? Is this just an approximation that's good enough?

2) This would destructively interfere the light reflected by the surfaces of the coating, but it may also be reflected off the back surface of the glass itself. Surely it wouldn't cancel out and still give reflection?

Thanks

$\endgroup$
  • $\begingroup$ Please read up on interference due to thin films $\endgroup$ – Tamoghna Chowdhury Oct 27 '15 at 12:10
0
$\begingroup$

You're missing the fact that there is, in the ray-tracing method of analysis, an infinite number of rays all entering in parallel. So the reflected light from ray 1 interferes with the incoming light from ray N and so on.

"Glare" is essentially light that is scattered or multiply-reflected so as to appear to come from a different source than the actual incoming light. What an antireflective coating does is to minimize the magnitude of these reflections so that most of the transmitted light is arriving at the same angle (i.e. from the same source point) as the actual source. You should be aware that both reflections and scatterers such as dust and scratches can contribute to glare. Eyeglasses, car windshields, etc. need to be kept clean for this reason; in fact coatings labelled "anti-glare" may be designed to minimize dust/dirt collection more than to minimize reflections.

$\endgroup$
  • $\begingroup$ So the light reflected off the back surface of the glass is not cancelled but it's negligible because the fact that there is less dust/dirt is more significant- is this what you're trying to say? $\endgroup$ – RelativisticDolphin Oct 27 '15 at 12:48
  • $\begingroup$ Wait - I'm not sure about the first one either. If ray 1 interferes with ray N, surely they aren't coherent so they won't interfere destructively will they? $\endgroup$ – RelativisticDolphin Oct 27 '15 at 13:37
  • $\begingroup$ Yeah, they're effectively coherent given the distance to the source and stuff like that. As to front vs. back face, take a look at the wikipedia page on Fabry-Perot interferometer. $\endgroup$ – Carl Witthoft Oct 27 '15 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.