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sliorbra Nov13-12 05:08 PM

reflection grating

I am bothered by some question for a long time.
From basic optics, i know that the the angle of incidence equals the angle of reflection [where we define the incident angle as the angle between the beam and the normal to the surface of course].

What I can't understand is why when a polychromatic beam reflects from the surface of a reflection grating [for example, with a periodic structure] the reflection angle of the beam depends on the wavelength? Why the reflection angle should not be the same for all the wavelengths that compose the polychromatic beam and equal the incident angle [which is, in my opinion, the same for the different wavelengths that compose the beam]??

I hope my question is clear.

sophiecentaur Nov13-12 05:29 PM

Re: reflection grating
Wouldn't the main beam (main maximum) be the same for a plane mirror as for a plane reflection grating? The path differences for all wavelengths would surely all be zero in the direction of the classical main beam (isn't that all about Fermat's Principle?).

sliorbra Nov14-12 03:10 AM

Re: reflection grating
I am not sure that I understood your answer. Can you please detail more?

Philip Wood Nov14-12 03:41 AM

Re: reflection grating
(a) A crude model of a plane reflection grating would be reflecting strips alternating with absorbing strips. The distance from centre of reflecting strip to centre of next reflecting strip (the grating element) is a few wavelengths. Thus each reflecting strip is, perhaps, two or three wavelengths wide. Such narrow reflectors don't reflect light approaching (in a plane of incidence at right angles to the strip lengths) according to the laws of reflection. There is a wide spread of angles for the light leaving the strip; we have scattering rather than 'specular reflection'. Even quite crude drawings of 'wavelets' arising from different points across the width of the strip will demonstrate this. The phenomenon is very similar to diffraction.

(b) What determines the angles at which light leaves the grating as a whole is interference between light scattered from the different strips.

(c) In practice, the intensity of the beams leaving the grating is 'tweaked' by modifying the simple grating I described at the beginning.

sophiecentaur Nov14-12 04:04 AM

Re: reflection grating

Quote by sliorbra (Post 4158649)
I am not sure that I understood your answer. Can you please detail more?

Have you looked up Fermat's Principle yet?
Have you looked up The Reflection Grating?

Where does the zeroth order beam emerge from a reflection grating? It will be in a direction where there is no path difference for any of the wavelets produced by the grating. What direction will this be?

Philip Wood Nov15-12 10:32 AM

Re: reflection grating
Sliorbra. Did you follow my attempt to answer your question?

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