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If light incident on a diffraction grating makes an angle $\alpha$ with respect to the normal to the grating, show how $$m \lambda = d\sin\theta$$ becomes $$m\lambda = d[\sin(\theta - \alpha) + \sin(\alpha)].$$

I have absolutely no idea what to do here. I've looked into the problem and I've gathered that for incident light $\theta_i$ you get $$m\lambda = d[sin(\theta_i + \theta)],$$ but I'm not sure why that is. No resource I've found has given me a good explanation for that.

If I could prove that $\theta = \alpha + \theta_i$, where $\theta_i$ is the zeroth-order line from the angled incident line, then I could easily substitute my values in, but my problems are that a) I don't understand how to prove the expression in my second paragraph and b) I'm not even sure if $\theta = \alpha + \theta_i$ is true.

I'm mostly looking for the theoretical knowledge to help me work through this example. Any help will be greatly appreciated.

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  • $\begingroup$ Apologies, question has been edited. d is multiplied by both $sin(\theta - \alpha)$ and $sin(\alpha)$ $\endgroup$
    – Alex
    Commented Nov 4, 2014 at 17:56
  • $\begingroup$ At the heart of it, this question is seeing if you understand how a diffraction grating works. Ponder upon momentum. $\endgroup$
    – Jon Custer
    Commented Nov 4, 2014 at 17:59
  • $\begingroup$ Are you saying that the light rebounds "elastically" such that the incident angle (relative to the zeroth-order diffraction line) will be $\theta = \alpha + \theta_i$ as I stated above? Or am I not making any sense whatsoever? $\endgroup$
    – Alex
    Commented Nov 4, 2014 at 18:04
  • $\begingroup$ A diffraction grating works by transferring quanta of momentum along the direction of the grating. The energy of the photon is conserved, the resolved momentum of the photon perpendicular to the grating remains constant, so the free parameter is the takeoff angle of the photon receiving $\pm n$ units of momentum from the grating (leading to the allowable $\pm n$ diffracted orders). $\endgroup$
    – Jon Custer
    Commented Nov 4, 2014 at 20:10

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The diffraction grating acts as an array of light sources. Each slit in the grating emits light, and you calculate the fringe positions by calculating the path lengths from all these sources/slits and identifying the points where the phase differences are multiples of $2\pi$.

However, as you usually do the calculation you assume the incident light is normal to the grating, and therefore that all the light sources/slits are in phase with each other. If the incident light is not normal to the grating then the incoming light at each slit in the grating is at a slightly different phase. Now the diffraction grating acts as an array of light sources but with each source/slit having a slightly different phase. You need to include this phase difference when you're calculating the phase differences for the light reaching your screen.

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  • $\begingroup$ Okay, following that I can understand how $m\lambda = d[sin(\theta) + sin(\alpha)]$, but what I don't understand is why the diffracted angle is still at the same angle relative to the normal (compared to when the incident light is normal). That seems to be what this, as well as a few other sources, are suggesting. If is thr case then I know how to finish answering my question, but I still don't understand WHY it's at the same angle. $\endgroup$
    – Alex
    Commented Nov 4, 2014 at 18:24
  • $\begingroup$ @Alex photons do not bounce off of things. They are absorbed followed by a new photon emitted in a random direction. These random directions include every point on the screen but because of Path Lengths there is constructive and deconstructive derivation of all the photons convoluting at the screen. $\endgroup$ Commented Dec 21, 2016 at 16:58

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