So, I'm given a certain wavelength $\lambda$ and the grating costant $d$ (distance between slits). I'm asked to find the maximum order of diffraction for this set of data. In general, when light falls upon the grating with angle $\theta_i$ and escapes (I don't know the right word in English, sorry) with an angle $\theta_s$, the total optical path difference is given by \begin{equation} \Delta=d(\sin \theta_i+\sin \theta_s) \end{equation} which combined with the maximum condition for bright zones $\Delta=m\lambda$ gives \begin{equation} d(\sin \theta_i+\sin \theta_s) =m\lambda \end{equation} where $m$ is the order of the maximum. So, the problem is very easy to solve when we have normal incidence (where $\theta_i=0$), but I can't manage to solve it in general where the two angles are involved. I have been told by colleagues that the maximum order is indeed $m=d/\lambda$ but I can't understand why. Every opinion is welcome. Thanks in advance for any response.

  • $\begingroup$ A photon has momentum and energy. A diffraction grating imparts specific momentum changes along the surface of the grating. Propagating orders imply that energy and momentum of the outgoing photon are preserved. Balance the two, and you have your answer. $\endgroup$
    – Jon Custer
    Commented Jun 8, 2020 at 13:24
  • $\begingroup$ Never thought of it like that, thanks for the comment! $\endgroup$ Commented Jun 8, 2020 at 15:47

1 Answer 1


As a first observation, there is no maximum order. There is however a maximum propagating order, for which $\sin \theta_s = 1$. Higher orders will not propagate but exponentially decay in the propagation direction.

The maximum propagating order propagates at 90$^{\circ}$ so parallel to the grating. It has $m=d/\lambda$ for perpendicular or $m=2d/\lambda$ for maximally oblique incidence.

  • $\begingroup$ Thanks for the comment. Does it make any physical sense maximally oblique incidence with $\theta_s=90º$? I'm struggling to see that $\endgroup$ Commented Jun 8, 2020 at 11:10
  • $\begingroup$ It is the theoretical limit. $\endgroup$
    – my2cts
    Commented Jun 8, 2020 at 11:33
  • $\begingroup$ I supposed it. Thanks again! $\endgroup$ Commented Jun 8, 2020 at 11:38
  • $\begingroup$ If you have an optical system with lenses and for example a limited projection screen, then of course you need to use the entry and exit NA of that system. Good luck ! $\endgroup$
    – my2cts
    Commented Jun 8, 2020 at 12:22

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