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The Law of Reflection states that the angle of reflection is equal to the angle of incidence.

However, consider the case of a diffraction grating with light incident at only one specific direction, how come the reflected light comes off as at range of different angles?enter image description here

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  • $\begingroup$ The elementary laws of reflection (Hero of Alexandria) and of refraction (Snellius) ignore diffraction, they are true only when the reflecting/refracting surfaces are smooth over scales of many wavelengths; an optical grating enhances diffraction. Toraldo di Francia generalized the laws of Hero and Snellius in his so-called para-geometrical optics to take account diffraction effects in reflection and refraction but these ideas were soon generalized and replaced by Keller's more general Geometrical Theory of Diffraction and is widely used in modern simulation packages of EM scattering. $\endgroup$
    – hyportnex
    Commented Dec 24, 2018 at 16:21
  • $\begingroup$ @hyportnex I thought diffraction are applicable when light passes through an aperture. Are there diffraction of light reflecting off a finite mirror? $\endgroup$
    – The Gypsy King
    Commented Dec 24, 2018 at 16:24
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    $\begingroup$ the diffraction (and of course interference, too) occurs here at the edges of the grooves or bevels (those sharp edges that break the smoothness of the surface) of the grating. $\endgroup$
    – hyportnex
    Commented Dec 24, 2018 at 16:28
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    $\begingroup$ Diffraction is in the nature of wave propagation and manifests itself whenever the propagating wave encounters an obstacle, any obstacle, not just a hole in a screen. Always remember Huygens' principle when thinking of propagation and you will not be misled... $\endgroup$
    – hyportnex
    Commented Dec 24, 2018 at 16:30
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    $\begingroup$ Related? $\endgroup$
    – Farcher
    Commented Dec 24, 2018 at 17:57

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Typically, diffraction from a reflective grating produces multiple diffraction orders: 0, -1, +1, -2, +2, and so on. The 0th order is the ordinary reflection. The other orders are offset from the 0th order according to the grating equation. The detailed shape of the grooves in the grating affects how much light goes into each order. This is explained here.

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  • $\begingroup$ My point is shouldn't all the light go in the direction of the 0th order, since the law of reflection tells us that the angle of reflection is always equal to the angle of incidence? $\endgroup$
    – The Gypsy King
    Commented Dec 24, 2018 at 16:37
  • $\begingroup$ The 0th diffraction order is not the same as the classical reflected ray. The former spreads out as a fan and its spreading angle is roughly the distance between the edges divided by the wavelength. The "classical" ray does not spread. $\endgroup$
    – hyportnex
    Commented Dec 24, 2018 at 16:57
  • $\begingroup$ @hyporonex, Please provide a reference; or at least an explanation to your assertion that the 0th order spreads out in a fan. It certainly does not conform to my experience in the lab. $\endgroup$
    – S. McGrew
    Commented Dec 24, 2018 at 17:23
  • $\begingroup$ @S.McGrew Can you explain to me why are there diffraction in this case? I thought it is applicable only when light passes through a hole/obstacle. $\endgroup$
    – The Gypsy King
    Commented Dec 24, 2018 at 17:36
  • $\begingroup$ Just think of how the reflected orders are formed: the incident plane wave is reflected from several finite pieces of glass, from one edge of the tilted piece until the next tilted piece without direct contribution from the sides, so it is restricted in extent over the tilt, hence it must show diffraction. You cannot constrain a plane wave without spreading; if the constrain is much wider than the wavelength it is more difficult observe the spread but it is there. $\endgroup$
    – hyportnex
    Commented Dec 24, 2018 at 18:19

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