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Imagine a grating with infinite number of slits, and the spacing D between slits is larger than the wavelength so that there are high order diffractions. In each of the diffraction directions the waves constructively interfere, but what decides the percentage of power that goes into each order?

My thinking is that each slit is a Huygens source, radiating cylindrical waves homogeneously in every direction, but due to interference, only those with constructive interference can exist. I guess the energy going into each order should be equal, which is not the case. So I am confused on how the light will distribute its energy to different orders. Thank you.

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  • $\begingroup$ The diffraction envelope of a single slit modulates the intensity of the diffraction grating maxima. $\endgroup$ – Farcher Apr 3 '17 at 14:08
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    $\begingroup$ This is a pretty complex question. You should start by looking at 'blazed gratings' and work from there. The shape of the grating profile influences the power in the various orders, and has been reasonably investigated for spectroscopy applications. $\endgroup$ – Jon Custer Apr 3 '17 at 14:42
  • $\begingroup$ @sammygerbil sorry for my bad description, by "infinite periods" I mean there are infinite number of slits, not just multiple slits, and D is the period, the separation between slits, hope it is more clear now. $\endgroup$ – Zhijie Ma Apr 3 '17 at 16:59
  • $\begingroup$ You specify an infinite number of slits. What about the cross-section area of the illumination? If it is finite, then that is the same as having a finite number of slits. If it is infinite, what exists that can modulate the intensity of the peaks? $\endgroup$ – garyp Dec 10 '19 at 11:42
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There are two steps only you need to calculate the intensity distribution behind a multi slit. Firstly you have to calculate the intensity distribution pattern behind a single slit. Secondly you has to calculate the aberration of the pointlike source to all the slits and to the observers screen and by this sumerize the intensities at all interesting you points.

Are you able to compare the calculations with the real experiment? Please share your results with us.

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The physical setup is:

$$ \text{light source} \quad \overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}} \quad \text{grating} \quad \overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}} \quad \text{observation screen} $$

where $\mathcal{F}$ denotes Fourier transform, as known in Fourier optics.

Let a single slit be $a(x)$, and denote $\delta_D(x)$ as the Dirac comb of spacing $D$. Denote $\star$ as convolution, the grating is:

$$ g(x) = a(x) \star \delta_D(x) $$

Assume we illuminate the grating with a point light source $\delta(x)$, it becomes plane wave when hitting the grating:

$$ \delta(x) \quad \overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}} \quad 1 \cdot g(x) \quad \overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}} \quad \text{sinc}(\pi D_1 x) \cdot \delta_{1/D}(x) $$

where $D_1$ ($< D$) is the width of a single slit.

We identify the intensity be $|\text{sinc}(\pi D_1 x)|^2 \cdot \delta_{1/D}(x)$, i.e. discretized $|\text{sinc}(\pi D_1 x)|^2$ sampled at $D$ period, which is also known as the diffraction orders.

Let fill-factor be defined as $D_1/D$.

  • If $D_1 = D$, i.e. 100 % fill-factor, there will be no higher-order diffractions but only $0^{\text{th}}$.

  • If $D_1 < D$, usually around 90 % fill factor, the intensity would contain difraction orders like this:

enter image description here

So the energy depends on the fill-factor $D_1/D$ of your grating. The larger $D_1/D$, more energy at $0^{\text{th}}$-order.

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