Diffraction grating with slits out of phase

[greenlight88]

Homework Statement

Describe how the Fraunhofer diffraction pattern of a grating depends on:

(a) the spacing between the grating lines;

(b) the overall extent of the grating;

(c) the amplitude distribution over an individual line;

(d) the phase distribution over an individual line.

A grating with 2500 lines per metre, each line 100 µm wide, is illuminated at normal incidence with light of 600 nm wavelength. Each line is divided lengthwise into halves, one of which introduces an additional half wavelength of path.

What is the form of the Fraunhofer diffraction pattern? Show that it is symmetrical about the centre and derive the relative intensities In of the diffracted beams for order n = 0, 1, 2, 3. (Answer should be I₀=0; I_±1=0.09; I_±2=0.25; I_±3=0.32;)

How might the phase distribution of each line be further modified in order to increase preferentially the intensity of beams on one side of the pattern?

Homework Equations

Convolution Theorem: $FT(f_{1} \times f_{2})=\frac{1}{\sqrt{2\pi}}FT(f_{1})\ast FT(f_{2})$, where FT stands for Fourier Transform.
Diffraction Pattern Intensity from a Single Slit experiment: $\frac{a}{\sqrt{2\pi}}\frac{\sin(qa/2)}{qa/2}$, where $a$ is the width of the slit and $q=\frac{2\pi}{\lambda}\sin(\theta)$.
Diffraction Pattern Intensity from a N-slit grating (no slit width): $I_{0}(\frac{\sin(NqD/2)}{\sin(qD/2)})^{2}$, where D is the spacing between the slits.

The Attempt at a Solution

(a) If the lines are further apart, there are more interference fringes
(b) The more slits there are, the sharper defined (narrower) the fringes will be
(c) I am not so sure about this - perhaps it will influence the maximum intensity and result in smudging of the fringes
(d) Perhaps again smudging of the pattern - anything more detailed here will be appreciated.

First, I consider the pattern from one of these binary slits. Note that the width is $a=100\mu m$ so the amplitude of the pattern will be:
$\frac{1}{\sqrt{2\pi}}(\int_{-a/2}^{0}(-1)\exp(-iqy)dy +\int_{0}^{a/2}(1)\exp(-iqy)dy) = \frac{a (-\imath)}{\sqrt{2\pi}}\frac{\sin(qa/4)\sin(qa/4)}{qa/4}$

This now should be multiplied by the amplitude from the N-slit diffraction pattern (because to obtain the aperture we convolve them in real space). Thus the overall intensity should be given by:
$I \propto \frac{\sin(qa/4)^{4}}{(qa/4)^2} (\frac{\sin(NqD/2)}{\sin(qD/2)})^{2}$
and since in our case the distance between the slits, $D =4a$, this becomes:
$I \propto \frac{(\sin(qa/4))^{4}}{(qa/4)^2} (\frac{\sin(2aNq)}{\sin(2aq)})^{2}$

Whenever I try to plot this against $q$ in Mathematica, especially for large $N$, this intensity does not appear symmetric. Also, I don't know how to get the relative intensities - I just don't seem to get the answer.

Any help will me much appreciated!

 

[Bill Simpson]

Since we have zero idea what any of the coefficients you are using in your Mathematica code it is likely very difficult to diagnose exactly why your plot does not appear symmetric and impossible to test whether any idea that might fix the problem actually works or not. Imagine trying to diagnose "My plot is funny. Why?"

Wild guess. Try experimenting with the plot options PlotPoints and MaxRecursion and see if either of those help.

If that doesn't fix it then post the simplest exact Mathematica code you have that demonstrates the problem and exactly what to look for in the plot to see the problem.

 

[greenlight88]

I am not worried about my Mathematica code but about the question from the problem - is this a symmetric function? Are my conclusions correct? How do I calculate the relative intensities? How does the phase distribution over the slit affect the pattern?