Diffraction grating with slits out of phase

In summary, the Fraunhofer diffraction pattern of a grating depends on the spacing between the grating lines, the overall extent of the grating, the amplitude distribution over an individual line, and the phase distribution over an individual line. The pattern is symmetrical about the center and the relative intensities for order n = 0, 1, 2, 3 are I0=0, I±1=0.09, I±2=0.25, I±3=0.32. Modifying the phase distribution of each line can increase the intensity of beams on one side of the pattern. However, the function used to calculate the intensity may not appear symmetric in some cases, so it is
  • #1
greenlight88
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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!
 
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  • #2
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.
 
  • #3
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?
 

Related to Diffraction grating with slits out of phase

1. What is a diffraction grating with slits out of phase?

A diffraction grating with slits out of phase is an optical device that consists of a series of parallel slits that are spaced at a specific distance and are out of phase with each other. This means that the slits are not perfectly aligned, resulting in the interference of light waves passing through them.

2. How does a diffraction grating with slits out of phase work?

A diffraction grating with slits out of phase works by causing different wavelengths of light to interfere with each other as they pass through the slits. This interference results in patterns of light and dark bands, known as diffraction patterns, which can be used to analyze the properties of light such as wavelength and intensity.

3. What is the purpose of using a diffraction grating with slits out of phase?

The purpose of using a diffraction grating with slits out of phase is to separate and analyze different wavelengths of light. This is useful in various scientific fields, such as spectroscopy, where the different wavelengths of light can provide information about the composition and properties of a sample.

4. How is a diffraction grating with slits out of phase different from other types of diffraction gratings?

A diffraction grating with slits out of phase is different from other types of diffraction gratings, such as those with evenly spaced slits, because it produces more complex and varied diffraction patterns. This is due to the interference of light waves passing through slits that are not perfectly aligned.

5. What are some real-life applications of a diffraction grating with slits out of phase?

Diffraction gratings with slits out of phase are commonly used in various scientific and technological applications, such as in optical instruments like spectrometers, lasers, and telescopes. They are also used in telecommunications, where they help separate and transmit different wavelengths of light for various purposes.

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