1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Plotting a far-field Intensity distribution

  1. Oct 30, 2017 #1
    Hey,

    I'm attempting to plot a far-field intensity distribution using theoretical values, however I'm having difficulty with calculating the intensity using the following equation:

    $$I = I_o \frac {sin^{2}{b}}{b^{2}} \frac {sin^{2}{Ny}}{sin^{2}{y}}$$

    where:

    $$y = \frac {kdX}{2f}$$

    $$b = \frac {kaX}{2f}$$

    $$X = f*theta$$

    I've got values for the focal length, f, the slit width, a, the slit separation (periodicity), d, the number of slits, N, the wavelength of the light beam and the original intensity which has been normalised to 1.

    I'm trying to find the maximal intensity, which I currently believe to be 400, given that N = 20, however I'm having difficulty finding the other maximal values (ideally up to 8 each side).

    I understand the nature of the convolution theory, but not entirely sure how I can utilise it to derive the necessary results.

    Any help would be much appreciated?
     
  2. jcsd
  3. Oct 30, 2017 #2

    Charles Link

    User Avatar
    Homework Helper

    For the interference part of the function ## I(\theta)=I_o \frac{\sin^2(N \phi /2)}{\phi/2} ## where ## \phi=\frac{2 \pi}{\lambda} d \sin(\theta) ## , the primary maxima occur when the denominator is equal to zero, and the limit is ##I(\theta_{max})=I_o N^2 ## as ## m \lambda=d \sin(\theta) ## for any integer ## m ##. This is the case for any ## N ##. The secondary maxima occur between the zeros of the numerator when the denominator is not equal to zero. The numerator is equal to zero when ## N d \sin(\theta)=k \lambda ## for some integer ##k ##. There are ## N-1 ## zeros of the numerator between adjacent principal maxima where the denominator is also zero. There are ## N-2 ## secondary maxima between adjacent primary maxima. One thing that makes doing this with a computer somewhat tricky is you need to simply assign the value of ## N^2 ## when the denominator is zero (at the primary maxima), because the computer will give an error there. ## \\ ## The other part of the function that gets multiplied by this is a single slit diffraction factor. You simply take a product of the two parts to get the complete result. It is not necessary to try to apply the convolution theorem from linear response theory and Fourier transform theory to graph this result. ## \\ ## The single slit diffraction pattern has a peak in the center (at ## \theta=0 ##), and has zeros at ## m \lambda=a \sin(\theta) ## for integers ## m \neq 0 ##. ## \\ ## Additional note: It is somewhat easy to sketch both of these functions by hand. The most important feature of the interference factor (the part described in detail above) is the primary maxima. The peaks are somewhat narrow with ## N=20 ## because the numerator drops to zero on either side of these peaks very quickly. The diffraction factor is also easily sketched.
     
    Last edited: Oct 30, 2017
  4. Oct 30, 2017 #3
    Thanks for the help, however I don't have any values for θ that I can use, all I've got that I can think may be of some help is the focal length of the lens, that I used to focus the diffraction pattern. Is there any other method of obtaining θ theoretically?
     
  5. Oct 30, 2017 #4

    Charles Link

    User Avatar
    Homework Helper

    Do you have any idea what ## d ## might be? If you used monochromatic light, with ## N =20 ##, (which is reasonably large), you should get mostly primary maxima as the main feature of the pattern. You'll get a central maxima at ## \theta=0 ##, (## m=0 ##), and other primary maxima at ## m \lambda=d \sin(\theta) ## for all integers ## m ## where ## \sin(\theta) \approx \tan(\theta)=x/f ##. (If ## d ## is on the order of the wavelength ## \lambda ##, ## m ## will be limited in that there will be a maximum ## m ## that ## |m \lambda/d| <1 ## in order to have ## |\sin(\theta)|<1 ##). ## \\ ## If you know the wavelength ## \lambda ##, you should be able to compute ## d ##. To get quantitative results for the slit width ## a ##, it usually takes considerably more effort. (For a single slit, ## a ## is more easily found. When there are many slits, e.g. when ## N=20 ##, ## a ## is more difficult to quantify from observing the diffraction pattern.)
     
    Last edited: Oct 30, 2017
  6. Oct 30, 2017 #5
    I've already calculated d and a, using a microscope, to be, d = 2.53x10^(-4)m, and, a = 5.4x10^(-5)m, respectively. Would I just have to then input integers from say, 1 to 8, to get the correct angle, θ, for both parts of the function, interference and diffraction?
     
  7. Oct 30, 2017 #6

    Charles Link

    User Avatar
    Homework Helper

    Basically yes. The single slit diffraction part should limit how wide the interference pattern is. After around ## m=2 ## or ## m=3 ##, for ## m \lambda=a \sin(\theta) ##, the diffraction factor becomes very small and is largest between ## m=1 ## and ## m=-1 ##. (The diffraction factor is equal to ## 1 ## at ## \theta =0 ##). (If ## a ## is very small, you get a diffraction factor that covers a wider range of ## \theta ## ). The diffraction factor is easily sketched. You can also graph it with a computer. ## \\ ## Just a quick guess at what you observed experimentally: You probably got about 10 bright fringes in a central region (with the fringes brightest in the center) and a second set(s) of about 5 fringes on either side of the 10 brightest ones, and perhaps a 3rd set of about 5 on either side of those that is considerably dimmer.
     
    Last edited: Oct 30, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Plotting a far-field Intensity distribution
  1. Far Field Radiation (Replies: 1)

Loading...