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Minima of a diffraction grating

  1. Feb 5, 2015 #1
    Note from mentor: This thread was originally posted in a non-homework forum, so it does not use the homework template.

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    We have n slits, however suppose half of the middle ones are covered. How could you go about finding the angles at which the minima occur at in the Fraunhofer limit? I've been told using phasors is the best approach to do this.

    All I can see is that the phase difference between adjacent slits will be kdsinθ with slit separation d, wavenumber k. However there is a jump in phase as we go from slit n/4 to 3n/4 of 0.5nkdsinθ. To obtain a minimum we need the remaining n/2 phasors to sum vectorially to give a closed loop. However converting this into some mathematical condition is proving very confusing.

    A few things I have thought:
    a) Consider the two groups of n/4 slits separately, and ask that their phasors alone form closed circles. This sets nkdsinθ/4=2mπ for integer m>0 and so θ≈4mλ/nd.
    b) If we consider all n slits, for the cases where the n phasors form circles that double/quadruple/octuple etc up on themselves, we can remove the middle n/2 phasors and still get closed circles. This gives nkdsinθ=2mπ for integer m>0, and so θ≈mλ/nd. Note this makes a) redundant.

    However these aren't right. Any help? Thanks.
     
    Last edited by a moderator: Feb 5, 2015
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  3. Feb 5, 2015 #2

    mfb

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    How does "half of the middle ones are covered" look like? Half of those in the middle (how many count?) are covered? Or do just have n/4 uncovered slits, then n/2 covered slits, then n/4 uncovered slits?

    Assuming the latter, there are are two ways you can get a minimum: each side can give zero intensity (then 0+0=0), or the two sides lead to opposite phases. Symmetry does not allow anything else.
     
  4. Feb 5, 2015 #3
    Thanks for your reply - yes you assumed right - n/4 uncovered then n/2 covered and then n/4 uncovered. Ok, I understand what you mean by your conditions, but then how would you go about converting them into mathematical expressions in order to get the angles of the minima? I'm having trouble doing this.
     
  5. Feb 5, 2015 #4

    jtbell

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    Suppose you start with 8 slits with equal widths (hence equal amplitudes for their phasors), and equally spaced. You add 8 phasors with equally spaced angles, say 0, Δ, 2Δ, 3Δ, 4Δ, 5Δ, 6Δ and 7Δ degrees, where Δ depends on the position on the viewing screen (proportional to distance outwards from the center). For certain values of Δ, the chain of phasors closes back on itself and the resultant is zero. More generally, you calculate the magnitude of the resultant for various values of Δ, and make a graph of magnitude versus Δ.

    Close the four slits in the middle. Now you add four phasors, with angles 0, Δ, 6Δ and 7Δ degrees. Again you can make a graph of magnitude versus Δ. You can also set up an equation that sets the magnitude to zero, and solve for Δ. The minima may not necessarily go all the way down to zero, however! (I haven't actually tried it yet, myself)

    Phasors are just vectors in two dimensions. Do you know how to add two (or more) vectors with given magnitudes and directions (angles), using their "x" and "y" components? Or you can do it graphically, of course, but that's tedious if you want to do a lot of different values of Δ.
     
    Last edited: Feb 5, 2015
  6. Feb 5, 2015 #5
    So for your n=8 case, if I draw the phasors for the case where the middle n/2 are covered, then we can split them up into two parts. The first two slits give two phasors, with overall phase Δ/2. The last two slts give two phasors with overall phase 6Δ+Δ/2=6.5Δ. Now I want the resultants of these pairs of phasors to be opposite, but I need to allow for 'going around more than once' and so I say Δ/2=2πm-6.5Δ for integer m. Now solving gives Δ=2πm/7 and then as Δ=kdsinθ, θ=mλ/7d.

    Now I'm pretty sure this is wrong. I know that the angular width of the principle maxima in the 'covered' form should be 2/3 that of those in the uncovered form. In uncovered form for n=8, we get minima at θ=mλ/8d. The factor of 2/3 doesn't work here.
     
    Last edited: Feb 5, 2015
  7. Feb 5, 2015 #6

    jtbell

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    For two angles to be in opposite directions, the difference between them needs to be...
     
  8. Feb 5, 2015 #7
    7

    Ok, I didn't think about that very well - so I believe we want Δ/2+(2m+1)π=6.5Δ for integer m, so then Δ=(2m+1)π/6, θ=(2m+1)λ/12d. These are equally spaced. Angular width is then λ/6d.

    Uncovered case, minima when nΔ=2mπ θ=mλ/8d for integer m but for m=8r for integer r. Again equally spaced, and so angular width around a principal maximum is λ/4d. The factor of 2/3 works nicely here.

    This raises a new question - you see in the uncovered case I have had to exclude the values of m=8r because then we will clearly get maxima if you think about how the phasors work out (phase between adjacent phasors is Δ=2rπ which gives a maximum). We seem to have zones with double then angular width where the principle maxima occur compared to where the subsidiary maxima occur - however I don't seem to have to do this in the covered case - why not?

    Aside from that, I just need to generalise now to n phasors, so I'll have a go at that. Thanks.
     
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