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Homework Help: Optics - Diffraction

  1. Dec 10, 2007 #1
    [SOLVED] Optics - Diffraction

    Hi. I was given an extra credit problem in class to derive equations related to multiple slit diffraction. I was wondering if someone wouldn't mind looking over what I have so far, and pointing me in the next direction. Thank you!

    1. The problem statement, all variables and given/known data

    2. Relevant equations


    3. The attempt at a solution


    For (a), I'm mostly worried about how to take into account the sum in the integral. I just sort of made up a way that I thought made sense. Having a sum and an integral in the same equation just feels strangely redundant. And, also for the second part of (b), I'm not sure how to go about the justification that it asks for. Thank you again!
  2. jcsd
  3. Dec 12, 2007 #2
    I'm bumping this because I finished parts (a) and (b), but now I'm having trouble with (c).

    I know that the irradiance is energy/area/time. But, I never understood how to derive it. The only equation our book (Hecht) has in it that defines irradiance is the time average of the poynting vector which doesn't help me.

    Do I start with the equation I derived in part (a)? And, since I'm deriving the equation for y=0, the [tex]\alpha[/tex] term drops out, right? Then, the [tex]\beta[/tex] term gets squared somehow, and the summation term (from part (b)) turns into the delta term. But, what and where does the [tex]I_{0}[/tex] term come from?

    And, I also tried to look at part d, but if z=0 and you plug it into the irradiance equation given in (c), [tex]\beta[/tex]=0 and [tex]\delta[/tex]=0, so I don't know what's happening there either. I briefly thought about taking the limit as z[tex]\rightarrow[/tex]0 and using L'Hopital's rule, but I don't think that's what I'm supposed to do.

    Here's my work for (c):

    I just started trying to convert that summation term over to the [tex]\delta[/tex] that's shown in the problem statement. I ran into a small snag with the N being in the top of the fraction.
    Last edited: Dec 12, 2007
  4. Dec 12, 2007 #3
    For part c I don't understand where the last equality you asserted came from. You've got the right idea there, but the last step isn't quite right. Ask yourself:
    How do you write sin(x) as a complex exponential?
    Is there anything I can do to [tex]\frac{e^{iNx}-1}{e^{ix}-1}[/tex] to get it in something close to the form of sin(x).

    Now you won't need the exact form for intensity in terms of energy - all you will need to know that the intensity is proportional to the modulus squared of the energy.

    I_0 is just a constant that covers this proportionality factor and any other constant factors. Yes the alpha term drops out because it's for y=0

    See if you can now get something close to what you're meant to derive - if you find that your delta doesn't quite match theirs then have a close look at part (b).

    And for part (d) why don't you think l'Hôpital's Rule is what you should do?
    Last edited: Dec 12, 2007
  5. Dec 12, 2007 #4
    I'm not sure how to get the exponentials into a sine function. Because exp(ix) = cos(x)-isin(x) and that doesn't seem to help me. I thought that exp(ix)-1=2i exp(ix) sin(x). Is that not right?

    As for (d), it was just the wording of the problem. It doesn't ask for the limit, but if no one can off any better insight, I'll probably do L'Hopital's rule. Since it seems like my best shot.
  6. Dec 12, 2007 #5
    exp(ix)=cos(x) + i sin(x) (!)

    Actually that's a clever identity, but not quite right. If you check it you find
    2i exp(ix) sin(x) = exp(2ix) - 1
    (check it!)

    Though another standard formula you should know is - since sin(x) is just i times the odd part of exp(ix)

    For part (d) - yeah, normally you'd just plug in the values, but e.g. sin(0)/0 isn't well defined, so you have to take limits via l'Hôpital.
  7. Dec 12, 2007 #6
    I used the right identity then, I just forgot to divide by 2. Because if you check the definition of delta in the problem statement, it's divided by 2R. I just don't know how to get rid of that extra exponential term.

    Also, I did a little thinking and thought that L'Hopital was the way to go because of the way the alpha term drops out. It's also 0/0, so taking the limit takes it to 1. So, the answer to (d) is I=I_0.
    I just haven't used L'Hopital's Rule since calc1 about 3 years ago, and I have never seen it used in the middle of a derivation like this. I didn't know that was a legal move.

    I see how I can use that sin(x) = ... Because the 1-exp(ix) is equal to the top of that fraction, then if I divide and multiply by 2i, I can get the sin(Nx)/sin(x) except I need to also have a one-half in each of the sines as well for delta to be definied correctly.
    Last edited: Dec 12, 2007
  8. Dec 12, 2007 #7
    Yeah, that was the main issue, the half. Note the form of the extra exponential term, it is a phase (exp(ik)).

    So given that [tex]I = c \left| E \right|^2[/tex] (c a constant), what does the phase contribute to the intensity?
  9. Dec 12, 2007 #8
    For part c, recall that irradiance is simply the time average of the modulus of the poynting vector. Now that might not make much sense if you haven't done any introductory electrodynamics yet, so in essence, all you really need to know is that:

    I = (1/2)\epsilon _0 c * |E|^2

    (going from memory, so you might want to check the coeffcient to that.. it's not important really though, since we'll end up setting it as I_0). Where e_0 is the permitivity of free space, c is the speed of light, E is your electric field.

    Now with this in mind then, all you really need is |E|^2. You've already found E, all that remains is this. Note that |E| is the modulus, and since E is complex, you must take the E*(E*). That is E times its complex conjugate.

    As for that factor you were widdling down, the trick is to factor out an:

    \ \ \ \ e^{ikz/2R}

    from the top and the bottom respectively.. you'll notice then, if you write down the definition of sin in complex exponentials, that your required sin fcns will follow directly.

    The outstanding factors of e^i... disappear when you take |E|^2.

    For part d, realize that sinx/x is well defined at x = 0, and is commonly called the sinc function. It's a continuous fcn at x=0, and is equal to unity at that point.
    Last edited: Dec 12, 2007
  10. Dec 12, 2007 #9
    I don't understand where the exp(ik) is coming from.

    And, by phase, do you mean phase shift. If that's the case, it doesn't effect it at all. E is the aplitude of the wave, and it doesn't depend on the phase as far as I can remember.
  11. Dec 12, 2007 #10
    Yes exactly!

    I wrote exp(ik) instead of exp(i*N*phi/2)/exp(i*phi/2)=exp(i*(N-1)*phi/2) for brevity (I just meant k was something I hadn't bothered working out).

    Can you get out the correct answer now?
  12. Dec 12, 2007 #11
    If that end part is true, then what I already did is correct, I just forgot to divide the arguments of the sine and exponential by 2.
  13. Dec 12, 2007 #12
    I did get it. Thank you to the both of you.
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