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Diffraction Grating and Intensity

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the maximum intensity for a diffraction grating of N slits at the first principal maximum is N^2 times bigger than for a single slit.

    2. Relevant equations

    I=I(0)[sin^2(NB)/sin^2(B)] where B is ([tex]\pi[/tex]dsin[tex]\phi[/tex])/[tex]\lambda[/tex]

    3. The attempt at a solution

    I feel i understand the concepts here but this question has stumped me. I've scoured the internet, read through the relevant chapters in my two textbooks and still i can't see how this would be proven. It is just stated in my lecture notes, is it that intuitive?

    I really would be grateful for any help.
  2. jcsd
  3. May 22, 2010 #2


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    Homework Helper

    The maximum is at Φ=0. You have to start with the limit sin(x)/x when x tends to 0. Find the limit of sin(Nx)/sin(x). Use that sin(Nx)/sin(x)=[sin(Nx)/Nx] N [x/sin(x)].

  4. May 23, 2010 #3
    Thanks for the help, but i don't really understand the bolded part. I haven't seen that trig identity before and i can't see how it would be found.
  5. May 23, 2010 #4


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    Homework Helper

    It is just multiplying both the numerator and the denominator with the same quantity and rearranging.

    [tex]\frac{\sin(Nx)}{\sin(x)}=\frac{Nx\sin(Nx)}{Nx\sin(x)}=\frac{\sin(Nx)}{Nx} N \frac{x}{\sin(x)}[/tex]

  6. May 23, 2010 #5
    Thankyou, and i apologise for my stupidity!

    I've got it now.
    Last edited: May 23, 2010
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