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PlEASE HELP: Diffraction by a single slit

  1. Jul 23, 2007 #1

    Two narrow slits of width a are separated by center-to-center distance d. Suppose that the ratio of d to a is an integer d/a=m.

    Show that in the diffraction patterns produced by this arrangement of slits, the mth interference maximum (corresponding to dsin(theta)=m[tex]\lambda[/tex]) is suppressed because of coincidence with a diffraction minimum. Show that this is also true for the 2mth, 3mth, etc., interference maxima.


    My attempt dsin(theta)=m[tex]\lambda[/tex] indicates the conditions for interference maxima, where d is the distance between the two slits. We know that diffraction minimum occurs when a*sin(theta)=[tex]\lambda[/tex], where a is the slit width. Divide by both equations and we can determine which interference maximum coincides with the first diffraction minimum: d/a=m


    I don't really understand what the question wants. Could someone please help me out? Thanks in advance.
     
    Last edited: Jul 23, 2007
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  3. Jul 23, 2007 #2

    Gokul43201

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    Please read the posting guidelines and use the template provided. We can not help you unless you first show some effort yourself.
     
  4. Jul 23, 2007 #3
    My bad. It's just that this question pisses me off.
     
  5. Jul 23, 2007 #4

    Gokul43201

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    Okay, you've almost answered the first part of the question, but you've gone about it backwards.

    Let [itex]d\cdot sin\theta = m \lambda [/itex] and [itex]d/a=m [/itex], then you can show that [itex]a \cdot sin \theta = (d/m)sin \theta =...[/itex]..which answers the first part of the question.

    For the second part, how would you proceed?
     
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