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Interference and Diffraction

  1. May 7, 2006 #1
    I have the follwoing question for prep for a lab, i have tried to work out the answer but i am struck with trying to work out the intensity at a certain point, in this case 4.1mm from the central maximum. I have tried using trig to calculate the angle to the point using trig then using
    I=I(sin(angle/2)/(angle/2) all squared. Am i on the right track?

    "..A diffraction pattern is formed on a screen 120 cm away from a 0.4 mm wide slit. Monochromatic light of 546.1 nm is used. Calculate the fractional intensity, I/I0 at a point on the screen 4.1 mm from the centre of the principal maximum..."

    could some please point me in the right direction, thanks
     
    Last edited: May 7, 2006
  2. jcsd
  3. May 7, 2006 #2

    nrqed

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    Welcome to the forums.

    This is just a matter of plugging in the numbers in an equation. The equation contains a sine squared over a certain quantity squared. Do you know this formula?
     
  4. May 7, 2006 #3
    sorry i forgot to add it in the first place, yes i think it is the one is added to my question?
     
  5. May 7, 2006 #4

    nrqed

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    well, it depends what you mean by ''angle over 2''. What is the expression for your angle?
    The angle over 2 should be equal to [itex]{\pi a y \over L \lambda}[/itex]
     
  6. May 7, 2006 #5
    I worked out that the angle is 0.191 degrees, substituting into the equation i get 0.0175*I, do i then multiply by 546.1nm?, I am confused about the Io
     
  7. May 7, 2006 #6

    nrqed

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    Your equation was incorrect. It should not be sin^2(angle/2)/(angle/2)^2...(this equation is the right one if by ''angle'' you mean the *phase difference* between the two waves but that is not what you are doing here).

    I gave you a formula that would not require to calculate the angle, but if you want to, then use the formula

    [tex] I = I_0 {sin^2( {\pi a sin( \theta) \over \lambda}) \over ({\pi a sin(\theta) \over \lambda})^2 [/tex]

    where theta is the angle you just calculated.
    A word of caution: you should put everything in radians and make sure your calculator is in radians mode when you use this formula.

    Since you want the ratio I/I_0, you just need to calculate the right side of the equation and divide by I_0 so it will disappear from your result.
     
  8. May 7, 2006 #7
    I used trig to work out the angle from the source to the point 4.1mm from the maximum. I then used this angle in the equation.
     
  9. May 7, 2006 #8

    nrqed

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    That's perfectly fine. But when y is much smaller than the distance to the wall, you can also use an equation in terms of y directly. Just replace the [itex] sin(\theta)[/itex] in the equation I gave before by y/L.

    In any case, if you use the expression I gave in my previous post you should get the answer you are looking for.
     
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