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Intensity at a point on the screen

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ = 0∘) is 3.00×10−5 W/m2 .
    What is the intensity at a point on the screen that corresponds to θ = 1.20∘?
    Express your answer with the appropriate units.

    2. Relevant equations
    $$I = I_0[\sin(\pi*a*\sin(\theta)/\lambda)/(\pi*a*\sin(\theta)/\lambda)]^2$$

    3. The attempt at a solution
    I used the formula and values and got an answer of 1.13*10^-9 W/m^2 but it says it's incorrect and i'm not sure why.
     
    Last edited: Nov 23, 2016
  2. jcsd
  3. Nov 23, 2016 #2

    haruspex

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    In these forums you need to double the dollar (or hash) signs:
    ##I = I_0[\sin(\pi*a*\sin(\theta)/\lambda)/(\pi*a*\sin(\theta)/\lambda)]^2##
     
  4. Nov 23, 2016 #3
    Thanks. Changed it
     
  5. Nov 23, 2016 #4

    haruspex

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    I get a number an order of magnitude greater. Any possible confusion between degrees and radians?
     
  6. Nov 23, 2016 #5
    Maybe but then do I need to change the 1.2 into radians or i don't know
     
    Last edited: Nov 23, 2016
  7. Nov 23, 2016 #6

    haruspex

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    I assume you are using a calculator. Most or all calculators allow you to enter the angle in either degrees or radians, but you have to select the right option. Whichever way you do the inner sine (sin(θ) in the quoted equation), the result of (πa sin(θ)/λ) is in radians.
     
  8. Nov 23, 2016 #7
    Ok, I tried it again and I'm now getting 3.47*10^-10
     
  9. Nov 23, 2016 #8

    haruspex

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    Then you need to post your full working.
    I get sin(1.2o)=0.021, πa/λ=154, multiplying gives 3.22, sine of that is -0.08, dividing by 3.22 gives -0.026, squaring to 0.00067, then multipying by I0 ends with 2 10-8W/m2.
     
  10. Nov 24, 2016 #9
    Thanks, that makes sense now. Who knows what I was doing in my calculations.
     
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