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Diffraction from a single slit.

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    You need to use your cell phone which broadcasts an 830 MHz signal but you're in an alley between two massive radio wave absorbing building that have only a 15m space between them. What is the angular width in degrees of the electromagnetic wave after it emerges from between the buildings.


    2. Relevant equations
    a sinO = (pd)/a
    a = slit width - 15m
    sinO = sine of theta, couldn't figure out how to do theta - unknown
    p = the number for the central maximum or minimum - 1?
    d = wavelength (didn't know how to do lambda either) - .361m

    3. The attempt at a solution
    There's a basic element I'm missing here. I determined wavelength by dividing the speed of light by the frequency. So I then divide wavelength by the slit width of 15m and get an even smaller number that leads to an extremely small angle after I take the arcsine that seems unreasonable given the dimensions involved in this problem.

    The book comes up with an answer of 2.9 degrees. There's got to be something basic to this problem I'm missing, but I can't figure out what. Can somebody give me a nudge in the correct direction?
     
  2. jcsd
  3. Feb 3, 2010 #2
    you are given the frequency .. cant you get the value of the wavelength?
     
  4. Feb 3, 2010 #3
    I already have the wavelength but thats not what the problem is asking for. Its asking for the angular width of the wavelength. I'm getting an angle with four decimal places which isn't anywhere near what the book gets or even what seems right.
     
  5. Feb 3, 2010 #4
    what is your final answer for this question?
     
  6. Feb 3, 2010 #5
    Sorry - I s'pose I shoulda given that.

    I divided .361/15 and took the arcsine of that. I'm getting 1.37 degrees. Looks like my previous answer (that was in thousandths of degrees) was because my calculator was still in rads. Still though, I'm really off by at least half.
     
  7. Feb 3, 2010 #6

    rl.bhat

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    Homework Helper

    In the formula θ is half the angle. And the angular width of the central maximum is 2θ.
     
  8. Feb 3, 2010 #7
    I have a question for you .. is the first equation you provided right? please check it once again ..
     
  9. Feb 3, 2010 #8
    I didn't recall the part about being half the angle. I'll consult the book again. I might have more questions though.
     
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