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Diffraction grating problem

  1. May 4, 2005 #1
    Hello everyone,

    My physics teacher assigned a whole bunch of homework that no one remembers how to do because we haven't taken light for a year and a half. I have gone through all of my old stuff but can't find the right unit, and what I did find isn't in enough detail. I am getting really frustrated with this so hopefully some of you guys can help me out.

    1) In a double slit experiment a student used a diffraction grating in which the slit separation was 1.00x10^-4m. A screen was placed at a distance of 4.00x10^-1m from the grating. The student measured the distance from the central maximum to the 2nd order minimum and found it to be 3.00 x 10^-3m. What was the frequency of the light used by the student?

    Answer ---> 6.00 x 10^14 Hz

    2) Calculate the frequency of monochromatic light used in a diffraction experiment if the 1st order maximimum is observed at an angle of deviation of 28 degrees when the slits of the diffraction grating are 1.00x10^-6m apart.

    6.39 x 10^14 Hz

    3) Light with a frequency of 4.28 x 10^14 Hz is diffracted by a diffraction grating ruled 5.00 x 10^5 lines/m. On a screen 2.00m from the grating maxima are 7.00 x 10^-1m apart. What is the distance between maxima if the frequency of the light was increased 1.50 times while all other variable remain constant?

    Answer ---> 4.66 x 10^-1 m
     
  2. jcsd
  3. May 4, 2005 #2
    Ok I finally figured out the last two on my own, but I am still having a lot of trouble with the first one. I think I need to find the wavelength for it and then use the wavelength and the speed of light to find the frequency but nothing will work out for me.
     
  4. May 4, 2005 #3
    Your intuition for part 1 was correct.

    First find the wavelength of light using the double slit equation.

    [tex]\lambda=\frac{dy}{mL}[/tex]

    where d is the separation between the slits
    y is the distance from the centre line
    m is the order number
    L is the distance between the grating and the screen

    then continue on with your thought process and find the frequency using:

    [tex]f=\frac{c}{\lambda}[/tex]
     
  5. May 4, 2005 #4

    Ouabache

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    Bross7's equations are correct. You may keep units in meters.
    I also tried your data and did not arrive at the solution you gave.
    (It did come out to the same magnitude though; ie 10^14 Hz).

    You may want to double check and see if your given data is correct.
    If so, then the expected solution you gave is very likely, incorrect.
     
  6. May 6, 2005 #5

    xanthym

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    The textbook's answer is CORRECT. Remember that the "2nd Order Minimum" occurs at m=(1.5). Using that value will result in the given answer.


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