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A 3rd method of finding wavelength in a double slit

  1. Mar 2, 2017 #1
    1. The problem statement, all variables and given/known data

    Upon using Thomas young’s double slit experiment to obtain measurements, the following data were obtain. Use these data to determine the wavelength of light being used to create the interference pattern. Do this in three different ways (6)
    • the angle to the eighth maximum is 1.12◦
    • the distance from the slit to the screen is 302cm
    • the distance from the first minimum to the fifth minimum is 2.95cm
    • the distance between the slit is 0.00025cm


    2. Relevant equations


    3. The attempt at a solution
    I found 2 ways of finding the wavelength using the formulas

    Change in x = L (Lambda) / d

    and

    m(Lambda)= (d)(sin Theta subscript m)

    For my third method i thought about using the formula

    m(Lambda)= (d)xm/ L

    the problem is I was not given xm sincexm is the distance from a point on a antinodal line to the right bisector.

    However, I was told by my friend that since the 8th antinodal is provided and that "change in x" is easily obtainable, xmcan be found by just multiplying the 2 together.

    which would look like

    change in x = 0.007375 (Converted to Meters from cm)
    8th antinodal = 8

    0.007375(8) = 0.059

    therefore
    xm= 0.059 meters

    Is my friend right?
     

    Attached Files:

    Last edited: Mar 2, 2017
  2. jcsd
  3. Mar 2, 2017 #2

    mfb

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    Staff: Mentor

    Please define the variables you use. L, ##\lambda## and d are clear, what is everything else?

    What is 0.059 meters? Certainly not the wavelength.

    The third method has to be very similar to one of the other two methods, you don't have enough given values for three completely independent methods. Distance between slits and distance to the screen alone doesn't tell you anything about the wavelength, the given angle leads to one method, the given separation between minima leads to the second method.
     
  4. Mar 2, 2017 #3

    Sorry, I was in a rush and I thought I attached my work when I posted. My work should be in the attachments now.
     
  5. Mar 2, 2017 #4

    mfb

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    Staff: Mentor

    Method 2 and 3 are nearly identical, they just differ by a factor 8 in both numerator and denominator.
     
  6. Mar 2, 2017 #5
    should I use a different one ?
     
  7. Mar 2, 2017 #6

    mfb

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    Staff: Mentor

    See post #2: No matter what you do, two methods will look very similar. I guess the shown "third method" is as good as it gets.
     
  8. Mar 2, 2017 #7
    Ahh ok.

    Thank you so much for reviewing my work! I really appreciate it.
     
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