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Young's Double Split Experiment

  1. Jan 16, 2007 #1
    1. In a Young's double-slit experiment, the seventh dark fringe is located 0.029 m to the side of the central bright finge on a flat screen, which is 1.1 m away from the slits. The separation between the slits is 1.5 x 10^-4 m. What is the wavelength of the light being used?



    2. wavelength = d sin(theta)
    Tan theta = y/x
    y= 0.029m
    x= 1.1m
    d = 1.5 x 10^-4 m




    3. Tan Theta = y/x = 0.029/1.1= 0.0264
    Theta = 1.51

    Wavelength = d sin (theta) = (1.5 x 10^-4) x 0.0264
    Wavelength = 3.95 x 10^-6


    This is not correct and I can't seem to place why.

    Any thoughts?
     
  2. jcsd
  3. Jan 16, 2007 #2
    ok, i see I forgot m, which equals 7.

    entering this into the equation:

    Sin Theta= 0.0264 = 7 x (Wavelength/(1.5 x 10^-4))

    working this through gave me an answer of 2.772 x 10^-5 which was also incorrect...
     
  4. Jan 16, 2007 #3
    i see i forgot +1/2 since it is a dark fringe.


    working now...
     
  5. Jan 16, 2007 #4
    Using 7.5 instead of 7 gave me an answer of 5.28 x 10^-7 which was still incorrect...


    I think i am now completely stuck
     
  6. Jan 16, 2007 #5

    hage567

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    Homework Helper

    I think I know what is wrong: you are using m=7.5. The first dark fringe does not occur at the centre, it is 1/2 shifted as the conditions for destructive interference specify. Your first dark fringe (m=0) is 1/2 shifted from the centre maximum, so if you count to the 7th dark fringe it isn't m=7.5. Try drawing it out. I hope this isn't confusing you but I think that might be the problem. Otherwise everything else looks OK.
     
  7. Jan 16, 2007 #6
    thank you very much hage. it was 6.5 and I now have it correct.


    do you think you could figure what I'm doing wrong on this problem?:



    A rock concert is being held in an open field. Two loudspeakers are separated by 6.00 m. As an aid in arranging the seating, a test is conducted in which both speakers vibrate in phase and produce an 80.0 Hz bass tone simultaneously. The speed of sound is 343 m/s. A reference line is marked out in front of the speakers, perpendicular to the midpoint of the line between the speakers. Relative to either side of this reference line, what is the smallest angle that locates the places where destructive interference occurs? People seated in these places would have trouble hearing the 80.0 Hz bass tone.


    d= 6.0
    v= 343 m/s
    f= 80.0Hz
    wavelength = 343/80 = 4.2875

    0.5 (4.2875) = 2.14375 = destructive interference

    I don't know how to figure out the angle without the distance "l".


    Any thoughts?
     
  8. Jan 17, 2007 #7

    hage567

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    Homework Helper

    Well, you know what the path difference has to be for destructive interference, and you have the info to calculate it. I would take that value and set it equal to the equation for dark fringes and solve for the angle. See if that helps.
     
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