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Double slit experiment spacing

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Light from a helium-neon laser (wave length 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.1 m behind the slits. Twelve bright fringes are seen, spanning a distance of 50 mm

    What is the spacing, in mm, between the slits?

    2. Relevant equations

    I believe it is:

    Ym=m(wavelength)/d

    Where L is the distance from the slits, m is an integer describing how many fringes up or down you are, d is the slit spacing and Ym is the distance to the mth bright fringe.

    3. The attempt at a solution

    Not sure if I'm interpreting the equation correctly. I plugged in 6 for m and 25x10^-3 meters for Ym and came up with the answer 0.469 mm. This is the wrong answer. My logical is that the distance from the center will give you the distance to the 6th fringe, and that would be half the total distance spanned.

    Is this the correct way to plug in numbers for that equation?

    Or I might be making some minor mistake because I don't have access to a scientific calculator (lost it today) and I'm also exhausted.
     
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 12, 2009 #2

    Delphi51

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    There is an error in the formula "Ym=m(wavelength)/d"
    It needs an L on the top.

    What is the question - what are you trying to find?
     
  4. Feb 12, 2009 #3
    oooh I forgot to give the question!

    What is the spacing, in mm, between the slits?
     
  5. Feb 12, 2009 #4

    Delphi51

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    Okay, that's the d in the formula.
    I got an answer just a wee tiny bit larger than yours . . .
    Recommend you solve the formula for d =, put in all the numbers and run them through your calculator all at once so no rounding.

    If you still have no calculator, use the one on your computer or just leave the 10^-9 out and remember your answer will be in nm.
     
  6. Feb 12, 2009 #5
    so the numbers I'm plugging in look correct?
     
  7. Feb 12, 2009 #6

    Delphi51

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    Yes, they do. I do wonder about one thing - there is a bright central maximum "spot" so the question should have said an odd number of fringes were seen. Likely that is something to ignore and your only problem is a slight inaccuracy in your calculator work. If you get the same answer again, perhaps you should write exactly what you are putting into the calculator.
    I have d = mL*lambda/x = .......
     
  8. Feb 12, 2009 #7
    Yeah I had a friend online plug my numbers into a calculator and it ended up as 4.70952x10^-4 meters which corresponds to 0.471 m. Still wrong.

    My algebra is the same as yours.
     
  9. Feb 12, 2009 #8

    Delphi51

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    I got 4.71 x 10^-4 meters.
    That's .000471 meters or 0.471 mm.
    Not 0.471 m.
     
  10. Feb 12, 2009 #9
    wow I'm tired I mistype again I'm sorry.

    The question does ask for the answer in mm, so 0.471 should be correct if everything else is right
     
  11. Feb 12, 2009 #10
    Thanks for your help so far it's my bed time so I won't be replying every seconds any more.
     
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