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Optic help

  1. Aug 7, 2009 #1
    1. Light from a helium-neon laser (l=650 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 2.12 m behind the slits. Twelve bright fringes are seen, spanning a distance of 49 mm. What is the spacing (in mm) between the slits?

    lamda=650x10^-9 m

    L=2.12 m

    Ym=49x10^-3

    m= 12

    d=solving for

    d=mL(lamda)/Ym= .03375 mm

    why doesnt that work?

    2. A 0.24-mm-diameter hole is illuminated by light of wavelength 476 nm. What is the width (in mm) of the central maximum on a screen 1.5 m behind the slit

    How do I do this since it has a diameter, I can't find it in the book anywhere



    3. A 578 line/mm diffraction grating is illuminated by light of wavelength 599 nm. How many bright fringes are seen on a 2.73-m-wide screen located 3 m behind the grating

    d=1/(578x 10^-3)

    lamda=599x10^-9

    Ym=2.73 m

    L=3 m

    m=Ym*d/(lamda*L)

    All these ways of solving them doesn't work why is that?
     
    Last edited: Aug 8, 2009
  2. jcsd
  3. Aug 8, 2009 #2

    Redbelly98

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    Please, please, please put different equation on separate lines in the future. It is very difficult to read the way you wrote this.

    Think about m more carefully:
    If there one bright fringe, that would be m=0.
    Two bright fringes would be m=0 and 1.
    Three bright fringes would be 0, 1, and 2.
    .
    .
    .
    Twelve bright fringes have m = 0 through ___ ?

    Retry with a different value of m

    Weird, it should be in your book or class notes. This is often referred to as a circular aperture.
    Here is more info:
    http://hyperphysics.phy-astr.gsu.edu/HBASE/phyopt/cirapp2.html#c2
    https://www.physicsforums.com/library.php?do=view_item&itemid=192

    d is incorrect here. Think about that one more carefully.

     
  4. Aug 8, 2009 #3
    Sorry about the equation problem I copied it from an email I had sent and it didn't excatly work as planned.

    1. m=0 with 12 bright fringes m=11 correct?

    2.I'll look at the sights you gave me and if I have anymore problems I will post on here, thank you

    3.I divde the d in half correct? I found an example, but I just don't understand why?
     
  5. Aug 8, 2009 #4

    Redbelly98

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    Okay. You can edit your post soon after posting it, so you can always check after you post and fix stuff like that ... next time.

    Yes.

    Okay.

    Uh, not quite.

    578 lines/mm, so d is (1/578) mm = ____ m?
     
  6. Aug 8, 2009 #5
    I did edit it, thank you for telling me I could.

    1. I got it now I just rushed things

    2. y=L*m*lamda/d

    where d is the diameter, L is the distance behind, lamda is the wavelength (obviously), m is bright fringes. How do I discover how many bright fringes there are there so I can do this I assumed one, but this is wrong. Oh also Y is the width because it is the displacement on the Y axis.

    3. 1.73 m?
     
  7. Aug 8, 2009 #6

    Redbelly98

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    No. Since a mm is 10-3 m,

    (1/578) mm = (1/578) x 10-3 m​

    Notice the answer is a lot smaller than a meter, because both (1/578) and 10-3 are small numbers.
     
  8. Aug 8, 2009 #7
    Oh what I was doing was 1/(578x10^-3)
     
  9. Aug 8, 2009 #8
    2. Can you offer so more advice please

    3. so doing it the correct way with
    d = 1/578 x 10^-3 m
    Ym=2.12 m
    lamda= 599x 10^-9 m
    L= 3 m

    all units cancel leaving basically bright fringes, plug and chug using the equation m=(Ym*d)/(lamda*L)
     
  10. Aug 8, 2009 #9

    Redbelly98

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    Correct except for the following:

    1. y is the radius of the diffraction pattern on the screen.
    2. m is a number you can get from the "m values" table at

    A hint: the diameter of the central bright fringe is really the diameter of the 1st minimum in the diffraction pattern.
     
  11. Aug 8, 2009 #10

    Redbelly98

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    I agree with the values for d, lambda, and L. But where does Ym=2.12 m come from?
     
  12. Aug 8, 2009 #11

    rl.bhat

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    Is it not true that in the double slit experiment, when m = 1, there will be two fringes on either side of the central bright fringe? Is it possible to get even number of fringes in the double slit experiment?
     
  13. Aug 8, 2009 #12

    Redbelly98

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    If the screen is off-center, there can be an even number of fringes appearing on it.
     
  14. Aug 8, 2009 #13

    rl.bhat

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    In such experiment usually the screen is placed perpendicular to the axis of the slits. In such case what is the position of the screen which is off -center? In this position is there no central maximum?
     
  15. Aug 9, 2009 #14

    Redbelly98

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    (Problem #1)

    The central maximum will be the bright fringe directly in front of the slits, no matter what the screen position is.

    I am (for now) going on the assumption that the problem was correctly copied when posted here.

    For the small angles being considered, it does not matter whether the screen contains the m=0 through m=11 fringes, or the m=-5 through m=6 fringes. As long as 12 fringes close to the central maximum are visible, the small-angle approximation holds and the answer will be the same.
     
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