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Distance shifted by the centre of pattern

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data

    In a Young's double slit experiment , the two slits are separated 0.5mm apart, Monochromatic light of wavelgnth 6.0x10^-7 m is used to illuminate the slits. When a piece of glass of thickness 0.024mm and refractive index of 1.5 used to cover one of the slits the whole intefrence pattern shifted in one direction. Find the distance shifted by the centre of pattern .

    2. Relevant equations



    3. The attempt at a solution
    OPD= (n-1)t = m λ
    (1.5-1)(0.024x10^-3) = m (6x10^-7)
    m = 20
    i am going to find the distance between central max and first min which i have x = λD/ a
    a= separation of slit
    but i dont have D (distance of the slit from the screen ) ....
    how to proceed with this?
    is the question wrong ? or ????
     
  2. jcsd
  3. Sep 13, 2014 #2

    Simon Bridge

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    You are not even told that there is a screen.
    What you are expected to do is locate the central maximum.
    Is there something other than position on a screen that gets used?
    Do you have some notes or coursework about what happens when a bit of glass is covering one slit?
     
  4. Sep 15, 2014 #3
    here's the notes
     

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  5. Sep 15, 2014 #4
    here's the notes:
     

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  6. Sep 15, 2014 #5

    BvU

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    Hi Des, and tell us: the second set of notes shows a worked example where you don't need the distance to the screen (because it's about two theta's and "the D divides out", so to say). Right ?

    I agree with Simon that you need D, so if it isn't given, the answer will be something like: shift = ... * D

    Why do you want to "to find the distance between central max and first min" when something quiet different is asked for?
     
  7. Sep 15, 2014 #6

    NascentOxygen

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    Perhaps the answer they are after is something like, "The central maximim moves to its left/right* to where was originally the 20th bright fringe."

    *You choose.
     
  8. Sep 16, 2014 #7

    BvU

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    :smile:
     
  9. Sep 16, 2014 #8
    so that i can find the distance shifted by the centre of pattern
    i am going to fing the distance between the central maximum and first min first, then i times it with 20 , to get the distance shifted by the centre of pattern . Am i wrong?
     
  10. Sep 16, 2014 #9

    BvU

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    There is no argument at all for 20 times the distance to the first minimum !

    The path straight ahead through the glass has become 20 wavelengths longer. For the central maximum (zero lambda difference), this now has to be compensated by some angle. The same angle that previously made the other path 20 wavelengths longer .... that angle pointed in the direction of the 20th ...
     
  11. Sep 16, 2014 #10
    i am really confused now. can you give me some guide what should i do in order to get the ans ? perhaps i can understand it much better
     
  12. Sep 16, 2014 #11

    NascentOxygen

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    I believe English is not your first language?

    In short: it is not possible to express the shift in cm because there is insufficient information.
     
  13. Sep 16, 2014 #12

    BvU

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    Well, you posted some notes about a single slit and an example exercise. Not the exercise we are spending time on.

    So far you've calculated that the optical path for the light that goes straight ahead is lengthened by 20 wavelengths if the glass plate is inserted.

    Perhaps it's time you make your own clear drawing. And post it, to help us help you.

    Then:
    Is it clear to you that if you want a distance shift in units of length, you need something to use as a scale ? Either the distance to the screen, the distance from the axis to the nth maximum (or minimum) in the pattern without the glass plate, or whatever. The alternative is an answer where you give an angle.

    And:
    I am rather convinced that "Find the distance shifted by the centre of pattern . " is not the exact wording of the exercise. What does it say, exactly ?
     
  14. Sep 16, 2014 #13
    here's my drwaing the original question
     

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  15. Sep 16, 2014 #14

    BvU

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    Good work: the second picture makes clear that the exercise isn't top quality didactics: we indeed have to "invent" a distance scale. And your rendering of the text is completely correct. It's the text itself that is cruddy (and I am not a native english speaking lingo expert).

    The first picture can now be completed (it doesn't show any light rays yet, does it!):
    Draw a ray from each slit to the top centre and mark the difference in geometrical length of the paths (the optical path difference is zero, remember?).
     
  16. Sep 16, 2014 #15
    how should I proceed from here?
     

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  17. Sep 16, 2014 #16

    BvU

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    difference in geometrical length of the paths is the length of the little piece from lower slit to the perpendicular. Has to be how long to make the two optical path lengths equal ?

    Same picture but without the glass would point to which order maximum or minimum ?

    1+1 = 2:
    Central maximum shifts to ... on the same/opposite side of the slit with the glass in front.
    And this is at an angle ... wrt the axis
     
  18. Sep 16, 2014 #17
    i am going to find the distance between the central maximum and first min first, then i times it with 20 , to get the distance shifted by the centre of pattern . Am i wrong?
    i have found out the central maximum of new fringe after the glass is added will move to 20th order of the fringe from the old central maximum (before the glass is added)...
    please help... i am confused now..
     
  19. Sep 16, 2014 #18

    NascentOxygen

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    Your method is right. But this question doesn't supply enough information for you to be able to determine a distance in cm. You have to leave your answer as "a distance of 20 fringes", or express as the equivalent angle deviation. You can't do cm here because you aren't told how far away is the screen.

    The answer consists of two parts: the number of fringes, and a direction. Draw these on a diagram to show your understanding.
     
  20. Sep 17, 2014 #19
    Thanks!
     
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