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Interference fringe width

  1. Dec 22, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-23_9-50-55.png

    2. Relevant equations


    3. The attempt at a solution
    upload_2017-12-23_9-35-1.png
    When the system is in air

    ## d \sin {\theta} = n \lambda ## .....(1)

    When the system is in water and slit separation is 2d,

    ## \mu 2d \sin {\theta} = n \lambda ## .....(2)

    ## \tan {\theta} = \frac y D ## .....( 3)

    Taking d<<D, ## \tan {\theta} \approx \sin {\theta} ## .....(4)

    ## \frac { n \lambda }{ 2 \mu d} = \frac { y} D ## .....(5)

    ## y = \frac { n \lambda D }{ 2 \mu d} ## .....( 6)

    Fringe width ## = \frac { \lambda D }{ 2 \mu d} ## .....(7)

    Now, ## \frac { \lambda D}{ d} =s ## .....( 8)

    Fringe width ## = \frac { s }{ 2 \mu } = \frac { 3s} 8 ##


    So, the correct option is (e).

    Is this correct?
     
  2. jcsd
  3. Dec 22, 2017 #2

    ehild

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    Correct. Nice work!
     
  4. Dec 22, 2017 #3
    Is there any easier way to solve it?
    Or one has to go through all of those steps for solving the above question.
     
  5. Dec 23, 2017 #4

    ehild

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    Was not it easy? You presented a nice solution, every step explained and clear.
     
  6. Dec 23, 2017 #5
    It was a multiple choice question. So, I was looking for a shorter way.
     
  7. Dec 23, 2017 #6

    ehild

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    Most steps were explained during the class to you, so it was not needed to write down. You have the formula for the distance between nearest interference fringes, and it is proportional to lambda and inversely proportional to d. s' / s = λ'/λ d/d' The wavelength in a medium is the vacuum wavelength divided by the refractive index. So s'/s=(3/4) x (1/2 ).
     
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