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Double slit experiment with a slit covered.

  1. Dec 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A thin flake of mica (n=1.58) is used to cover one slit of a double-slit arrangement.The central point on the screen is occupied by the 7th brigth fringe.If lamda = 550nm, what is the thickness of the mica?

    2. Relevant equations
    path difference d=(n-1)*thickness
    d=7*lamda
    snell's law?

    3. The attempt at a solution
    first, the paths above the glass must be the same.So they are not in consideration.
    let l be the thickness of the glass,t be the length travelled by light in mica,x in air
    The mica:
    sinθ=nsin
    sin∅=sinθ/n....(1)
    sin∅=l/t..(2)

    (1)/(2),
    sinθ=nl/t...(3)

    In the air:
    sinθ=l/x...(4)
    compare 4 and 3,
    i came out, t=xn.

    Then i started to consider the difference of wavelength,
    because the wavelength in mica is shorter than in air.
    I tried to subtract the sum of the pulses in air and mica and set it equals to 7 lamda,
    but what came out is peculiar,
    let m , n be the number of pulses in mica and air respectively,f=frequency
    m=tf/v...(5)
    n=xf/c..(6)
    so....
    m/n=nt/x
    i suppose it being m/n = t/nx ......so that they can be cancelled out......
    Please help for this:((
     
  2. jcsd
  3. Dec 12, 2015 #2

    nrqed

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    I have the feeling that they do not expect you to take into consideration refraction through the mica (how could you, you do not have an angle of incidence). I think you can safely neglect that and just consider an incident angle of zero in which case what you wrote in part 2 is all you need.
     
  4. Dec 13, 2015 #3
    I don't quite like the way that this question is worded.
    I think that it means that what was the 7th bright fringe becomes shifted into the central bright maximum.
    Overall, I don't think that the appearance of the screen pattern would change.
    Then the problem becomes to calculate the thickness of the mica in order for the
    mica to contain 7 more wavelengths than would be contained in an equivalent thickness of air.
    This should not be too difficult since you know that the wavelength in the mica = 5.5 * 10E-7 / 1.58.
    as compared to 5.5 * 10E-7 in air.
     
  5. Dec 14, 2015 #4
    thank you all :)) i have solved the problem :D
     
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