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Homework Help: Diffraction immersed in water

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Light of wavelength 676 nm in vacuum is incident on a single slit whose width is 1.60 × 10-5 m. The setup is then immersed in water whose index of refraction is 1.33. What is the angle θ that locates the third dark fringe with respect to the central bright fringe?


    2. Relevant equations
    1. sin[tex]\alpha[/tex] = (m +1/2)[tex]\lambda[/tex]/d
    where [tex]\alpha[/tex] is the angle of diffraction, [tex]\lambda[/tex] is the wavelength, and d is the width of the slit
    2. sin[tex]\angle[/tex]i / sin[tex]\angle[/tex]r = n
    where i is the angle of incidence, r is the angle of refraction, and n is the refractive index

    3. The attempt at a solution
    sin[tex]\alpha[/tex] = (3 + 1/2)(676)(10^-9)/(1.6)/(10^-5)
    sin[tex]\alpha[/tex] = .147875
    [tex]\alpha[/tex] = 8.5037997

    sin[tex]\alpha[/tex] = sin[tex]\angle[/tex]i ( i presume, so...)
    .147875/sin[tex]\angle[/tex]r = 1.33
    sin[tex]\angle[/tex]r = .11118421
    [tex]\angle[/tex]r = 6.383584609

    i'm not sure is theta = [tex]\angle[/tex]r
     
  2. jcsd
  3. Mar 18, 2010 #2
    i figured it out. i was using the wrong formula for diffraction. instead of m + 1/2, i should have been using m

    sin(theta) = m(lambda)/(d) instead of m+1/2
     
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