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Optics qn

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data
    fringes are observed when a parrallel beam of light of wavelength 500nm is incident perpendiculary onto a wedge-shaped film with an index of refraction of 1.5. What is the angle of the wedge if the fringe separation is 1/3cm?


    2. Relevant equations
    t(x)=xtan(theta)
    2nt=m*lamda (m+.5) for deconstructive
    x=(lamda/2)/2n*tan(theta)
    t=lamda/2n
    3. The attempt at a solution
    ok im really struggling to understand this qn, i thought simple trig, t=1.66x10^-9
    but now completely stuck... coz i think i need the x value..... but no idea how to find it??? can anyone help? pic-1.jpg
     
  2. jcsd
  3. Mar 26, 2009 #2
    anyone?
     
  4. Mar 26, 2009 #3
    anyone know how?
     
  5. Mar 26, 2009 #4

    mgb_phys

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    Sorry - if you reply to your own question is shows as answered so people ignore it!

    With[tex]\theta[/tex] in radians t = x [tex]\theta[/tex]
    As you say you need 2t to be an odd number of half wavelenghts
    t = n(m+0.5) [tex]\lambda[/tex]

    Now you have two values of x for m and m+1
     
  6. Mar 26, 2009 #5
    ok thankyou
     
  7. Mar 26, 2009 #6
    hi again... srry, ive been looking at wat u said, andi dont understand how t=x(theta) isnt it xtan(theta)?? and from then i thought tan(theta)=t/x?? but it doesnt work because i have 2 unknowns??? theta and x. t=lamda/2n... plz can u tell me the process to solve this problem?
     
  8. Mar 26, 2009 #7
    but then where does this d=1/3cm come from???
     
  9. Mar 26, 2009 #8

    mgb_phys

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    For small angles, angle = sin angle = tan angle (assuming radians)
    d (in your diagram) doens't equal 1/3cm, that is the spacing of the fringes along the wedge - essentially it's the extra length of wedge needed to get one wavelenght of extra 't'
     
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