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Diffraction immersed in water

  • Thread starter whiteman
  • Start date
  • #1
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Homework Statement


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?


Homework 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

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
 

Answers and Replies

  • #2
9
0
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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