Light of wavelength 680 nm falls on two slits and produces an interference pattern in which the fourth
  1. A Michelson interferometer is set up using white light. The arms are adjusted so that a bright white spot appears on the screen (constructive interference for all wavelengths). A slab of glass (n = 1.46 ) is inserted into one of the arms. To return to the white spot, the mirror in the other arm is moved 6.73 cm. (a) Is the mirror moved in or out? (b) What is the thickness of the slab of glass?
    Solution.
    (a) Since the wavelength of light is shorter in glass than in vacuum, it will take a larger number of wavelengths to pass through the glass. To increase the number of wavelengths in the other arm, the mirror must be move out.
    (b) Let T be the thickness of the glass. If the vacuum wavelength of one of the components of the white light is λ0, then the wavelength in glass is λ = λ0/n. The number of wavelengths traveled in the glass is
               2T/λ = 2T/(λ0/n) = 2Tn/λ0
         Let d be the distance the mirror moved. The number of extra wavelengths traveled in leg 2 is 2d/λ0
         The number of wavelength in both cases must be equal
                2Tn/λ0 = 2d/λ0
                T = d/n = 6.73 cm/1.46 = 4.61 cm.
  2. Light of wavelength 680 nm falls on two slits and produces an interference pattern in which the fourth-order fringe is 48 mm from the central fringe on a screen 1.5 m away. What is the separation of the two slits?
    Solution. for constructive interference, the path difference is a multiple of the wavelength:
       dsin θ = mλm = 0, 1, 2, . . .
    We find the location on the screen form
       y = Ltan θ
    For small angles, we have  sin θ = tan θ, which igves
      y = L(mλ/d) = mLλ/d
    For the fourth order we have
      48 x 10-3 m = (1.5 m)(680 x 10-9 m)(4)/d, which gives d = 8.5 x 10-5 m = 0.085 mm.
  3. How wide is the central diffraction peak on a screen 2.50 m behind a 0.0348-mm-wide slit illuminated by 589-nm light?
    Solution.
    We find the angle to the first minimum from 
        sin θ1min = mD/λ= (1)(589 x.10-9 m)/(0.0348 x 10-3 m) = 0.0169 m, so θ1min = 0.970.
    We find the distance on the screen from
          y1 = Ltan θ1 = (2.50 m)tan0.970 = 4.23 x10-2 m = 4.23 cm
    Thus the width of the peak is
        Δy = 2y1 = 2(4.23 cm) = 8.46 cm.
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