Light of wavelength 680 nm falls on two slits and produces an
interference pattern in which the fourth

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 λ_{0}, then the
wavelength in glass is λ = λ_{0}/n. The
number of wavelengths traveled in the glass is
2T/λ = 2T/(λ_{0}/n) = 2Tn/λ_{0}
Let d be the distance the mirror moved. The number of extra
wavelengths traveled in leg 2 is 2d/λ_{0}
The number of wavelength in both cases must be equal
2Tn/λ_{0} = 2d/λ_{0} T = d/n = 6.73 cm/1.46 = 4.61
cm.

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 = mλ,
m = 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.

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 θ_{1min} =
mD/λ= (1)(589 x.10^{-9} m)/(0.0348 x 10^{-3} m)
= 0.0169 m, so θ_{1min} = 0.970°. We find the distance
on the screen from y_{1} = Ltan
θ_{1 }= (2.50 m)tan0.970° = 4.23 x10^{-2} m = 4.23
cm Thus the width of the peak is Δy =
2y_{1}= 2(4.23 cm) = 8.46
cm.