INTRO:
A double-slit experiment is performed with light of wavelength 630nm. The bright interference fringes are spaced 1.6mm apart on the viewing screen.
Part A:
What will the fringe spacing be if the light is changed to a wavelength of 420nm ?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
givens:
λA= 630 nm =
ΔyA = 1.6 mm
λB= 420 nm
using the formula Δy = λL⋅1/d we can determine the ratio of the slits' distance from the screen and their distance from each other, or L/d, which would be constant for both wavelengths
using the first wavelength,
L/d = Δy/λ = ΔyA/λA = 1.6 mm / 630 nm = 2.54⋅10-3 mm/nm
now we can use this to determine the fringe spacing for the second wavelength,
ΔyB = λB⋅L/d = (420 nm)⋅(2.54⋅10-3 mm/nm)
→ΔyB = 1.067 mm ≈ 1.1 mm
The answer sheet at the back of the textbook for this question says 1.2 nm. You got 1.1 mm and I also got 1.2 mm. Unit error?
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