22: Problem 22.6

INTRO:
Light from a sodium lamp (λ=589nm)illuminates two narrow slits. The fringe spacing on a screen 110 cm behind the slits is 3.6mm.

Part A:
What is the spacing (in mm) between the two slits?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens: 
λ = 589 nm = 5.89⋅10^-7 m
L = 110 cm = 1.1 m
Δy = 3.6 mm = 3.6⋅10^-3 m

we can rearrange the formula Δy = λ⋅L/d to give us the distance between the slits, d
d=λ⋅L/Δy → d = (5.89⋅10^-7 m)⋅(1.1 m/3.6⋅10^-3 m) = 1.8⋅10^-4 m = 0.18 mm

22: Problem 22.4

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: 
λ_A= 630 nm =
Δy_A = 1.6 mm
λ_B= 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 = Δy_A/λ_A = 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,
Δy_B = λ_B⋅L/d = (420 nm)⋅(2.54⋅10^-3 mm/nm) 
→Δy_B = 1.067 mm ≈ 1.1 mm