Young Double Slit Experiment (Determine wavelength of light source)

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


Upon using Thomas Young's double slit experiment to obtain measurements, the following data was obtained. Use this data to determine the wavelength of light being used to create the interference pattern. DO this in 3 different ways!
  • The Angle to the Eighth maximum is 1.12 deg.
  • The distance from the slits to the screen is 3.02 m
  • The distance from the first minimum to the fifth minimum is 2.95 cm
  • The distance between the slits is 0.00025 m


Homework Equations


sin∅n=(n-1/2)λ/d
sin∅m=mλ/d
Δx=Lλ/d
Xn/L=(n-1/2)λ/d
Xm/L=mλ/d

The Attempt at a Solution


sin∅n=(n-1/2)λ/d
λ=(sin1.12)(0.00025)/(7.5)
λ=651 nm


The distance from the first minimum to the fifth minimum is 2.95 cm .. therefore
4Δx=2.95 cm
Δx=0.0074 m ( i converted it)

Im unsure about 4 x if anyone can shed any light on that that would be cool


Δx=Lλ/d
λ=Δxd/L
λ=(0.0074) (0.00025)/ (3.02)
λ= 612 nm

as for this one I dont know if I should use the Xm or Xn equation..I used

Xm/L=mλ/d

(0.0295)/3.02=4λ/(0.00025)
λ= 610 nm

Number seems plausible but it also does when I use the Xn equation

thanks for any help in advance
 

Answers and Replies

  • #2
1,506
18
Using Sinθ = 8λ/d for the 8th max I got λ = 6.1 x 10^-7m
Using 4x for the separation of 5 minima then using x = λL/d I got λ = 6.1 x 10^-7m
 
  • #3
56
0
Using Sinθ = 8λ/d for the 8th max I got λ = 6.1 x 10^-7m
Using 4x for the separation of 5 minima then using x = λL/d I got λ = 6.1 x 10^-7m

since its a max shouldn't you use sin∅=(n-1/2)λ/d?
 
  • #4
1,506
18
For a max the path difference from the slits must be a whole number of wavelengths.
For the 8th max the path diff = 8 wavelengths.
For minima the path diff must be an odd number of half wavelengths (n+1/2) but the SEPARATION of max and min is given by increases in path diff of whole numbers of wavelengths.
Hope that sounds OK
 

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