Young Double Slit Experiment (Determine wavelength of light source)by HarleyM Tags: diffraction, double slit, light wave, young experiment 

#1
Dec3011, 12:19 PM

P: 56

1. The problem statement, all variables and given/known data
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!
2. Relevant equations sin∅_{n}=(n1/2)λ/d sin∅_{m}=mλ/d Δx=Lλ/d X_{n}/L=(n1/2)λ/d X_{m}/L=mλ/d 3. The attempt at a solution sin∅_{n}=(n1/2)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 X_{m} or X_{n} equation..I used X_{m}/L=mλ/d (0.0295)/3.02=4λ/(0.00025) λ= 610 nm Number seems plausible but it also does when I use the X_{n} equation thanks for any help in advance 



#2
Dec3011, 12:57 PM

P: 1,506

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
Dec3111, 09:39 AM

P: 56





#4
Dec3111, 10:27 AM

P: 1,506

Young Double Slit Experiment (Determine wavelength of light source)
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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