Double Slits - Measuring Wavelength from the number of bright fringes

  • Thread starter Glorzifen
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  • #1
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Homework Statement


A double-slit interference pattern is observed on a screen 1.0m behind two slits placed on 0.3mm apart. Ten bright fringes span a distance of 1.65cm. What is the wavelength of the light?

Homework Equations


Since we're looking at bright fringes:
(m+1/2)[tex]\lambda[/tex] = dsin[tex]\theta[/tex] = d(x/L)

The Attempt at a Solution


d = 3E-4m
m = 4 (I feel like I'm supposed to use just one half of the bands, excluding the middle band. Is this right? Why do I do this?)
L = 1.0m
y = 8.25E-3m (this is the spread of the 10 fringes divided by 2 since I'm using half of them)

(m+1/2)[tex]\lambda[/tex] = d(x/L)
(4+1/2)[tex]\lambda[/tex] = 3E-4(8.25E-3/1.0)
[tex]\lambda[/tex] = 5.5E-7m

On my formula sheet there is an 'x'...the variable for the spread of the fringes is traditionally 'y'...I figured they were just the same thing so used y for x in the above example. If that's wrong then could someone explain where I get 'x' from? Any explanation would be appreciated.

Thanks!
 

Answers and Replies

  • #2
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Just realized that (m+1/2) is for DARK fringes. Would that be correct for dark fringes at least?
 
  • #3
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If you use m=4 that is 4+4+1 which only gives you 9 fringes. The question doesn't say these are all the possible fringes, maybe only a part of all fringes. What I would do is use m= 9 (that is 10 because we include m=0) and use distance y as 1.65cm. And yes you are supposed to use (m)LaTeX Code: \\lambda
 

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