Two Slit Interference Question =] Thanks

AI Thread Summary
The discussion revolves around a two-slit interference problem involving monochromatic light with a wavelength of 656 nm. The first intensity minimum (y1) is calculated to be 4.4608 mm above the central maximum, while the third intensity minimum (y3) is found at 22.3045 mm. The challenge arises in determining the wavelength of a second monochromatic beam, which has its second maximum coinciding with the third minimum of the first beam. Attempts to equate the equations for the two beams have led to incorrect results, indicating a misunderstanding in the application of the interference formulas. The participant is seeking clarification and assistance to resolve the calculation for the second beam's wavelength.
chipmunk951
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Homework Statement


Monochromatic light of wavelength 656 nm is normally incident on a vertical plate with two slits at positions +-d/2 where d=.25mm, the resultant interference pattern is observed on a screen at distance D= 3.4m away, let y denote vertical position on the screen with y=0 at the same height as the midpoint between the two slits. Picture attached
http://img66.imageshack.us/img66/5378/showmeplmx4.gif

a) What is the location y1 of the first intensity minimum on the screen, measured up from the central maximum?

b) what is the location y3 of the third intensity minimum on the screen?

c) A Second monochromatic beam is normally incident on the same slits. The second maximum pof the second beam occurs at y3, the third minimum of hte first beam. What is the wavelength of the second beam?

Homework Equations


y=L tan theta
d*sin theta = m lambda
d*sin theta = (m-1/2) lambda

The Attempt at a Solution



Part a) It's a dark fringe so dsin theta = (m-1/2) lambda and m=1 because it's the first one, and solve for theta, use y= L tan theta, take the difference which is y-0 and you get y1, whihc in my case is 4.4608 mm

b) Same thing except m is now 3, ends up being 22.3045 mm

c) here's where I got stuck, It said the second max of the second beam occurs at y3, the third minimum of the first beam, here's what I thought I Could do,

d*sin theta = (3-1/2)lambda1 This is for the y3, third minimum
d*sin theta = 2lambda2 This is for the second beam, set them equal to each other
(3-1/2)lambda1 = 2 lambda2
And solve for lambda2, but this didn't work, i got a value of 8.2e-7 and it's not correct :(, I also tried going backwards with my location y3, solving for the angle and doing (d*sin theta)/2 but i go the same answer.
 
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Second maximum is for m=1 (first maximum is in the center, at m=0)
 
nasu said:
Second maximum is for m=1 (first maximum is in the center, at m=0)

So it will be like

(3-1/2)lambda1 = lambda2

?
 
chipmunk951 said:
So it will be like

(3-1/2)lambda1 = lambda2

?

+1 Bump
 
chipmunk951 said:
+1 Bump

Bump, that still didn't work and I'm kind of out of ideas >.<
 
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