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Interference fringes between rectangular pieces of glass

  • Thread starter Yroyathon
  • Start date
hello again folks, this is the last problem that I haven't gotten in this week's homework. see attached image.

1. Homework Statement
Two rectangular pieces of glass (see attached image, I painstakingly made it with a very fussy/crash-ey program) are laid on top of one another on a plane surface. A thin strip of paper is inserted between them at one end, so that a wedge of air is formed. The plates are illuminated by perpendicularly incident light of wavelength 595 nm, and 18 interference fringes per centimeter-length of wedge appear. What is the angle of the wedge?


2. Homework Equations
x/y = L/d (similar triangles)
2y = (m + 1/2) * lambda
Tan theta = (y / x)

3. The Attempt at a Solution
calculating x' from x(m+1) - x(m), I get x'=(L/d) * (lambda/2) . the change in x, x', I think is the space between the fringes. but whether this is 1/ (18/(10^(-2))) or 18/(10^(-2)) I'm not sure. The latter is way too big, and resulted in an incorrect answer. The former is ultimately 1/1800, which seems intuitively right.

so 1/1800 = (L/d) * (lambda/2).
L/d = 2/(1800*lambda)
d/L = (1800*lambda)/2
theta = Arctan (d/L) =Arctan ((1800*lambda)/2)

does this look right? I haven't calculated or submitted this answer, as i wanted to wait and think it over before wasting any more of my last 3 chances to get it right.

Thanks for any tips or suggestions.
,Yroyathon
 

Attachments

Delphi51

Homework Helper
3,407
10
I think you have it right.
You get a bright fringe when the thickness is nL/2 (using L instead of lambda).
thickness = x*tan(A) so x = nL/(2*tan(A)) and the horizontal distance between fringes is
dx = .5*L/tan(A)
Your dx is .01/18 = .5*L/tan(A)
tan(A) = 900L
 

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