Interference fringes between rectangular pieces of glass

[Yroyathon]

hello again folks, this is the last problem that I haven't gotten in this week's homework. see attached image.

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?

Homework Equations

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

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

 

[Delphi51]

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

 

[nlightner]

Hello Yroyathon,

Thank you for sharing your attempt at solving this problem. Your approach is on the right track, but there are a few things that need to be clarified.

Firstly, the equation you have used, x/y = L/d, is not applicable in this scenario. This equation is used to calculate the magnification of an image formed by a lens or mirror, where x is the size of the object, y is the size of the image, L is the distance between the object and the lens/mirror, and d is the distance between the lens/mirror and the image.

In this problem, we are dealing with interference fringes, which are formed due to the superposition of waves. The equation that relates the distance between fringes (y) to the wavelength (λ) and the angle of the wedge (θ) is y = (m + 1/2) * λ * tan(θ), where m is the number of fringes.

Secondly, the value of x' that you have calculated is incorrect. It should be x' = (L/d) * λ, where L is the length of the wedge and d is the distance between the plates.

Now, coming to your approach, you have correctly identified that the change in x is equal to the distance between fringes (y). However, the equation you have used to calculate x' is incorrect. It should be x' = (L/d) * λ, which is the same as y = (m + 1/2) * λ * tan(θ).

Using this equation, we can calculate the angle of the wedge (θ) as follows:

θ = arctan(y / ((m + 1/2) * λ))

Substituting the given values, we get:

θ = arctan((1/18) / ((18 + 1/2) * 595 * 10^(-9))) = 0.0033 radians

Therefore, the angle of the wedge is 0.0033 radians or approximately 0.19 degrees.

I hope this helps clarify your doubts. Keep up the good work!