Interference fringes between rectangular pieces of glass

In summary: A very small angle.In summary, the problem involves two rectangular pieces of glass with a thin strip of paper inserted between them, creating a wedge of air. The plates are illuminated with perpendicular light of wavelength 595 nm and 18 interference fringes per centimeter appear. The question asks for the angle of the wedge, which can be calculated using the formula tan(A) = 900L.
  • #1
Yroyathon
42
0
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
 

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  • #2
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
 
  • #3
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!
 

1. What causes interference fringes between rectangular pieces of glass?

The interference fringes between rectangular pieces of glass are caused by the phenomenon of interference, which occurs when two or more waves of light overlap and combine to form a new wave. In this case, the waves are created by the reflection and refraction of light through the glass.

2. How do interference fringes form?

Interference fringes form when light waves from a single source travel through two or more glass surfaces that are parallel to each other. The waves interfere with each other, creating regions of constructive and destructive interference, which appear as bright and dark fringes, respectively.

3. What determines the spacing of interference fringes?

The spacing of interference fringes is determined by the wavelength of the light and the distance between the glass surfaces. As the distance between the surfaces increases, the spacing of the fringes decreases. Additionally, shorter wavelengths of light will result in smaller fringe spacing.

4. How can interference fringes be used in scientific research?

Interference fringes have many practical applications in scientific research, such as measuring the thickness of thin films, determining the refractive index of materials, and studying the properties of light. They are also used in interferometers, which are devices used to make precise measurements and detect small changes in distance or wavelength.

5. Can interference fringes be observed with other types of materials besides glass?

Yes, interference fringes can be observed with other materials as well, such as thin films of different substances or even air. As long as there are two or more surfaces that allow light to pass through and the conditions for interference are met, interference fringes can be observed.

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