Wavelength used in interference

In summary: The full formula is Δϕ=2π\frac{nd}{λ} we know constructive occurs when Δϕ (phase difference is m*2π where m is fringe number)With Δϕ=m*2π and a simple rearrangement of formula we get λ=\frac{2nd}{m}. n in this case is refractive index. Because the gap is air we can assume a n value of 1.
  • #1
songoku
2,293
325

Homework Statement


Two flat glass plates are stacked and a thin paper strip is inserted between them at one side, creating a narrow wedge-shaped space between the plates. As shown in the figure below, monochromatic light that has a wavelength in air is directed at the plates from above. When viewed from above, parallel bright and dark fringes appear. This is the result of interference between light reflected from the bottom surface of the upper plate and light reflected from the top surface of lower plate.
1-4.jpg

When the angle between the glass plates is θ, N number of bright fringes are observed between points P and Q, whose distances from O, the point of contact between the plates, are x1 and x2 respectively. What is the value of λ? (N is sufficiently large and θ is sufficiently small to that tan θ= θ

a. (x2 - x1)θ / N
b. 2(x2 - x1)θ / N
c. (x2 - x1)/ Nθ
d. 2(x2 - x1)/ Nθ

Homework Equations


d sin θ= nλ
d y / D = nλ
w = λD/d


The Attempt at a Solution


Actually I don't really understand the question. I think the interference pattern occurs at the bottom surface of the upper plate, not at PQ because the light reflected at the bottom part of the upper plate never reaches PQ, so the light reflected by both surfaces will superimpose each other at the bottom surface of the upper plate.

I don't know how to use all the formulas I wrote above for this question. I don't have d, D, w. I even don't know how to use the distance of PQ, angle, and N given by the questions because in my opinion they can't be put in the equation
 
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  • #2
bump again
 
  • #3
songoku said:
bump again

Use the following formula, λ=[itex]\frac{2nd}{m}[/itex]
You've been told P→Q=X[itex]_{2}[/itex]-X[itex]_{1}[/itex] but you want the gap distance which is P→Q*θ. You also know N is max number of constructive interference (max possible value for m in above equation)

The new formula becomes 2(X[itex]_{2}[/itex]-X[itex]_{1}[/itex])θ/ Nif you want to read more about this look into Thin film optical coating. (this is what happens when you get a multicoated lens on glasses)

Hope this helps
 
  • #4
bayan said:
Use the following formula, λ=[itex]\frac{2nd}{m}[/itex]
You've been told P→Q=X[itex]_{2}[/itex]-X[itex]_{1}[/itex] but you want the gap distance which is P→Q*θ. You also know N is max number of constructive interference (max possible value for m in above equation)

The new formula becomes 2(X[itex]_{2}[/itex]-X[itex]_{1}[/itex])θ/ N

Hope this helps

Sorry I can't follow you. I never see that kind of formula. Can I know what is the formula about?

In your formula, what is n and m?

I assume d is the gap distance; and by gap distance do you mean that it is the distance between slits such as in double slit experiment?

I still don't get how to use the information given from the question. Let say PQ is the screen; in the formula d sin θ = nλ, d should be the distance where the light comes from ,which is above, but there is no slits drawn in the figure. θ in the formula also should be angle between the center bright and certain orde of interference and in this case should be measured vertically, not horizontally as given. n is the orde observed so if N is the number of bright fringes observed between PQ, I think n = (N/2 - 1) ??

Thanks
 
  • #5
songoku said:
Sorry I can't follow you. I never see that kind of formula. Can I know what is the formula about?

In your formula, what is n and m?

I assume d is the gap distance; and by gap distance do you mean that it is the distance between slits such as in double slit experiment?

I still don't get how to use the information given from the question. Let say PQ is the screen; in the formula d sin θ = nλ, d should be the distance where the light comes from ,which is above, but there is no slits drawn in the figure. θ in the formula also should be angle between the center bright and certain orde of interference and in this case should be measured vertically, not horizontally as given. n is the orde observed so if N is the number of bright fringes observed between PQ, I think n = (N/2 - 1) ??

Thanks

The full formula is Δϕ=2π[itex]\frac{nd}{λ}[/itex] we know constructive occurs when Δϕ (phase difference is m*2π where m is fringe number)

With Δϕ=m*2π and a simple rearrangement of formula we get λ=[itex]\frac{2nd}{m}[/itex]. n in this case is refractive index. Because the gap is air we can assume a n value of 1.

With gap distance I mean the gap between two glass pieces because that is what causes the constructive/destructive interference. tan is [itex]\frac{Opposite}{Adjacent}[/itex] and you have been told that tanθ=θ therefore the gap distance (or the thickness of Thin paper strip is P→Q*θ. Subbing these into our formula to find λ we get λ= [itex]\frac{2*(X_{2}-X_{2})θ}{m}[/itex]


I think you are assuming that there is a slit experiment when in fact this is only the light interfering with itself as reflected from two different surfaces.

Hope this clears some of it up for you :)
 
  • #6
bayan said:
The full formula is Δϕ=2π[itex]\frac{nd}{λ}[/itex] we know constructive occurs when Δϕ (phase difference is m*2π where m is fringe number)

With Δϕ=m*2π and a simple rearrangement of formula we get λ=[itex]\frac{2nd}{m}[/itex]. n in this case is refractive index. Because the gap is air we can assume a n value of 1.

With gap distance I mean the gap between two glass pieces because that is what causes the constructive/destructive interference. tan is [itex]\frac{Opposite}{Adjacent}[/itex] and you have been told that tanθ=θ therefore the gap distance (or the thickness of Thin paper strip is P→Q*θ. Subbing these into our formula to find λ we get λ= [itex]\frac{2*(X_{2}-X_{2})θ}{m}[/itex]


I think you are assuming that there is a slit experiment when in fact this is only the light interfering with itself as reflected from two different surfaces.

Hope this clears some of it up for you :)

Sorry for taking long time to reply.

I don't know there is formula for light interfering itself..:redface: I still have several things that not clear yet:
1.
Δϕ = 2π[itex]\frac{nd}{λ}[/itex]
m*2π = 2π[itex]\frac{nd}{λ}[/itex]
λ=[itex]\frac{nd}{m}[/itex]

How can you get λ=[itex]\frac{2nd}{m}[/itex] ?

2.
Based on the figure, which one is gap distance? The vertical distance between Q and the upper plate? If we look at triangle OQA, where A is the point on the upper plate which is directly above Q, tan θ = AQ / x2 and leads to AQ = x2 . θ ??
I don't know how to get d = (X2-X1

Thanks
 

What is the definition of wavelength used in interference?

Wavelength is the distance between two consecutive peaks or troughs of a wave. In interference, it refers to the specific wavelength of light that is used in experiments to create interference patterns.

How is the wavelength used in interference determined?

The wavelength used in interference is determined by the type of light source being used, such as a laser or a monochromatic light source. It can also be calculated using the equation λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency of the wave.

Why is the wavelength used in interference important?

The wavelength used in interference is important because it determines the spacing between the interference fringes and the overall pattern that is created. It also affects the intensity and color of the resulting pattern.

What happens if two different wavelengths are used in interference?

If two different wavelengths are used, the resulting interference pattern will be a combination of both wavelengths. This may result in a more complex pattern or a pattern with varying intensities.

Can any wavelength be used in interference?

No, not any wavelength can be used in interference. The light source must emit a single wavelength or a very narrow range of wavelengths in order to create a clear and distinct interference pattern. This is why lasers are commonly used in interference experiments.

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