How many bright fringes are seen in the reflected light

AI Thread Summary
The discussion focuses on calculating the number of bright fringes seen in the reflected light between two glass plates separated by a wire. The air gap thickness varies from 0 to 0.210 mm, and the light wavelength is 570.0 nm. The key equation for constructive interference is 2t = (m + ½)λ, where 't' is the thickness of the air wedge. Participants explore how to derive the fringe width and relate it to the number of fringes using the dimensions provided. Ultimately, the calculation involves substituting values into the equations to find the number of bright fringes observed.
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Homework Statement



Two optically flat plates of glass are separated at one end by a wire of diameter 0.210 mm; at the other end they touch. Thus, the air gap between the plates has a thickness ranging from 0 to 0.210 mm. The plates are 15.0 cm long and are illuminated from above with light of wavelength 570.0 nm. How many bright fringes are seen in the reflected light?

Homework Equations



2t = (m + ½)λ (constructive)
m = 0, 1, 2, …
t = thickness of the air wedge
λ = wavelength in vacuum (air)

The Attempt at a Solution



Pretty darn clueless on this one. I don't think the above equation is even proper. I'm lost on what to do with the plethora of values they give you, and can't understand how to fit them into ANY of the equations I've been mulling through. Since we are looking for bright spots, then clearly we are dealing with constructive interference, but I'm quite lost after that. Any help greatly appreciated. Thanks in advance!
 
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msk172 said:

Homework Statement



Two optically flat plates of glass are separated at one end by a wire of diameter 0.210 mm; at the other end they touch. Thus, the air gap between the plates has a thickness ranging from 0 to 0.210 mm. The plates are 15.0 cm long and are illuminated from above with light of wavelength 570.0 nm. How many bright fringes are seen in the reflected light?

Homework Equations



2t = (m + ½)λ (constructive)
m = 0, 1, 2, …
t = thickness of the air wedge
λ = wavelength in vacuum (air)

At a point distance x from the point of contact of the plates can be found by comparing similar triangles formed at x and at extreme end.
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.
Find the value x for m and m + 1. From that you can find the fringe width.
 
rl.bhat said:
At a point distance x from the point of contact of the plates can be found by comparing similar triangles formed at x and at extreme end.
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.
Find the value x for m and m + 1. From that you can find the fringe width.

I am trying like heck to follow the explanation you've provided, but have not gotten anywhere. I see what you mean about substituting the 2t equation in for t, but don't know where to go from there. Perhaps you could simply re-word in a fashion that would be slightly easier to follow? I do greatly appreciate your feedback!
 
Plus, you reference fringe width, but the question asks for number of fringes seen, not what their width is. Perhaps these two are connected, but if they are I am not seeing it...
 
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.

The equation reduces to x = l/2*h*(m + ½)λ
When m = 1, you get fringe width between two bright fringes.
Number of fringes = l/fringe width.
 
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