How many bright fringes are seen in the reflected light

msk172

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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!

rl.bhat

Quote by msk172

1. The problem statement, all variables and given/known data

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?

2. Relevant 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.

msk172

Quote by rl.bhat

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!

msk172

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...

rl.bhat

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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msk172	How many bright fringes are seen in the reflected light 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...

rl.bhat Recognitions: Homework Help	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/2h(m + ½)λ When m = 1, you get fringe width between two bright fringes. Number of fringes = l/fringe width.