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Old Jun 26th 2010, 02:41 AM   #1
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Air Wedge

A ray of light consisting of 2 monochromatic wavelengths, 550 nm and 672 nm, is incident on an air wedge. It is found that, at the nth bright fringe from the vertex of the air wedge for the 550 nm wave, a bright fringe for the
672 nm wave is also formed. Determine the smallest value of n.

Equations involved in this problem is
2t=(m-0.5)λ
x= λ/(2 tan θ)

I tried solving this question:

For the wave of 550 nm,
2t = (n-0.5)x , where t is the thickness at a point of the air wedge

For the wave of 672 nm,
2t= (m-0.5)x'

Since it's the same air wedge i.e. t is the same for both equations,we can equate the 2 equations.
(n-0.5)x=(m-0.5)x'
(n-0.5)(550)=(m-0.5)(672)

But I can't find another equation relating m and n.

The solution given is:

Distance of nth bright fringe of wavelength λ from the vertex = (n-0.5)x

Distance of (n-1)th bright fringe of wavelength λ' from the vertex
=(n-1.5)x'

(n-0.5)x=(n-1.5)x'
(n-0.5)(550)=(n-1.5)(672)
n=6

My question is, can we consider the fringes of both waves that coincide at a point of the air wedge, nth and (n-1)th fringes from the vertex? Shouldn't it be nth and mth fringes from the vertex since they have different wavelengths?

Is my approach to the question correct?
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Old Jun 26th 2010, 05:17 AM   #2
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Let d be the distance of the point from the vertex where the thickness of the air wedge is t.
If a bright fringe is formed at that point due to smaller wavelength, number of wavelengths between the vertex and the point will be n. If a slightly larger wave forms a bright fringe on the same place, the maximum number of wavelengths between the vertex and the point will be (n-1).
Now apply the condition for the bright fringe.

Last edited by sa-ri-ga-ma; Jun 26th 2010 at 05:20 AM.
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Old Jun 27th 2010, 05:09 AM   #3
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Originally Posted by sa-ri-ga-ma View Post
Let d be the distance of the point from the vertex where the thickness of the air wedge is t.
If a bright fringe is formed at that point due to smaller wavelength, number of wavelengths between the vertex and the point will be n. If a slightly larger wave forms a bright fringe on the same place, the maximum number of wavelengths between the vertex and the point will be (n-1).
Now apply the condition for the bright fringe.
Hi,thanks for answering my question.
Is there any proof that the slightly larger wave forms (n-1) wavelengths at that particular point?
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Old Jun 27th 2010, 06:17 PM   #4
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Suppose two waves start prom a point. One has the wave length 1.2 m and the other has 1 m. After five waves first one travels 6m. In that distance second one will have six waves.
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