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=(m0.5)λ
x= λ/(2 tan θ)
I tried solving this question:
For the wave of 550 nm,
2t = (n0.5)x , where t is the thickness at a point of the air wedge
For the wave of 672 nm,
2t= (m0.5)x'
Since it's the same air wedge i.e. t is the same for both equations,we can equate the 2 equations.
(n0.5)x=(m0.5)x'
(n0.5)(550)=(m0.5)(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 = (n0.5)x
Distance of (n1)th bright fringe of wavelength λ' from the vertex
=(n1.5)x'
(n0.5)x=(n1.5)x'
(n0.5)(550)=(n1.5)(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 (n1)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?
