Thin Film interference: air wedge

[Abelard]

Homework Statement

A pair of very flat glass plates, 7.41 cm long, touch at one end and are separated at the other end by a small piece of 44 gauge copper wire, 5.08×10−5 m in diameter. An air wedge is formed between the glass plates by this supporting wire. Light of wavelength 631 nm illuminates the apparatus from above. How many bright fringes will be seen from above, along the 7.41 cm distance?

Homework Equations

m* wavelength= 2nt

The Attempt at a Solution

struggling with picturing the situation here. Can anyone draw a situation of this?

Thanks

 

[Abelard]

So I assume n=1.00029 roughly equal to 1. so the equation simplifies to 2t= m*wavelength.
There will be a constructive interference as the ray is reflected off of the glass into the air since air acts as a sliding ring.

 

[ehild]

One part of the light beam reflects from glass into air with no phase change, the other part enters the air wedge and then reflects from glass, with pi phase change and interferes with the direct reflected ray. The phase difference between the two rays is
(4π/λ) N t + π/2 =2 m π for the bright fringes, that is

2 t N= (m-1/2)λ.

Find those spots along the wedge where the thickness fulfils this equation. How many such place are there along the length of the wedge? (You can consider N=1.)

ehild

 

[Abelard]

Actually, the equation for the constructive interference is 2t=(m+1/2)*wavelength, not really m-1/2. But I guess it could be right. OK, I will try.

 

[ehild]

It is m-1/2 for m=1,2,... m+1/2 for m=0, 1,2

ehild