How Many Interference Fringes Are Observed in a Wedge-Shaped Film of Air?

[Malby]

I'm having trouble understanding this problem. I think I'm missing something or misunderstanding the question:

A wedge-shaped film of air is made by placing a small slip of paper between the edges of two pieces of glass as shown below. Light of wave-length 600nm is incident normally on the glass, and interference fringes are observed by reection. If the angle Theta made by the plates is 3x10^-4 rad, how many interference fringes per centimetre are observed? (Hint: Use the small-angle approximation Theta = t/x.)

My thinking is that there is a path difference of 3x10^-4 rad and it is similar to a double slit experiment (or single slit experiment) causing interference on a screen some distance L away. However we haven't been given any such distance.

Here is a diagram: http://i.imgur.com/KX1C4.png

 

[rl.bhat]

Ηι Malby, welcome to PF.

The condition for the constructive interference is

2*t = m*λ, where t is the thickness of the air wedge at a distance x from the point of contact of the glass plates.

t/x = θ. So 2*x*θ = m*λ.

Hence the number of fringes per unit length = m = 2*x*θ/λ.

 

[Malby]

In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is away from the source.

 

[rl.bhat]

In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is away from the source.

2*x*θ = m*λ.

If β is the fringe width ( distance between successive bright or dark fringes) then

2*(x+β)*θ = (m+1)λ

So β = λ/(2θ)

And number of fringes per unit length is

1/β = 2θ/λ