Calculating Angle of Wedge in Interference Setup

[ephdub]

Physical Optics: Interference & Diffraction

Here's my question:

Two identical rectangular pieces of glass are laid on top of one another on a plane surface. A thin strip of paper is inserted between them at one end, so that a wedge of air is formed. The plates are illuminated by perpendicularly incident light of wavelength 670nm, and 15 interference fringes per 5cm of wedge appear. (i.e. 15 maxima and 15 minima per 5cm of wedge.)
What is the angle of the wedge?

So.. I sort of have a clue on how to do this but help would be nice.
Since I'm calculating the angle of the wedge, I'm trying to setup my question for trig.. I picture it like this:
/|
/ | <-- H
/ Y |
----
^
horizon = L

Where Y is going to be my angle of the glass wedge. (the picture is not to scale of course and not as extreme as in my diagram in real life).

So to calculate my H, I was assuming that the 15th bright fringe occurs where the separation between the plates was a maximum.. so..

1/2 + 2d/(lamba) = m
solve for d and plug in values
where lambda = 6.7E-7m
m = 15
... churn out the numbers and d = 4.8575E-6 m

Here's where I'm stuck. The way that I set up my question is that I would need to solve for the horizontal (L - in my diagram) and use Trig to find the angle between the glass wedge. How can I do this? Am I on the right track or.. am I just doing something really stupid?

I'm fooling around with the 15 maxima per 5cm of wedge. I'm sure it has to do something with that.. Any help is much appreciated.

 

[Doc Al]

Here's where I'm stuck. The way that I set up my question is that I would need to solve for the horizontal (L - in my diagram) and use Trig to find the angle between the glass wedge. How can I do this? Am I on the right track or.. am I just doing something really stupid?

L is given: it's 5 cm.

 

[wais]

Your approach is on the right track. To calculate the angle of the wedge, we can use the following formula:

θ = tan^-1 (H/L)

Where H is the height of the wedge and L is the length of the wedge. We already have H calculated as 4.8575E-6 m, but we need to find the value of L. We know that there are 15 maxima and 15 minima per 5cm of wedge, which means there are 30 total fringes per 5cm. We can use this information to find the length of the wedge, L, in terms of the wavelength λ:

L = (30λ)/2 = 15λ

Now we can plug in our values for H and L to calculate the angle of the wedge:

θ = tan^-1 (4.8575E-6 m / 15λ)

Since we know that λ = 670nm, we can convert it to meters and calculate the angle:

θ = tan^-1 (4.8575E-6 m / 15 * 670E-9 m) = 0.0258 radians

Thus, the angle of the wedge is approximately 0.0258 radians, or about 1.48 degrees.