OPTICS: Thin Film Interference

[n0_3sc]

Please help:

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed.
The plates are illuminated at normal incidence by 546-nm light from a mercury-vapor lamp.
Interference fringes are formed, with 15.0 fringes per centimeter.
Find the angle of the wedge.

I have a lot of trouble with Thin Film Interference so a detailed answer will be highly appreciated.

(Answer: 0.0235 degrees)

 

[rayjohn01]

phase

Close to the point of touching there are 4 reflected rays ( one from each surface), since the glass is nearly transparent each reflected ray is veryweak BUT about the same intensity.
They occur in pairs two from the upper surfaces (which are in phase) and two from the lower sufaces ( which are also in phase) -- but the second pair are 180 degrees out of phase wrt. the firstpair. ( this is the phase reversal which occurs to to the direction of media density change ).
This means that close to touching the ray pairs cancel each other .

As one departs from the point of contact the difference in the distance traveled by the ray pairs increases and is twice the separation (d) at any point .
The first fringe will occur when the phase shift has increased by 180 degrees = lambda/2 so we can now write

lambda / 2 = 2 . d = 2 . ( x / 2) . theta

where theta is the angle in radians ( assumed small so that tan(theta) ~ theta ) , d is the separation at a point and x/2 is the distance from the touch point . x is the separation of fringes ( and x/2 is the separation of the first fringe from the touch point -- 1/2 a fringe ).

so we have theta = lambda / x / 2 radians = 546 . 10^-9 . 1500 / 2

= .0004095 radians or .02346 degrees.

The reinforcements will fade with distance due the the reflections each only being a few percent each time .
Sorry I could not be of more help.
ray.