Question

A piece of paper is placed at the end of an air wedge 4.0 cm long. Interference fringes appear when light of wavelength 639 nm is reflected from the wedge. A dark fringe occurs both at the vertext of the wedge and at its paper end, and 56 bright fringes appear between. Calculate the thickness of the paper.

Answer 1

Following the ray diagram, we can see the path difference of the ligt occurs in the air wedge, which is approximately 2 times the distance of the air wedge. At the edge though, this value is 2d, or 2 times the thickness of the paper.

As light reflects from a material of lower index off a higher index, there is a 180 degree phase shift.

For constructive interference, it's usually a integer value of wavelength, but at the air-paper boundry at the bottom, there is a 180 degree phase shift, to correct that we end up having something similar to destructive intereference for most cases.
\[2d=(m+\frac{1}{2})\lambda\]
where m= 0,1,2,3...
Since m=0 is one fringe, the last fringe is m=55
Thus:
\[d=\frac{55.5*639nm}{2}\]
\[d=0.018cm\]