Wedge fringes
If two glass
plates are placed face to face with one end separated by a piece of tissue paper or thin metal
foil an air wedge will be formed between them. If monochromatic light is shone on the plates a
series of straight-line fringes will be seen parallel to the line along which they touch (Figure 1).
This is due to interference by division of amplitude, as with Newton's rings. Some light is
reflected from the bottom surface of the top plate and some from the top surface of the bottom
plate.
To see the fringes clearly the angle must be small, something like 4 minutes of
arc. You should also look for fringes close to the join of the plates where the air gap is smallest,
since the fringes are not well defined for path differences of more than some hundred
wavelengths (0.058 mm for sodium light - compare this with the thickness of a sheet of
paper).
Consider a point a distance x from the join. Path difference = 2e = 2xq
where q is the angle between the
plates in radians (this angle is small, so tan q = q in radians).
For an air wedge there is a phase change on reflection
at the top surface of the lower plate and so:
2e = 2xq = ml for a dark fringe
2e = 2xq = (2m + 1)l/2 for a bright fringe
The travelling microscope or the eye must be focused close
to the upper surface of the air wedge since this is where the fringes are
localised.
Pressing down gently with your finger on the plates will move the interference
pattern, since only a very small movement is needed to alter the path difference
significantly.
The vertical soap film is a good example of wedge fringes. As the soap
drains to the bottom of the film a wedge of very small angle is formed. When the top part goes
black the film is about to break. The flatness of a glass surface may be tested by placing it
on a test surface which is known to be flat and illuminating them with monochromatic light; any
imperfections will show up as loop-shaped interference fringes around bumps or depressions on
the surfaces.
Example problem
An air wedge is formed by placing a sheet of foil between the edges of two glass plates 75 mm from their point of contact. When the wedge is illuminated with light of wavelength 5.8 x 10-7 m the fringes are 1.30 mm apart.
Calculate the thickness of the foil.
Number of bright bands in the air wedge = 75/1.30 = 57.7 = ml.
Change in vertical height from one fringe to the next = l/2.
Therefore vertical height for 57.7 fringes = [5.8x10-7x57.7]/2 = [3.34x10-5]/2
The foil thickness is therefore 1.67x10-5 m.