High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > Wave Optics

Air Wedges

Diffraction Grating Previous Section
Circular Apertures and Resolution Next Section
Popular Courses
Find My Course

Air Wedges


For air wedges, the basics are exactly the same as thin film coating.












The path difference comes from the different lengths traveled by ray #1 and ray #2. The thickness dd of the wedge is traveled by ray #2 twice. Therefore we have Δx=2d\Delta x=2d and the phase difference is given by:

 Δϕ=kΔx=k⋅2d=4πdλ \boxed { \ \Delta\phi=k\Delta x =k \cdot2d=\dfrac{4\pi d}{\lambda} \ }

  • Δx\Delta x is the path difference between the two interfering rays
  • kk is the wave number of light in air
  • λ\lambda is the wavelength of light in air
  • dd is the thickness of the air wedge

PAGE BREAK

Wize Concept
  • Note that in this case the path difference happens in the air since that's the "material" sandwiched between the other two layers.
  • In general, make sure to use the index of refraction n\bf n of that particular material when finding the wavelength, which is λn\dfrac{\lambda}{n}.



Now let's look at the phase shifts that are happening for the two rays:
  1. When ray #1 reflects off the air wedge into glass, there is no phase shift since we have a soft boundary condition.
  2. When ray #2 reflects off the glass into the air wedge, there is a π\bcfi{\pi} phase shift since we have a hard boundary condition.
Therefore we have a net phase shift of π\bct{\pi} and we need to swap the constructive / destructive equations once.




Constructive interference:

The phase shift is 2π(m+12)2\pi \bigg(m+\dfrac{1}{2}\bigg) for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ... . This can also be written as πm\pi m for m=±1,±3, ...m=\pm1, \pm3, \ ... (odd only).

Combining this with the equation at the top we get:
 d=mλ4\boxed{ \ d=\dfrac{m\lambda}{4}}
m=1,3, ...m=1, 3, \ ... (odd only)



Destructive interference:

The phase shift is 2πm2\pi m for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ....

Combining this with the equation at the top we get:
 d=mλ2\boxed{ \ d=\dfrac{m\lambda}{2}}
m=0,1,2, ...m=0,1, 2, \ ...

PAGE BREAK


For wedges of small angle we can assume that the thickness forms the little portion of an arc length of radius RR (which is the location of the bright or dark spot), so we get:

 d=θR \boxed{\ d=\theta R \ }
where θ\theta is measured in radians.

Example: Water Wedge


An air wedge is formed between two glass plates in contact along one edge and slightly separated at the opposite edge. When the plates are illuminated with monochromatic light from above, the reflected light has 7878 dark fringes. Calculate the number of dark fringes that appear if water (n=1.334n=1.334) replaces the air between the plates.



For dark fringes we have to use 2d=mλ2d=m\lambda.


First, the material is air:

2d=mairλ2d=m_{air}\lambda


Then, the material is water (index of refraction nn):

d=mwaterλ2nd=\dfrac{m_{water}\lambda}{2n}

Isolate mm and combine with the equation for air to get:

mwater=2d⋅nλm_{water}=\dfrac{2d\cdot n}{\lambda}

=mairλ⋅nλ=\dfrac{m_{air}\cancel{\lambda}\cdot n}{\cancel\lambda}

=mairn=m_{air}n

Put the numbers in:

=78×1.334=104.052=78\times1.334=104.052

Therefore we have 104104 fringes.

Practice: Air and Oil Wedge


Two glass plates form an air wedge of θ=1.1°\theta=1.1\degree. The approximate number of fringes in the air at distance 1010 mm from the corner of the wedge is 865865.

a) What is the wavelength of the incident light?

b) By what factor does the number of fringes change if you replace the air by silicone oil (n=1.4n=1.4)? Answer without finding the new number of fringes.