Wize University Physics Textbook (Master) > Wave Optics

Air Wedges

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

Image illustrating Air Wedges

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

\boxed { \ \Delta\phi=k\Delta x =k \cdot2d=\dfrac{4\pi d}{\lambda} \ }

$\Delta x$ is the path difference between the two interfering rays
$k$ is the wave number of light in air
$\lambda$ is the wavelength of light in air
$d$ is the thickness of the air wedge

Now let's look at the phase shifts that are happening for the two rays:

When ray #1 reflects off the air wedge into glass, there is no phase shift since we have a soft boundary condition.
When ray #2 reflects off the glass into the air wedge, there is a $\bcfi{\pi}$ phase shift since we have a hard boundary condition.

Therefore we have a net phase shift of $\bct{\pi}$ and we need to swap the constructive / destructive equations once.

Constructive interference:

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

Combining this with the equation at the top we get:

\boxed{ \ d=\dfrac{m\lambda}{4}}

$m=1, 3, \ ...$ (odd only)

Destructive interference:

The phase shift is $2\pi m$ for $m=0,\pm1, \pm2, \ ...$ .

Combining this with the equation at the top we get:

\boxed{ \ d=\dfrac{m\lambda}{2}}

$m=0,1, 2, \ ...$

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

\boxed{\ d=\theta R \ }

where $\theta$ 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 $78$ dark fringes. Calculate the number of dark fringes that appear if water ( $n=1.334$ ) replaces the air between the plates.

Image illustrating Example: dark fringes

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

First, the material is air:

$2d=m_{air}\lambda$

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

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

Isolate $m$ and combine with the equation for air to get:

$m_{water}=\dfrac{2d\cdot n}{\lambda}$

$=\dfrac{m_{air}\cancel{\lambda}\cdot n}{\cancel\lambda}$

$=m_{air}n$

Put the numbers in:

$=78\times1.334=104.052$

Therefore we have $104$ fringes.

Practice: Air and Oil Wedge

Two glass plates form an air wedge of $\theta=1.1\degree$ . The approximate number of fringes in the air at distance $10$ mm from the corner of the wedge is $865$ .

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 ( $n=1.4$ )? Answer without finding the new number of fringes.

Wavelength is (in nm, 3 significant digits):Number of fringes changes by a factor of:

I don't know