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Phase Change on Reflection
To understand interference caused by multiple reflections it is
necessary to consider what happens when a light wave moving in air
hits a material such as glass.
The reflected pulse is said to
undergo a phase change of 180°
or π radians. The reflected pulse
is 180° out of phase with the
incident pulse.
If these two pulses were to meet
they would momentarily cancel as
they passed one another.
This happens whenever light waves are reflected from a material with a
higher refractive index.
If the waves are reflected from a lower refractive index material there is no
phase change.
The importance of phase change is that the optical path of the
reflected ray is changed. A phase shift of π radians is equivalent
to having travelled an extra λ/2 distance compared to a wave that
had not been reflected.
Thin (parallel) film interference
When we look at a thin film of oil on water we see colours.
This is caused by interference produced by reflection from the air/oil
boundary and the oil/air boundary.
1
air
oil
water
2
n=1
d
n=1.45
n=1.33
At each boundary some light is
reflected and some refracted. This
is called division by amplitude.
Someone looking at rays 1 and 2
would see an interference pattern.
This is caused by path difference
between the rays.
1
2
air
n=1
oil
d
n=1.45
If we assume the angle of incidence to
be 0 degrees, then the extra distance
travelled by ray 2 will be 2d.
This means that the optical path
difference will be equal to 2nd
However, there will be a phase change
n=1.33 of λ/2 at the first boundary (ray 1),
since the ray is being reflected by a
layer of greater refractive index
water
The total optical path difference will be equal to?
= 2nd + λ/2
For constructive interference the path difference must be equal to a
whole number multiple of wavelengths
2nd +λ/2 = mλ
2nd = mλ – λ/2 = (m + ½) λ
1
d
(m  )λ
2
2n
This equation gives the thickness at which light of wavelength, λ will
produce constructive interference.
For destructive interference the path difference must be equal to an
odd number of half wavelengths.
2nd +
λ/2 = (m + ½ ) λ
2nd = mλ + λ/2 – λ/2
mλ
d
2n
For destructive interference, m=1 for minimum thickness.
Example
Calculate the minimum thickness of oil which will produce
destructive interference in green light of wavelength 525nm.
d
mλ
2n
1 525  109
d
2  1.45
 1.81 107 m