Phase Change on ReflectionPhase Change on ReflectionTo understand interference caused by multiple reflections it is To understand interference caused by multiple reflections it is necessary to consider what happens when a light wave moving in air necessary to consider what happens when a light wave moving in air hits a material such as glass.hits a material such as glass.

The reflected pulse is said to The reflected pulse is said to undergo a undergo a phase change of 180°phase change of 180° or or π radiansπ radians. The reflected pulse . The reflected pulse is 180° out of phase with the is 180° out of phase with the incident pulse.incident pulse.

If these two pulses were to meet If these two pulses were to meet they would momentarily cancel as they would momentarily cancel as they passed one another.they passed one another.

This happens whenever light waves are reflected from a material with a This happens whenever light waves are reflected from a material with a higher refractive index.higher refractive index.

If the waves are reflected from a If the waves are reflected from a lower refractive indexlower refractive index material there is material there is no no phase change.phase change.

The importance of phase change is that the optical path of the The importance of phase change is that the optical path of the reflected ray is changed. A phase shift of π radians is equivalent 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 to having travelled an extra λ/2 distance compared to a wave that had not been reflected.had not been reflected.

Thin (parallel) film interferenceThin (parallel) film interference

When we look at a thin film of oil on water we see colours.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 This is caused by interference produced by reflection from the air/oil boundary and the oil/air boundary.boundary and the oil/air boundary.

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.

doil

air

water

n=1

n=1.45

n=1.33

1 2

doil

air

water

n=1

n=1.45

n=1.33

1 2 If we assume the angle of incidence to If we assume the angle of incidence to be 0 degrees, then the extra distance be 0 degrees, then the extra distance travelled by ray 2 will be 2d.travelled by ray 2 will be 2d.

This means that the This means that the optical path optical path differencedifference will be equal to will be equal to 2nd2nd

However, there will be a However, there will be a phase changephase change of of λ/2 λ/2 at the first boundary (ray 1), at the first boundary (ray 1), since the ray is being reflected by a since the ray is being reflected by a layer of greater refractive indexlayer of greater refractive index

The The total optical path differencetotal optical path difference will be equal to? will be equal to?

= 2nd + λ/2 = 2nd + λ/2

For constructive interference the path difference must be equal to a For constructive interference the path difference must be equal to a whole number multiple of wavelengthswhole number multiple of wavelengths

2nd +λ/2 = mλ2nd +λ/2 = mλ

2nd = mλ – λ/2 = (m + ½) λ2nd = mλ – λ/2 = (m + ½) λ 2n

)λ(md 2

1

This equation gives the thickness at which light of wavelength, λ will This equation gives the thickness at which light of wavelength, λ will produce constructive interference. produce constructive interference.

For destructive interference the path difference must be equal to an For destructive interference the path difference must be equal to an odd number of half wavelengths.odd number of half wavelengths.

2nd + λ/2 = (m + ½ ) λ2nd + λ/2 = (m + ½ ) λ

2nd = mλ + λ/2 – λ/22nd = mλ + λ/2 – λ/2

2n

mλd

For destructive interference, m=1 for minimum thickness.For destructive interference, m=1 for minimum thickness.

ExampleExample

Calculate the minimum thickness of oil which will produce Calculate the minimum thickness of oil which will produce destructive interference in green light of wavelength 525nm.destructive interference in green light of wavelength 525nm.

2n

mλd

m101.81

1.452

105251d

7

9