Interference of waves

Coherent waves must have a constant phase difference.

This means they must have the same frequency. Often the phase difference is zero but it doesn't have to be.

Light is emitted by electron transitions within individual atoms. Thus waves from the same source will have the same frequency but will not be coherent as the phase difference will vary. (Except for laser light)

For coherence the wave must come from the same position on the light source – division of wavefront.

Or from the same wave which is split in two by reflection – division of amplitude

For interference in air (vacuum)

Constructive path difference = n n = 0,1,2…

Destructive path difference = ( n + ½)

For interference in another medium we have to use the optical path difference.

Optical path length is the distance for the same number of wavelengths in a vacuum

water

m m

(m

The two rays have different geometrical path lengths but have the same optical path length. Ie they contain the same number of wavelengths.

This means a lens does not affect the phase difference between rays

A B

Refractive index n = vacuum

material

For interference in any material

Constructive optical path difference = m m = 0,1,2…

Destructive optical path difference = (m + ½)

So vacuum = nmaterial

Optical path length = n x geometrical path length

Phase difference = 2 x optical path difference

vacuum

Phase Change on Reflection

air glass glass air

Optically more denseie. at a higher refractive index radians phase change

Optically less denseie. at a lower refractive index0 radians phase change

Examples of interference by division of amplitude

1. Parallel sided thin film

doil

air

water

n=1

n=1.45

n=1.33

1 2 At each boundary some light is reflected and some refracted.

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 If we assume the angle of incidence to If we assume the angle of incidence to be near 0 degrees, then the extra be near 0 degrees, then the extra distance travelled by ray 2 will be 2d.distance 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/2λ/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/22nd + 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

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

2n

mλd

m101.81

1.452

105251d

7

9

Blooming – non-reflective coatingsIf no light is reflected it will all be transmitted.

air

coating

glass

For glass with a refractive index of 1.5 you would need a film of refractive index 1.22. Also this film must be durable.

As the refractive index of a material varies with frequency, the coating will only be non–reflective to one frequency.

The non-reflective frequency is usually the middle of the visible spectrum so bloomed surfaces look purplish as some blue and red light will be reflected.

2. Thin air wedge – used to measure small thicknesses

Angle for the wedge is very small.

x

t

Monochromatic light source

D Eye or travelling microscope

A

BC

The path difference between the two rays ABD and ACD is 2t.

3 Newton’s Rings

Glass plate at 45o

Glass block – optically flatx

A

B

Extended source

Fringes are concentric circles.Reflection at A zero phase change.Reflection at B radians phase change

Destructive interference 2t = m m = 0,1,2…Constructive interference 2t = (m + 1/2)

Fringe at centre is dark.

tB

Division by wavefront

Monochromatic

Source eg sodium lamp

Narrow slit gives circular waves

Diffraction grating where coherent waves hit

dsin = m from Higher

Young's Slits ExperimentThe diagram below shows light from a single source of monochromatic light incident on a double slit. The light diffracts at each slit and the overlapping diffraction patterns produce interference.

The first bright fringe is observed at P. Angle PMO is θ

N is a point on BP such that NP = AP

Since P is the first bright fringe BN = λ

For small values of θ, AN cuts MP at almost 900 giving angle MAQ = θ and hence angle ΒΑΝ = θ

Again providing θ is very small, sin θ = tan θ = θ in radians

From triangle BAN: θ = λ / dand from triangle PMO: θ = Δx / D

So λ / d = Δx / D

Therefored

λDΔx

Two points to note:

1. This formula only applies if x<<D, which gives θ small. This is likely to be true for light waves but not for microwaves.

2. The position of the fringes is dependent on the wavelength. If white light is used we can expect overlapping colours either side of a central white maximum. The red, with the longer wavelength, will be the furthest from this white maximum

(Δxred > Δxviolet since λred > λviolet).

ExampleA laser beam is incident on two narrow slits of separation (0.250 ± 0.002)mm. An interference pattern is produced on a screen (2.455 ± 0.005)m from the slits. The spacing across ten fringes is measured as (62.0 ± 0.5)mm.(a) Calculate the wavelength of the light.

(b) State the uncertainty in the wavelength and express the wavelength as (value ± absolute uncertainty).

Δx = 62.0x10-3/10 Δx= λD/d

62.0x10-3/10 = λx2.455/0.250x10-3

λ=631nm

Using % uncertainties: Uncertainty in d =(0.002/0.250)x100=0.8% Uncertainty in D=(0.005/ 2.455)x100=0.2% Uncertainty in Δx=(0.5/62.0)x100=0.8%

The uncertainty in D can be ignored since it is less then one third of the others.

Total % uncertainty is = √( 0.82 + 0.82) = 1.1%

1.1% of 631nm = 7nm so wavelength = (631 ± 7)nm

(d) State, with a reason, if the fringe spacing increases, decreases or remains the same if red laser light is replaced by a Helium-Cadmium laser with a wavelength of 422nm.

(e) State one way in which you might be able to improve the accuracy of the experiment, give a reason.

Measure across more fringes, for example twenty fringes. This would reduce the percentage uncertainty in Δx.

Since the wavelength has decreased the fringe spacing decreases, since Δx λ