3. interference

7/27/2019 3. Interference

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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 phasedifference 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 avacuum

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 dense

ie. at a higher refractive

index radians phase change

Optically less dense

ie. at a lower refractive

index

0 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 and2 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

be near 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

of λ/2 at the first boundary (ray 1),

since the ray is being reflected by alayer of greater refractive index

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 2n

)λ(m

d 2

1



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

2nmλd

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.

2n

mλd

m101.81

1.452

105251d

7

9



Blooming – non-reflective coatings

If 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

B

C

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

Narrowslit

gives

circular

waves

Diffraction

grating

where

coherent

waves hit

dsin = m from Higher



Young's Slits Experiment The diagram below shows light from a single source ofmonochromatic light incident on a double slit. The lightdiffracts at each slit and the overlapping diffraction patternsproduce 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 givingangle MAQ = θ and hence angle ΒΑΝ = θ

Again providing θ is very small, sin θ = tan θ = θ inradians



From triangle BAN: θ = λ / d and from triangle PMO: θ = Γx / D

So λ / d = Γx / D

Therefore

d

λDΔx



Two points to note:

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

for microwaves.

2. The position of the fringes is dependent on thewavelength. 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).



Example

A 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 acrossten 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 thewavelength 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

+ 0.8

2

) = 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

3. interference

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