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Chapter 35 Interference of Light The concept of optical interference is critical to understanding many natural phenomena, ranging from color shifting in butterfly wings to intensity patterns formed by small apertures. These phenomena cannot be explained using simple geometrical optics, and are based on the wave nature of light. In this chapter we explore the wave nature of light and examine several key optical interference phenomena. •  Huygen’s Wave propagation •  Coherence of light •  Constructive and Destructive Interference •  Two slit Interference •  Thin film interference •  Wedge Interference •  Phasor Analysis

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Huygen’s proposed that points on a wavefront serve as point sources of spherical secondary wavelets. After time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets.

Huygen’s Principle

There is some deeper mathematics here; that a traveling wave front is mathematically equivalent to a superposition of spherical waves.

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Bending of Light: a wave property

Based on Huygen’s view it is easy to see how light bends around a corner. At the corner point a spherical wavelet projects Itself around the corner. This behavior supports the wave view view of light.

The scattering of light from objects is called difffraction.

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Diffraction

Diffraction is the scattering of light from objects or orifices. For plane waves entering the single slit, the waves emerging from the slit start spreading out, or diffracting.

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Young’s Experiment on Interference of Light

For waves entering a two slit, the emerging waves interfere and form an interference (diffraction) pattern.

Interference The pattern on the screen is explained by a second wave property of light , that of the Interference of light waves.

Coherent The spherical waves emerging from slits S1 and S2 start at in phase and remain coherent.

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Coherence of Light

Coherence light sources are those which provide a stable phase relationship that does not change with time: E(t) = E0 cos(ω t + ϕ). Interference can only occur with coherent light sources.

•  Coherent sources: Phase ϕ must be well defined and constant. When waves from coherent sources meet, stable interference can occur. Sunlight is coherent over a short length and time range.

Since laser light is produced by atom emissions in lock phase. It is coherent over long length and time ranges. •  Incoherent sources: ϕ jitters randomly in time, no stable

interference occurs. coherent incoherent

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Constructive and Destructive interference Consider wave-1 and wave-2. The amplitudes of waves add. (a)  Purely constructive interference is obtained when identical waves are in

phase. (b)   Purely destructive interference occurs when identical waves are exactly 1800

out of phase, or shifted by half a wavelength.

wave1+wave2=wave12Asin(ω t)+Asin(ω t)=2Asin(ω t)

wave1+wave2=wave12Asin(ω t)+ Asin(ω t +π )Asin(ω t)− Asin(ω t) = 0

Total destructive interference Constructive interference

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The phase difference between two waves can change if the waves travel paths of different lengths.

Locating Fringes by path length method

What appears at each point on the screen is determined by the path length difference Δr = r2 - r1 of the rays reaching that point.

Path Length Difference:

Δr = d sinθ = mλ for m = 0,±1,±2,… max ima= (m+1/ 2)λ for m = 0,±1,±2,..min ima

Δr

Δφ = 2π (Δr

λ)

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Q1: Two narrow slits on a plate are positoned a distance 0.5mm apart. Red light of wavelength λ=650 nm from a coherent light source is incident on the plate. Find the location of the first minima on a screen a distance L=1maway.

d sinθ = mλ for m = 0,1,2,… max ima= (m+1/ 2)λ for m = 0,1,2,..min ima ←

d sinθ = λ / 2 m = 0 min ima

sinθ = λ / 2d = 650nm / 1mm = 6.5x10−4

θ = sin−1(6.5x10−4 ) = 6.5x10−4

Δy = L tanθ = (1m)(6.5x10−4 ) = 0.65mm

θ L

d sinθ

Δy

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Intensity in Double-Slit Interference

E = E0 sin(ωta!)

slitt−1" #$ %$+ E0 sin(ωt +φ

b!)

slit−2" #$$ %$$φ = 2π (

Δrλ

) = 2π d sinθλ

U sin g trig identity sin(a)+ sin(b) = 2sin(a + b

2)cos(

a − b2

)

E(t) = 2E0 sin(ωt + ϕ2

)cos(ϕ2

) = [2E0 cos(ϕ2

)

Amplitude" #$ %$

]sin(ωt + ϕ2

)

I = 4| E0 |2

µ0ccos2(

ϕ2

) I = 4I0 cos2(ϕ2

) where I0 =| E0 |2

µ0c

I = 4I0 cos2 12φ

I = 1

µ0c|!E1 +

!E2 |2

φ = 2π d sinθ

λ

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We may want to combine more than two waves. For example, there may be more than two slits.

Prodedure:

1.  Construct a series of phasors representing the waves to be combined. Draw them end to end, maintaining proper phase relationships between adjacent phasors.

2.  Construct the sum of this array. The length of this vector sum gives the amplitude of the resulting phasor. The angle between the vector sum and the first phasor is the phase of the resultant with respect to the first.

Combining More Than Two Waves by Phasors (extra)

E1 E2

E3 E4

E

Φ

E1=E10sin(ωt+φ1)

E2=E20sin(ωt+φ2) E3=E30sin(ωt+φ3)

E4=E40sin(ωt+φ4)

φ1

Mechanical Wave Action Upon Changing Media

R T

R T

As a wave moves from a denser media to less dense media Z0>Z1, both the reflected (R) and transmittted (T ) waves retain their phase, Δϕ=0.

As a wave moves from a less dense media to a more dense media Z0>Z1, The reflected wave (R) inverts (Δϕ=π) but the transmitted (T ) wave retains its phase.

Z0<Z1

Z0>Z1

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Phase Shifts upon Reflection (R) and Transmission (T)

n1 n2

n1 > n2

n1 n2 n1 < n2

Rays traveling with n1 > n2: Reflection Δϕ=0 noflip Transmission. Δϕ=0 noflip

Rays traveling with n1 < n2: Reflection Δϕ=π flip Transmission. Δϕ=0 noflip

Thin film

n=1.3 n=1.0 n=1.0

i

r2

r1

In a thin film the π phase shift occurs at the r1 rellection. There is no phase shift at the r2 reflection.

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Interference from Thin Films Light incident on a thin film will reflect from the front face and rear face. Paths r1 and r2 will recombine in region I. The light remains coherent within the thin film, making the interference possible.

Reflected light in region I will be a combination rays r1 and r2. Ray-1 has undergone a π phase

shift upon reflection at n2. Δφ= 2π (Δrλn

)

λn = λ / n2 isthe wavelength in media n where

Δr has occurred. L = thickness of the filmMaxima and minina conditions:

2L = m+ 12( )λn for m = 0,1,2,...(constructive)

2L = m( )λn

n2

for m = 0,1,2,… (totaldestructive)

I II III

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Ultra Thin-Film Interference

For L << λ ultra − thin filmsΔφ = π phase shift only. L ~ 0The film will appear dark whenviewed from ilumin ated side atall wavelengths.

Bright and dark fringes are inter-mixed with colors in the photo,with many colors dispersed from white light.

Near the top where gravity has made the film ultra-thin we see a dark reflection due to the Δφ=π phase shift. Colors appear below due to the λ /n path length dispersing the differnent colors. The filmthickness is also changing awith height.

L<<λ

air

film

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Newtons Rings: Interference of a Concave ense on a Glass Plate

R2 - r2

The phenomenon is caused by the interference between ray 2 and 3, reflecting from the plano-convex lens at A and B, respectively. Analysing the phases:⋅ ray − 2 no phase shift at A⋅ ray − 3 Δφ = π flip at B

2t = (m+ 12 )λair m = 0,1,2.. max ima mth order

2t = (m)λair m = 0,1,2.. min ima mth order

t = R − R2 − r 2 or r = R2 − (R − t)2 1)

For light of wavelength λ illumin ating thetop ::1)Adark spot (ring) occurs at the center m = 0

2) t = 0λdarkm=0

! ,12λ

brightm=0

!, 1λ

darkm=1

! ,32λ

brightm=1

!,2λ

darkm=2

! ,52λ

brightm=2

!,......

Theradii can be calculated from 1)

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Fringe Pattern from an Air Wedge

Two very flat glass plates separated on the edge by a smallobject of height t will form an air wedge. When illuminated from the top with monochromatic light wedge fringes occur. By counting the fringes the thickness of the object can bemeasured. Let t be a variable ranging t = 0...tmax

⋅ ray −1 Δφ = 0 flip at A⋅ ray − 2 Δφ = π flip at B

2t = (m+ 12 )λair m = 0,1,2.. max ima mth order

2t = (m)λair m = 0,1,2.. min ima mth order

Adark fringeoccursat t = 0 where the plates touch.

The number of fringes is N = 2tλ

.

λ

LΔL

Δt ΔL

Δt = λ / 2 m = 0 min ima

θ ~ tL= ΔtΔL

= λ / 2L / N

= Nλ2L

= tL

Nλ = 2t

Δt

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39. Light of wavelength λ=680nm illuminates normally two glass plates 120 mm long that touch at one end, and are separated by a wire 0.048mm diameter at the other end. How many fringes appear over the 120 mm distance?

N =

2tλ

= 0.096mm680nm

= 141

40. An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?

Looking for max ima (colors) :

2t = m+ 12( ) λ

n2

for m = 0,1,2,... (maxima of order m)

λ = 2ntm+ 1

2( ) =2(1.4)120nm

m+ 12( ) = 336nm

m+ 12

λm=0 = 672nm (red visibel) ←

λm=1 = 224nm (ultraviolet not visible)

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STOP HERE