Lecture 15 Interference Chp. 35

• Topics– Interference from thin films– Due to the wave nature of light– Change in wavelength and phase change in a medium – Huygen’s principle, Coherence– Young’s Interference Experiment– Intensity– Diffraction

• Demos• Polling

Interference of light from different layers of thin films such as air between glass, soap

bubble, and nail polish on water

An important question to consider is once light is in phase what are the ways it can get

out of phase

1. The selected rays can travel different distances.

2. Rays go through different material with different index of refraction

3. Reflection from a medium with greater index of refraction

First what do we mean by out of phase or in phase

In Phase Out of Phase by 180 degrees or radians or/2

In between

Destructive interferenceConstructive interference

n2

n1

air water

Explain using wave machine

Demonstrate using rope

The wave changes phase by 180 degrees when reflection from a medium with greater index of refraction. See cartoon below.

air 1.0

air 1.00

soap 1.30 L

eye

21

Reflection 180 deg phase change for Ray 1when reflecting from the soap film surface but not ray 2

What are the conditions for Constructive Interference in reflection from a soap bubble?

€

Suppose the soap thickness L is such that

2L =1

2

⎛

⎝ ⎜

⎞

⎠ ⎟λ n

€

2L = m +1

2

⎛

⎝ ⎜

⎞

⎠ ⎟λ n 2

where m = 0, 1, 2, ...

We must add a half wavelength to account for the 180 deg phase change for constructive interference. Path difference must = integral number of wavelengths plus 1/2 a wavelength.

Now consider the path length differences

€

n =λ

n

First consider phase change upon reflection

Answer

soap 1.30

What are the conditions for Constructive Interference in transmission from a soap bubble?

air 1.0

air 1.00

L

eye

Transmission

2

1€

2L = m( )λ

n λ =

2nL

m

m = 1,2,3,4….

For constructive interference path differencemust = integral number of wavelengths

No phase changes upon reflection

n =1.30

Demonstrate with soap bubble

Why does it look black on top?

What are the conditions for destructive interference?

Chapter 35 Problem 55. A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.30).

(a) If you are looking straight down from an airplane while the Sun is overhead at a region of the slick where its thickness is L=460 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference?

air 1.0

Water 1.30

Kerosene 1.20 L

2

1

180 deg phase change

(b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest? (Hint: use figure (a) with appropriate indices of refraction.)

Scuba diver

air 1.0

Water 1.30

Kerosene 1.20 L

2

1

We note that only the 441.6 nm wavelength (blue) is in the visible range,

€

2L = m +1

2

⎛

⎝ ⎜

⎞

⎠ ⎟λ

n2

Solving for λ

λ =4n2L

2m +1

€

λ =4n2L

2m +1=

4(1.2)(460)

1= 2208 nm m = 0

€

λ =4n2L

2m +1=

4(1.2)(460)

3= 736 nm m =1

€

λ =4n2L

2m +1=

4(1.2)(460)

5= 441.6 nm m = 2

Visible spectrum is 430 nm - 690 nm

Example S1 and S2 are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power.

(a) If a detector is moved to the right along the x-axis from source S1, at what distances from S1 are the first three interference maxima detected?

detector

Consider what the path difference is between the detector and S1

and the detector and S2

€

path difference = d2 + x 2 − x = mλ m =1,2,3..

The solution for x of this equation is

For constructive interference we have

xdetector

.22 xd +

Solve for x

€

x =d2 −m2λ2

2mλ for m =1, 2, 3,..

€

d2 + x 2 − x = mλ

d2 + x2 = mλ + x Now square both sides

€

d2 + x 2 = ( mλ + x)2

d2 + x 2 = m2λ2 + 2mλx + x 2 Now cancel x2

€

d2 = m2λ2 + 2mλx solve for x

.2

222

mmd

x−

=

€

For m = 3 x =16 − 3( )

2

2( ) 3( )=1.17m.

m=3

What about m = 4 ? This corresponds to x=0. Path difference =4 meters.

€

x =16 −m2

2m

m=2

€

For m = 2 x =16 − 2( )

2

2( ) 2( )= 3.0m.

m=1

€

For m =1 x =16 − 1( )

2

2( ) 1( )= 7.5m.

Although the amplitudes are the same at the sources, the waves travel different distances to get to the points of minimum intensity and each amplitude decreases in inverse proportion to the square of the distance traveled. The intensity is not zero at the minima positions.

€

I1 =P0

4πx 2 I2 =

P0

4π (d2 + x 2)

I1I2

=x 2

d2 + x 2=

(0.55)2

42 + (0.55)2~

1

64

m=3 m=2 m=1

€

x =d2 − (m + 1

2)2λ2

2(m + 12)λ

.

Where do the minima occur?

€

x =16 − (m + 1

2)2

2(m + 12)

.

€

path difference = d2 + x 2 − x = (m + 12)λ m = 0,1,2,3

m=0 x=15.75 mm=1 x =4.55 mm=2 x=1.95 mm=3 x= 0.55 m

m=0m=1

Demo with speakers using sound waves

Set oscillator frequency to 1372 Hz,Then wavelength of sound is 343/1372 =0.25 m

Set speakers apart by 1m. Then maxima occur at .2

222

mmd

x−

=

x =1−m2 (1 /16)2m(1 / 4)

x=16 −m2

8 −m

m =4x=0

m=3x=16-924

=724

m≈0.33m

m=2x=16-416

=1216

m≈0.75m

m=1x=16-17=157

m≈2.0m

Huygen’s Principle, Wavefronts and Coherence

E =Emsin(2

x−2 ft)

k

€

E = Em sin(kx −ωt)

Examples of coherence are: Laser light Small spot on tungsten filament Wavefront

Most light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherent

E =Emsin(2(x−ct))

Youngs Double Slit Interference Experiment

D

m=0

m=1

m=1

m=2

m=2

y

In order to form an interference pattern, the incident light must satisfy two conditions: (i) The light sources must be coherent. This means that the plane waves from the sources must maintain a constant phase relation. For example, if two waves are completely out of phase with φ = π , this phase difference must not change with time. (ii) The light must be monochromatic. This means that the light consists of just one wavelength λ = 2π/k .

Light emitted from an incandescent lightbulb is incoherent because the light consists of waves of different wavelengths and they do not maintain a constant phase relationship. Thus, no interference pattern is observed.

Constructive interference Constructive interference Destructive interference

If you now send the light from the two openings onto a screen, an interference pattern appears, due to differing path lengths from each source

• we have constructive interference if paths differ by any number of full wavelengths

• destructive interference if difference is half a wavelength longer or shorter

Constructive interference

Destructive interference

δ =path length difference

δ=d sinθ

We want to find the pathlength difference

How do we locate the vertical position of the fringes on the screen?We want to relate y to m, L and the angle

1) L >> d2) d >> λThese tell us that θ is smallTherefore,

tan =yL

y=L tan

y is the distance to the m’th fringe

ym =mL

d

d sin =mMaxima

d sin =(m+12)m=0,1,2,3..

ym =L tan

m +/- ym

0

1

2

3

0

L/d

2L/d

3L/d

Maxima

m +/- ym

0

1

2

3

L/2d

3L/2d

5L/2d

7L/2d

Minima

ym =(m+1 / 2)L

d

sin=md

tan ; sinym=Lsin ym

Chapter 35 Problem 19.Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes?

From the table on the previous slide we see that the separation between bright fringes is S =L

d

S=Ld=(5.4m)

500 ×10-9m0.0012m

S=0.00225m=2.25mm

How far off the axis is the m=3 bright fringe? y =3L/d = 3S=6.75 mm

What is the Intensity Distribution along the screen

The electric field at P is sum of E1 and E2. The intensity is proportional to the square of the net amplitude.

represents the correlation between the two waves. For incoherent light, as there is no definite phase relation between E1 and E2 and cross term vanishes

and incoherent sum is

I

E2

E1

E

For coherent sources, the cross term is non-zero. In fact, for constructive interference E1 = E2 and I = 4I1

For destructive interference E1 = -E2 and and correlation term = - I1, the total intensity becomes I = I1 -2I1 + I1 = 0

Suppose that the waves emerged from the slits are coherent sinusoidal plane waves. Let the electric field components of the wave from slits 1 and 2 at P be given by

We have dropped the kx term by assuming that P is at the origin (x =0) and we have acknowledged that wave E2 has traveled farther by giving it a phase shift (ϕ) relative to E1

For constructive interference, with path difference of δ = λ would correspond to a phase shift of ϕ=2π. This then implies

E1

E0

E0

E

t

E2

E1

E

For constructive interference, with path difference of δ = λ would correspond to a phase shift of ϕ=2π. This then implies

The intensity I is proportional to the square of the amplitude of the total electric field

I ∝ E2

E2 =4E02 cos2 (

φ2)

Let I0 =E02

Then.. I =4 I 0 cos2 12φ

What about the intensity of light along the screen?

€

I = 4I0 cos2 12 φ

φ =2πd

λsinθ

You see the intensity is not a pure cosine wave. There are additional bright and dark spots. So what is going on?

In reality this is what you see