1

i2

=1

f2

!1

p2

=1

!15cm!

1

!30cm = -

1

15 +

1

30 i

2= !30cm.

Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It isvirtual (since i2 < 0).

30

The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the imagehas the same size orientation as the object.

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20

-15

40

Actual ray diagram purple.

Dashed lines are virtual rays

Lecture 15 Interference Chp. 35

• Topics– Interference is due to the wave nature of light– Huygen’s principle, Coherence– Young’s Interference Experiment and demo– Intensity in double slit experiment– Change in wavelength and phase change in a medium -

soap bubble– Interference from thin films

• Demos• Polling

Huygen’s Principle, Wavefronts and Coherence

E = Em sin(2!

"x # 2! f t)

k

�

E = Emsin(kx !"t)

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 = Em

sin(2!

"(x # c t))

In order to form an interference pattern, theincident light must satisfy two conditions:

(i) The light sources must be coherent. This means that theplane waves from the sources must maintain a constant phaserelation. For example, if two waves are completely out ofphase with φ = π , this phase difference must not change withtime.

(ii) The light must be monochromatic. This means that the lightconsists of just one wavelength λ = 2π/k .

Light emitted from an incandescentlightbulb is incoherent because thelight consists of waves of differentwavelengths and they do notmaintain a constant phaserelationship. Thus, no interferencepattern is observed.

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

out of phase

1. The selected rays 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 by180 degrees or πradians or λ/2

In between

Destructive interferenceConstructive interference

1) Lets consider the first case. How does light get out ofphase when selected rays travel different distances.

This leads to the famous Young’s double slit experiment

Young’s Double SlitInterference Experiment

m=0

m=1

m=1

m=2

m=2

θ

D

y

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 lengthsfrom each source• we have constructive interference if paths differ by any numberof full wavelengths• destructive interference if difference is half a wavelengthlonger or shorter

Constructive interference

Destructive interference

! = path length difference

!=d sin"

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! =y

L

y = L tan!

y is the distance to the m’th fringe

ym =m!L

d

d sin! = m" Maxima

d sin! = (m +1

2)" m=0,1,2,3..

ym= L tan!

0Lλ/d2Lλ/d3Lλ/d

0123

+/- ymm

Maxima

Lλ/2d3Lλ/2d5Lλ/2d7Lλ/2d

0123

+/- ymm

Minima

ym =(m +1 / 2)!L

d

sin! =m"

d

tan! ! sin!

ym= L sin! ym

Young’s Double SlitInterference Experiment

m=0

m=1

m=1

m=2

m=2

θ

D

y

Problem 13E.Suppose that Young’s experiment is performed with blue-green lightof 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4m 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=L!

d=(5.4 m)

500 "10-9

m

0.0012 m

S=0.00225 m=2.25 mm

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

What is the Intensity Distribution along the screen

The electric field at P is sum ofE1 and E2. The intensity isproportional to the square of thenet amplitude.

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

and incoherent sum is

I

For coherent sources, the cross term is non-zero. In fact, forconstructive 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 coherentsinusoidal plane waves. Let the electric field components of thewave 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 bygiving it a phase shift (ϕ) relative to E1

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

E1

E2

E0

E0

E

ωt

φ

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

The intensity I is proportional tothe square of the amplitude of thetotal electric field

I ! E2

E2= 4E

0

2cos

2(!

2)

Let I0= E

0

2 Then.. I = 4I0cos

2 1

2!

What about the intensity of light along the screen?

�

I = 4I0cos

2 1

2!

! =2"d

#sin$

2) How does a wave get out of phase when passing through differentmedia with different indices of refraction? You must consider thethickness of the two materials in numbers of wavelength N1 and N2.Note that the velocity and wavelength changes but not the frequency.

Path length difference =

c = f λ

Vacuum Vacuum

vn= f λn =f λ/n n=c/v

N2=

L

!n2

=Ln

2

!

N1=L

!n1

=Ln

1

!

( )!

12nnL "

( )!

12

12

nnLNN

"="

Concept of path length difference, phase and index ofrefraction

c = f λ

Vacuum Vacuum

vn= f λn =f λ/n n=c/v

Rays are in phase if where m=1, 2, 3..

Rays are out of phase if where m=1,2, 3

( )!

!m

nnL=

"12

( )!

! "#$

%&'

+=(

2

112m

nnL

n2n1

air water

EExplain using wavemachine

Demonstrate using rope

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

E stands for the Electric field of the wave

air 1.0

air 1.00

soap 1.30 L

eye

21

Reflection 180 deg phase change forRay 1when reflectingfrom the soap film surfacebut not ray 2

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

�

Suppose the soap thickness L is such that

2L =1

2

!

" #

$

% & 'n

�

2L = m +1

2

!

" #

$

% & 'n 2

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

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

Now consider the path length differences

�

!n

=!

n

First consider phase change upon reflection

Answer

soap 1.30

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

air 1.0

air 1.00L

eye

Transmission

21

�

2L = m( )!

n ! =

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?

39. 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 isoverhead at a region of the slick where its thickness is L=460 nm, forwhich wavelength(s) of visible light is the reflection brightestbecause of constructive interference?

air 1.0

Water 1.30Kerosene 1.20 L

21

180 deg phasechange

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

Scuba diver

air 1.0

Water 1.30Kerosene 1.20 L

2

1

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

�

2L = m +1

2

!

" #

$

% & 'n2

Solving for '

' =4n2L

2m + 1

�

! =4n2L

2m + 1=

4(1.2)(460)

1= 2208 nm m = 0

�

! =4n2L

2m + 1=

4(1.2)(460)

3= 736 nm m =1

�

! =4n2L

2m + 1=

4(1.2)(460)

5= 441.6 nm m = 2

Visible spectrum is 430 nm - 690 nm

27. S1 and S2 in Fig. 36-29 are pointsources of electromagnetic waves ofwavelength 1.00 m. They are in phaseand separated by d = 4.00 m, and theyemit at the same power.

(a) If a detector is moved to the rightalong the x-axis from source S1, atwhat distances from S1 are the firstthree interference maxima detected?

xdetector

Consider what the path difference is between the detector and S1 and the detector and S2

�

path difference = d2

+ x2! x = m" m = 1,2,3..

The solution for x of this equation is

For constructive interference we have

xdetector

.22xd +

Solve for x

�

x =d

2!m

2"

2

2m" for m =1, 2, 3,..

�

d2

+ x2! x = m"

d2+ x2 = m" + x Now square both sides

�

d2

+ x2

= ( m! + x)2

d2

+ x2

= m2!

2+ 2m!x + x

2 Now cancel x

2

�

d2

= m2!

2+ 2m!x solve for x

.2

222

!

!

m

mdx

"=

�

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 !m

2

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 traveldifferent distances to get to the points of minimum intensity and eachamplitude decreases in inverse proportion to the square of the distancetraveled. The intensity is not zero at the minima positions.

�

I1 =P0

4!x2

I2 =P0

4! (d2

+ x2)

I1

I2

=x

2

d2

+ x2

=(0.55)

2

42

+ (0.55)2

~1

64

m=3 m=2 m=1

�

x =d2! (m +

1

2)2"2

2(m +1

2)"

.

Where do the minima occur?

�

x =16 ! (m +

1

2)2

2(m +1

2).

�

path difference = d2

+ x2! x = (m +

12)" 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

!

!

m

mdx

"=

x =1! m

2(1 /16)

2m(1 / 4)

x =16 ! m

2

8 ! m

m = 4 x = 0

m = 3 x=16-9

24=

7

24m ! 0.33m

m = 2 x=16-4

16=

12

16m ! 0.75m

m = 1 x=16-1

7=

15

7m ! 2.0m