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• Lecture 15 Interference Chp. 35

Opening Demo

Topics

Interference is due to the wave nature of light

Huygens principle, Coherence

Change in wavelength and phase change in a medium

Interference from thin films

Examples

Youngs Interference Experiment and demo

Intensity in double slit experiment

Warm-up problem

Demos

• Huygens Principle, Wavefronts and Coherence

E = Em sin(2!

"# 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

• Interference is the combination of two or more waves to forma composite wave, based on the principle of superposition

• 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 _ = ! , this phase difference must not change withtime.

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

• Once light is in phase there are three ways to

get light out of phase

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

2. Reflection from a medium with greater index of refraction

3. The selected rays travel different distances.

Now lets look at examples

• In Phase Out of Phase by

180 degrees or !

In between

• Concept of path length difference, phase and index of

refraction

Path length difference = Phase difference =

c = f "

Vacuum Vacuum

vn= f "n =f "/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

1 " is the same as 2! radian (rad), "/2 is the same as ! rad, etc.

!!

!!

2

2

2

1

1

1

LnLN

LnLN

n

n

==

==

( )!

12

12

nnLNN

"="

( )!

12nnL "

( )!

!m

nnL=

"12

( )!

! "#\$

%&'

+=(

2

112m

nnL

• Wave reflects 180 degrees out of phase when

n1 < n2

n2n1

air water

E

• air 1.0

air 1.00

soap 1.30 L

eye

2

1

Reflection

180 deg phase change

Thin film Interference Phenomenon: Reflection

E = Emsin(kx !"t)

E1 = Em sin(kx !"t + # )

E1 = !Em sin(kx !"t)

E2 = Em sin(k(x + 2L) !"t)

Suppose the soap thickness L is such that

2L =1

2

! " #

\$ % & 'n

E2 = Em sin(kx + k2L !"t) =

Emsin(kx +

2#

\$

\$

2!"t) =

E2 = Em sin(kx !"t + # )

2L = m +1

2

! " #

\$ % & 'n 2 where m = 0, 1, 2, ...

Constructive Interference

Now consider the path length differences

!n

=!

n

First consider phase change upon reflection

Show wave

demo

• soap 1.30

Thin film Interference Phenomenon

Transmission

air 1.0

air 1.00

L

eye

Transmission

2

1

2L = m( )!

n ! =

2nL

m

m = 1,2,3,4.

Constructive interference

No phase changes upon reflection

n =1.30

• 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 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?

Path difference between ray 1and ray 2 = 2L.

Phase changes cancel out

For constructive interference path difference

must = integral number of wavelengthsair 1.0

Water 1.30

Kerosene 1.20 L

2

1

180 deg phase

change

! =2n2L

m=

2 1.20( ) 460nm( )

m= 1104, 552, 368 nm

for m =1, m = 2, and m = 3 respectively

We note that only the 552 nm wavelength falls

within the visible light range.

2L = m( )!

n2

! =2n

2L

m

• (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

For transmission, ray 2 undergoes 180 deg phase shift upon reflection at the

Kerosene-water interface. Therefore, for constructive interference 2L= integral

number of wavelengths in n2 plus half a wavelength.

air 1.0

Water 1.30

Kerosene 1.20 L

2

1

2L = m +1

2

! " #

\$ % & 'n

2

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

Solve for '

2L = m +1

2

! " #

\$ % & 'n 2 where m = 0, 1, 2, ...

'n

='n

2

• We note that only the 441.6 nm

wavelength (blue) is in the visible range,

2L = m +1

2

! " #

\$ % & 'n2

Solving for '

' = 4n2L2m + 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

• The wave from S1 travels a distance x and the wave from S2 travels a

distance .22xd +

27. S1 and S2 in Fig. 36-29 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?

xdetector

.22xd +

The path difference is

d2

+ x2! x

• d2

+ x2

x

path difference = d2 + x 2 ! x = m" m = 1,2,3..

The solution for x of this equation is

For constructive interference we have

• 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 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

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 + 12)"

.

Where do the minima occur?

x =16 ! (m + 1

2)2

2(m + 12).

path difference = d2 + x 2 ! x = (m + 12)" m = 0,1,2,3

m=0 x=15.75 m

m=1 x =4.55 m

m=2 x=1.95 m

m=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