Lecture 2: Interference, Exercises of French

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Lecture 2: Interference

S

1

S

2

d

λλλλ

Today

Interference of sound waves

Two-slit interference

Superposing sine waves

If you added the two sinusoidal waves shown,what would the result look like?

• 1 . 0

0

• 0 . 5

0

0

. 0

0

0

. 5

0

1

. 0

0

0

100

200

300

400

500

600

700

800

900

1000

• 0 .5 0- 1 .0 0- 1 .5 0 - 2 .0 0

2 .0 01 .5 01 .0 00 .5 00 .0 0

0

100

200

300

400

500

600

700

800

900

1000

The sum of two sines having the same frequency is another sinewith the same frequency. Its amplitude depends on their relative phases.

Let’s see how this works.

Adding Sine Waves with Different Phases

Suppose we have two sinusoidal waves with

the same A

1

ω

and

k

y

1

= A

1

cos(kx -

ω

t)

and

y

2

= A

1

cos(kx -

ω

t +

φ

One starts at phase

φ

after the other:

1

1

cos

cos

2

cos

cos

2

2

A

A

β

α

β

α

α

β

−









=

















1

2

y

y

(

)

(

)

/ 2

kx

t

ω

φ

−

1

2

cos(

/ 2)

cos(

/ 2)

y

A

kx

t

=

−

Use this trig identity:

Spatial dependenceof 2 waves at t = 0:Resultant wave:

Amplitude

Oscillation

y = y

1

+y

2

φ

Example: Changing phase of the Source

Each speaker alone produces an intensity of I

1

= 1 W/m

2

at the listener:

Drive the speakers in phase. What is the intensity I at the listener?

Now shift phase of one speaker by 90

o

.What is the intensity I at the listener?

φφφφ

I = I

1

= A

1

2

= 1 W/m

2

I =

I =

Example: Changing phase of the Source

Each speaker alone produces an intensity of I

1

= 1 W/m

2

at the listener:

Drive the speakers in phase. What is the intensity I at the listener?

Now shift phase of one speaker by 90

o

.What is the intensity I at the listener?

φφφφ

I = I

1

= A

1

2

= 1 W/m

2

I = (2A

1

)

2

= 4I

1

= 4 W/m

2

I =

ACT 1:

Noise-cancelling Headphones

What must be the phase of the signal from the speaker relative to the

external noise? a.

b.

c.

π

d.

e.

π

What must be the intensity I

s

of the signal from the speaker relative to the

external noise I

n

a.

I

s

= I

n

b.

I

s

< I

n

c.

I

s

> I

n

Noise-canceling headphones work usinginterference. A microphone on theearpiece monitors the instantaneousamplitude of the external sound wave,and a speaker on the inside of theearpiece produces a sound wave tocancel it.

Solution

Destructive interference occurs when the waves are ±180˚ out of phase.180º =

π

radians!

What must be the phase of the signal from the speaker relative to the

external noise? a.

b.

c.

π

d.

e.

π

What must be the intensity I

s

of the signal from the speaker relative to the

external noise I

n

a.

I

s

= I

n

b.

I

s

< I

n

c.

I

s

> I

n

Noise-canceling headphones work usinginterference. A microphone on theearpiece monitors the instantaneousamplitude of the external sound wave,and a speaker on the inside of theearpiece produces a sound wave tocancel it.

Lecture 2, p.

Interference Exercise

The two waves at this point are“out of phase”. Their phasedifference

φ

depends on the path

difference

δ ≡

r

2

• r

1

.

The relative phase of two waves also depends on the relative distancesto the sources:

δ

φ

0 λ/4 λ/2 λ

Path

difference

Phase

difference

A = 2A

1

cos(

φ

r

2

r

1

λ

δ

=

π

φ

2

Each fraction

of a wavelength of

path difference gives

that fraction of 360º

(or 2

π

) of phase

difference:

Lecture 2, p.

δ

φ

I

0 λ/4 λ/2 λ

Path

difference

Phase

difference

A = 2A

1

cos(

φ

Solution

The two waves at this point are“out of phase”. Their phasedifference

φ

depends on the path

difference

δ ≡

r

2

• r

1

.

r

2

r

1

The relative phase of two waves also depends on the relative distancesto the sources:

λ

δ

=

π

φ

2

Each fraction

of a wavelength of

path difference gives

that fraction of 360º

(or 2

π

) of phase

difference:

Reminder: A can be negative.“Amplitude” is the absolute value.

2A

1

4I

1

2π

2A

1

4I

1

π/

√

A

1

2I

1

π

Summary

Interference of coherent waves

Resultant intensity of

two equal-intensity waves

of the same wavelength

at the same point in space:

I = 4 I

1

cos

2

For

unequal intensities

, the maximum and minimum intensities are

I

max

= |A

1

+ A

2

2

I

min

= |A

1

• A

2

2

The phase difference between the two waves may be due to a differencein their source phases or in the path difference to the observer, or both.The difference due to path difference is:

φ = 2π(δ/λ) where

δ

= r

2

• r

1

Note: The phase differencecan also be due to an index ofrefraction, because that willchange the wavelength.

In order to calculate I,we need to know

φ

.

Light - Particle or Wave?

Diffraction of light played an important historical role.

: French Academy held a science competition

Fresnel proposed the diffraction of light.

One judge, Poisson, knew light was made of particles, and thoughtFresnel’s ideas ridiculous; he argued that if Fresnel ideas werecorrect, one would see a bright spot in the middle of the shadow of adisk.

Another judge, Arago, decided to actually do the experimentN(our lecture demo)

Conclusion:Light

must

be a wave, since particles don’t diffract!

Transmission of Light

through Narrow Slits

Monochromaticlight sourceat a great distance,or a laser.

Slit pattern

Observationscreen

Lecture 2, p.

Double

slit interference

Light (waveleng

th

λ

) is i

ncident on a

two-slit (two narrow, rectangularopenings) apparatus:If either one of the slits is closed, aspread-out image of the open slit willappear on the screen. (The image isspread due to

diffraction

. We will

discuss diffraction in more detail later.)If both slits are open, we seeinterference “fringes” (light and darkbands), corresponding to constructiveand destructive

interference

of the

wave passing through the two slits.

Monochromatic light

(wavelength

λ

)

S

1

S

2

screen

Diffraction

profile

I

1

I

S

1

S

2

screen

Interference

fringes

Note: In the laser demo, there is littlevertical spread, because the laserspot is small in that direction.