Lecture physics a2 interference huynh quang linh

Lecture 2 Interference S3 S2 P Incident wave (wavelength l) y L d S1 Content  Interference of Sound waves  Two Slit Interference of Light Young Interference  Phasors  Multiple Slit Interference In[.]

Lecture 2: Interference

S3

S2

P

Incident wave (wavelength l )

y

L

d

S1

Content:

Interference of Waves:

and time (single w ), they lead to interference:

 Add amplitudes (e.g., pressures or electric fields).

 What we observe however is Intensity (absorbed power).

I = A 2

Stereo speakers: Listener:

y 2 = A 1 cos(kx - wt + )

y 1 = A 1 cos(kx - wt)

y +y = 2A cos( / 2) cos(kx   w   t / 2)

A

 = 0:waves add “in phase” (“constructive”)  I = |2 A 1 | 2 = 4|A 1 | 2 = 4I 1

 = p:waves add “out of phase” (“destructive”)  I = |2 A1*0|2 = 0

Interference Exercise

The two waves at this point are

“out of phase” Their phase difference  depends on the path difference d  r2 - r1.

The relative phase of two waves also depends on the relative

distances to the sources:

0

l/4

l/2

l

Path

difference difference Phase

A = 2A1cos(/2)

r1

l

d

= p



2

Each fraction

of a wavelength of path difference gives that fraction

of 360º (or 2 p ) of phase difference:

d 

I

0

l/4

l/2

l

Path

difference differencePhase

A = 2A1cos(/2)

Solution

The two waves at this point are

“out of phase” Their phase difference  depends on the path difference d  r2 - r1.

r1

The relative phase of two waves also depends on the relative

distances to the sources:

l

d

= p



2

Each fraction

of a wavelength of path difference gives that fraction

of 360º (or 2 p ) of phase difference:

Reminder: A can be negative.

“Amplitude” is the absolute value.

0 2A1 4I1

2p 2A1 4I1

p/2 2A1 2I1

Amplitude vs Intensity (for 2 interfering waves)

cos(/2) cos 2 (/2)

Note: What is the average intensity?

Plot here as a function of 

0 l 2l 3l 4l 5l



A = 2A 1cos(/2)

I = 4A 1 2cos2(/2)



0 2p 4p 6p 8p 10p

d

Constructive Interference

Destructive Interference

2A1

4A12

Iave = 4I1*0.5 = 2I1

1) Compute path-length difference: d = r 2 - r 1 =

2) Compute wavelength: l =

3) Compute phase difference (in degrees):  =

4) Write a formula for the resultant amplitude:

5) Compute the resultant intensity, I =

Sound velocity in air: v = 330 m/s

r 1

3 m

4 m

Each speaker alone produces intensity I 1 = 1 W/m 2 at the listener, and f = 900 Hz Drive the speakers in phase Compute the

intensity I at the listener:

 = 2p(d/l)

with d = r 2 – r 1

l

d

= p



2

Sound wave example:

1) Compute path-length difference: d = r 2 - r 1 = 1 m

2) Compute wavelength: l = v/f = (330 m/s)/(900 Hz) = 0.367 m

3) Compute phase difference (in degrees):  = 360 (d/l) = 360(1/0.367) = 981

4) Write a formula for the resultant amplitude: A = 2A 1 cos(/2) , A 1 = I 1

5) Compute the resultant intensity, I = 4 I 1 cos 2 (/2) = 4 (1 W/m 2 ) (0.655) 2

Sound velocity: v = 330 m/s

r 1

3 m

4 m

 = 2p(d/l)

with d = r 2 – r 1

l

d

= p



2

Sound wave example:

Each speaker alone produces intensity I1 = 1 W/m2 at the listener,

and f = 900 Hz Drive the speakers in phase Compute the intensity

I at the listener:

What happens to the intensity at the listener if we decrease the frequency f? (Recall the phase shift was 981.)

a decrease

b stay the same

c increase

Exercise 2 : Speaker interference

r 1

What happens to the intensity at the listener if we decrease the

frequency f? (Recall the phase shift was 981.)

a decrease

b stay the same

c increase

Exercise 2 : Speaker interference

r 1

f decreases

 l increases

 d/l decreases

  decreases  I decreases (see figure)

I



0 360 720 900

 = 981 

waves at the same point in space is:

I = 4 I1cos2(  /2)

minimum intensities are

Imax = |A1 + A2|2 Imin = |A1 - A2|2

be due to a difference in their source phases or a difference in the path lengths to the observer

In the latter case:

 = 2p(d/l)