Chapters 11, 12, 24

The Principle of Linear Superposition
When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.

The Principle of Linear Superposition
When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.

The Principle of Linear Superposition
THE PRINCIPLE OF LINEAR SUPERPOSITION
When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves.

Constructive and Destructive Interference of Sound Waves
When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference.
λ = 1 m

Constructive and Destructive Interference of Sound Waves
When two waves always meet condensation-to-rarefaction, they are said to be exactly out of phase and to exhibit destructive interference.
λ = 1 m

Constructive and Destructive Interference of Sound Waves
Noise-cancelling headset
Outside noises are picked up by a microphone and electronics produces an out-of-phase version of the noise which cancels out the original noise.

Constructive and Destructive Interference of Sound Waves
If the wave patterns do not shift relative to one another as time passes, the sources are said to be coherent .
For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3, . . ) of wavelengths leads to constructive interference; a difference in path lengths that is a half-integer number
( ½ , 1 ½ , 2 ½ , . .) of wavelengths leads to destructive interference.

Constructive and Destructive Interference of Sound Waves
Example: What Does a Listener Hear?
Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at C, which is 2.40 m in front of speaker B.
Both speakers are playing identical 214 Hz tones, and the speed of sound is 343 m/s.
Does the listener hear a loud sound, or no sound?

Constructive and Destructive Interference of Sound Waves
Calculate the path length difference.
( 3 .
20 m ) ( 2 .
40 m ) 2
− 2 .
40 m = 1.60
m
Calculate the wavelength.
= f v
=
343 m
214 Hz s
= 1 .
60 m
Because the path length difference is equal to an integer (1) number of wavelengths, there is constructive interference, which means there is a loud sound.

Constructive and Destructive Interference of Sound Waves
Conceptual Example: Out-Of-Phase Speakers
To make a speaker operate, two wires must be connected between the speaker and the amplifier. To ensure that the diaphragms of the two speakers vibrate in phase, it is necessary to make these connections in exactly the same way. If the wires for one speaker are not connected just as they are for the other, the diaphragms will vibrate out of phase. Suppose in the figures (next slide), the connections are made so that the speaker diaphragms vibrate out of phase, everything else remaining the same. In each case, what kind of interference would result in the overlap point?

Constructive and Destructive Interference of Sound Waves

Young ’ s Double Slit Experiment
In Young ’ s experiment, two slits acts as coherent sources of waves, e.g. light waves.
Light waves from these slits interfere constructively and destructively on the screen.

Young ’ s Double Slit Experiment
The waves coming from the slits interfere constructively or destructively, depending on the difference in distances between the slits and the screen.

Young ’ s Double Slit Experiment
Δ ℓ = d sin θ order of the fringe
Bright fringes of a double-slit
Δ l = m λ ⇒ sin θ = m
λ d m = 0,1, 2, 3,....
Dark fringes of a double-slit
Δ l = ( m + 1
2
) λ ⇒ sin θ = ( m + 1
2
)
λ d m = 0,1, 2, 3, …

Young ’ s Double Slit Experiment
Example: Young ’ s Double-Slit Experiment
Red light (664 nm) is used in Young ’ s experiment with slits separated by 0.000120 m. The screen is located a distance 2.75 m from the slits.
Find the distance on the screen between the central bright fringe and the third-order bright fringe.

Young ’ s Double Slit Experiment
= sin − 1
⎛
⎜
⎝ m
d
⎞
⎟
⎠
= sin − 1
⎛
⎜⎜
⎝
3
664
1.20
× 10
× 10
− 9
− 4 m m
⎞
⎟⎟
⎠
= 0 .
951 !
y = L tan
= ( 2 .
75 m ) tan ( 0 .
951 !
)
= 0 .
0456 m

Beats
Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats.

Beats
The beat frequency is the difference between the two sound frequencies. f beat
= f
1
f
2
In this case, f beat
= 440
438 = 2 Hz

Refraction Wave transmitted from one medium to another à Refraction
In this figure, v
1
> v
2 f is same in each medium but v and λ are different
λ = v f
⇒ sin θ
1
=
λ
1 h
= v
1 fh sin θ
1 v
1
=
1 fh
= sin θ
2 v
2
λ
1
= v
1 f and sin θ
2
=
λ
2 h
= v
2 fh
λ
2
= v
2 f
v
2 sin θ
1
= v
1 sin θ
2
Snell’s
Law

Example: A sound wave in air is incident on a pool of water at an angle of 10 o with respect to the normal to the surface. Find the angle of the wave with respect to the normal to the surface after it is transmitted into the water. v
2 sin θ
1
= v
1 sin θ
2
⇒ v water sin θ air
= v air sin θ water sin θ water
= v water v air sin θ air
=
1482 sin10 o
343
= 0.750
∴ θ water
= sin − 1 0.750
= 49 o

Chapters 11 and 24

17.3 Diffraction
The bending of a wave around an obstacle or the edges of an opening is called diffraction .

Diffraction
Diffraction is the bending of waves around obstacles or the edges of an opening.
Huygens ’ principle
Every point on a wave front acts as a source of tiny wavelets that move forward with the same speed as the wave; the wave front at a latter instant is the surface that is tangent to the wavelets.

Diffraction
Approximate expression for angular bending of waves of wavelength λ around an object or opening of size D
D

Shadow region infringed upon by diffracting waves
Diffraction around a hill allows radio reception in the valley

Stickies
Example of diffraction of electromagnetic waves around a hill.
An AM radio station broadcasts at a frequency is 1230x10 3 Hz, and a microwave transmitter broadcasts waves at 2.50 GHz. Approximately how much will each of the waves be bent when they encounter a hill that is 1000 ft high? How far behind the hill must a receiver antenna be located to pick up the AM radio broadcast? Will it pick up the microwave signal?
4/12/14, 12:04 PM
Page 1 of 1

Stickies 4/12/14, 12:04 PM
AM
} h
µ W
=
= f
AM v v f
µ W
=
3.00
10 8 m s
1230
10 3 Hz
= 244 m
=
3.00
10 8 m s
2.5
10 9 Hz
= 0.12 m
θ
AM
≈
λ
AM h d
244 m
=
1000 ft × 0.305 m ft
= 0.80 rad → 46 o A lot of diffraction!
θ
µ W
≈
λ
µ W h
0.12 m
=
1000 ft × 0.305 m ft
= 3.9
× 10 − 4 rad → 0.023
o hardly any diffraction!
AM
= h d h
d =
AM
=
=
o
Page 1 of 1

Diffraction
W

Diffraction
Light waves also exhibit diffraction effects.

Diffraction
This top view shows five sources of Huygens ’ wavelets.

Diffraction
These drawings show how destructive interference leads to the first dark fringe on either side of the central bright fringe.
Dark fringes for single-slit diffraction sin
= m
W m = 1, 2, 3......

Example for single-slit diffraction. Light passes through a slit and shines on a flat screen that is located L = 0.40 m away. The wavelength of the light in a vacuum is λ = 410 nm. The distance between the midpoint of the central bright fringe and the first dark fringe is y . Determine the width 2y of the central bright fringe when the width of the slit is (a) W = 5.0 x 10 -6 m and (b) W = 2.5 x 10 -6 m. first dark fringe
W
θ
θ y midpoint of central bright fringe y first dark fringe
L
( ) θ = sin − 1
$
"
#
λ
W
%
&
' = sin − 1
$
"
#
410 × 10 − 9
5.0
× 10 − 6
%
&
' = 4.7
o
2 y = 2 L tan θ = ( ) tan 4.7
o
= 0.066
m
( ) 2 y = 0.13
m à The w idth of the central bright fringe is greater for smaller W

D
Stickies
4/12/14, 2:36 PM
Diffraction pattern of light incident on a circular aperture sin
= 1 .
22
D
θ
Page 1 of 1

Diffraction
Example.
A 1500 Hz sound and a 8500 Hz sound each emerges from a loudspeaker through a circular opening that has a diameter of 0.30 m. Find the diffraction angle θ for each sound (assume v sound
= 343 m/s).
λ = v f
sin θ = 1.22
λ
D
= 1.22
v fD
1500 Hz:
8500 Hz:
θ = sin − 1
$
#
"
1.22
v fD
%
' = sin
&
− 1
*
)
(
1.22
*
343
( 1500 ) ( 0.30
)
+
-
-
= 68 o
θ = sin − 1
$
#
"
1.22
v fD
%
' = sin
&
− 1
*
)
(
1.22
*
343
( 8500 ) ( 0.30
)
+
-
-
= 9.4
o

Diffraction
Diffraction pattern formed by an opaque disk.