Other Applications of Two-Point Source Interference
As is often the case, physics is not a science that is
restricted to the sterile confines of a laboratory. Physics is naturally and
frequently seen by any observer who glances at the world around them. An
informed physics student should see physics in action on a daily basis and should be
able to exclaim without embarrassment - "that happens because of
physics." Indeed, the physics concepts and principles that we study in the
Physics Classroom Tutorial simply emerge from the phenomenon that are in our
world of touch and see and feel. These physics concepts and principles are
simply humankind's attempt to explain the observable. So where in this world do
we observe two-point source interference? Where can
we experience the phenomenon that light taking two paths from two locations to
the same point in space can undergo constructive and destructive interference?
There are several answers to these questions and they will be discussed in this
last section of Lesson 3.
The
Big Idea
Before identifying and explaining real-world examples of
two-point source interference, the big idea behind it ought to be reviewed. In
the second part of Lesson 3, it was noted that two coherent waves
traveling along two different paths to the same point will interfere
constructively if there is a difference in distance traveled that
is equivalent to a whole number of wavelengths. And similarly, two coherent waves
traveling along two different paths to the same point will interfere
destructively if there is a difference in distance traveled that
is equivalent to a half number of wavelengths. The difference in distance traveled by the waves from the two sources to the
single point is referred to as the path difference. That is,
|
Constructive Interference: |
PD = m • λ |
where m = 0, 1, 2, 3, 4, ... |
|
|
|
|
|
Destructive Interference: |
PD = m • λ |
where m = 0.5, 1.5, 2.5, 3.5,
... |
These principles were presented to explain the two-point
source interference patterns that are characteristic of Young's experiment and
a wavelength measurement. Yet, these principles are more general in the sense
that they can explain any physical situation in which waves take two different
paths from two coherent sources to the same point. Such coherent waves will undergo interference. And if the
difference in distance traveled is a whole
number of wavelengths, then the interference will occur in such a way that the
two waves will constructively reinforce each other. Likewise, if the difference
in distance traveled is a half number of
wavelengths, then the interference will occur in such a way that the two waves
will destroy each other.
The
Applications
Given the above principle, the clue to finding two-point
source interference in the real world would be to look for situations in which
waves from two coherent sources travel along two different paths to the same point. Since the
two waves must be coherent, it is best that they can be traced to the same
source, but separated into two paths at some point due to passage through two
openings or reflection off a barrier. A common example of this involves the
interference of radio wave signals that occur at the antenna of a home when
radio waves from a very distant transmitting station take two different paths
from the station to the home. This is relatively common for homes located near
mountain cliffs. In such an instance, waves travel directly from the
transmitting station to the antenna and interfere with other waves that reflect
off the mountain cliffs behind the home and travel back to the antenna.
In this case, waves are taking two different paths from the
source to the antenna - a direct path and a reflected path. Clearly, each path
is represented by a different distance traveled from
the source to the home, with the reflected pathway corresponding to the longer
distance of the two. If the home is located some distance d from the
mountain cliffs, then the waves that take the reflected path to the home will
be traveling an extra distance given by the expression 2•d. The 2 in this expression is due to the fact that the waves taking the
reflected path must travel past the antenna to the cliffs (a distance d) and then back to the antenna from the cliff (a second distance d). Thus, the path difference of 2•d results in destructive interference whenever it
is equal to a half number of wavelengths.
|
Destructive: |
PD = 2 • d = m • λ |
where m = 0.5, 1.5, 2.5, ... |
Since radio stations transmit their signals at a specific and
known frequency, the wavelengths of these light waves can be determined by
relating it to the transmitted frequency and the light speed (3 x 108 m/s).
This same principle of destructive interference of radio
signals can be observed when waves from the transmitting source reflect off
airplanes that are flying overhead. In this case, there will be a difference in
the distance traveled by the wave moving along
the direct path to the antenna and the wave that travels along the reflected
path off the plane to the antenna. While the interference is momentary (the
plane does not remain in a stationary location), it is nonetheless observable.
If we suppose that the plane is directly overhead and that the distance from
the antenna to the transmitting station is relatively large, then the path
difference is simply the height of the plane above the house.
There are a variety of potential heights that lead to
destructive interference. Each height satisfies the criteria that destructive
interference will occur when the path difference is equal to a half
number of wavelengths.
|
Destructive: |
PD = height = m • λ |
where m = 0.5, 1.5, 2.5, ... |
By substituting various values of m into the equation, the
variety of potential heights can be determined.
A final application of two-point source interference that is
discussed here involves the interference of sound waves. All waves, whether
light waves, sound waves or water waves, exhibit the same characteristics
properties. Waves refract, reflect, diffract and interfere in the same manner
according to the same rules. And as such, two coherent sound waves traveling
along different paths to the same point will destructively interfere provided
that the path difference is equal to a half number of
wavelengths.
A relatively common demonstration of sound wave interference
can be performed with two speakers in a large room such as an auditorium. If
both speakers are hooked up to the same sound source producing a monotone
sound, then a sound interference pattern can be observed within the room. If
one were to walk along a line parallel to the line connecting the speakers,
there would be clear locations of destructive and constructive interference. At
locations of destructive interference, the sound intensity would become weak,
perhaps even barely noticeable. At locations of constructive interference, the
sound intensity would be amplified. These locations would be observed along the
line at which one walks at nearly regular intervals of distance.
Imagine that there were simply two speakers in a large auditorium set up
so that there were certain seats that were located along nodal lines for
particular frequencies. When those particular frequencies were sounded out by
the speakers, the people in the seats along those nodal lines would be at
locations of destructive interference and would not hear the sound from the
speakers (at least during the time that those particular frequencies were being
sounded out). Acoustic engineers must take these factors into account when
designing auditorium walls and ceilings. The walls and ceilings must act as
additional "sources" of sound as they reflect sound waves to all
parts of the room so that even when waves traveling directly from the speakers
to a seat undergo destructive interference, there is still sound reflecting off
walls and ceilings to the same seats. In this sense, the walls and ceilings of
a well-designed auditorium serve to reflect sound in such a way as to fill in those
locations where destructive interference might be occurring.
Why
Does Light From Two Light Bulbs Not Form an
Interference Pattern?
When the topic of Young's experiment and interference is
discussed, the question is often raised: Why doesn't light from two light bulbs
undergo interference to produce a two-source interference pattern? Why do I not
observe bright and dark fringes along my living room wall from the interference
of light from two lamps? The explanation pertains to the nature of ordinary
incandescent bulbs and to the necessity of coherent sources. Recall from earlier in
this lesson that the importance of coherent light sources was emphasized. Coherent
light sources are sources of light that produce waves that have a constant
phase difference between them over a significant duration of time. Two waves
may have the same frequency and wavelength but be offset from each other in
that they are at different points in a complete cycle. For instance, one wave
could be at a crest position just prior to the moment in time when the other
source is creating a crest. They are said to be out of phase. Yet if they
maintain the same difference in phase, they are considered coherent light
sources. Even if the sources of light do not stay in step with each
other, as long as the amount by which they are out of step remains the same
over time, the light sources are said to be coherent. To be coherent, two waves
must have the same frequency and there must not be any disruption of their
cycle over the course of time.
In an incandescent bulb, the vibrations of electrons within
the filament lead to the production of the electromagnetic wave. Several
million times in a second, there are small disruptions of the filament that
result in discontinuities in the waves that they produce. Ultimately, two
different incandescent bulbs are unable to produce light waves that maintain a
constant difference in phase over time. As a result, any interference pattern
that does occur will not endure for a sufficient length of time to allow the
human eye to observe the pattern.
Check Your Understanding
1. Anna Litical is
listening to FIZX - 1040 kiloHertz on the
dial. FIZX broadcasts from a location of about 78 kilometers from
her home. Regrettably for Anna, the presence of a long mountain range with
steep cliffs reflects the signal to her home and causes destructive
interference. The mountain cliffs are located directly behind her home relative
to a line drawn from the broadcasting station to her home. What is the minimum
distance that the mountain cliffs are located behind her home? (Assume that the
reflected wave does not undergo a phase change upon reflection off the
mountain.)
Answer: 72.1 m
An important first step is to determine the wavelength of the
radio waves using the v = f • λ where the v value is the speed of light (3
x 108m/s). Using 1040 kiloHertz or 1.04 x 106 Hz
as the frequency, the wavelength is calculated as
λ = v / f = (3 x 108 m/s) / (1.04 x 106 Hz)
= 288.46 m
Destructive
interference occurs when the path difference is equal to a half-number of
wavelengths. The wave traveling the greatest distance must travel past the
house to the mountain a distance of d and then back to the house another distance of d. Thus, the path difference
is d + d or 2•d.
Destructive interference will occur if the difference in distance traveled for the direct path compared to the reflected
path is some half number of wavelengths.
|
PD = 2•d = m • λ |
where m = 0.5, 1.5, 2.5, ... |
By substituting values of wavelength and m = 0.5, 1.5, 2.5, ... into the above equation, various possible
values for the separation distance between the house and the mountain cliffs
can be determined. The minimum distance occurs when m = 0.5 is used:
2 • d = 0.5 •
(288.46 m)
d = 0.5 • (288.46 m) / 2
d = 72.1 m
2. Anna Litical is
listening to WFIZ when she observes that destructive interference occurs due to
reflection of radio waves off an overhead plane. Suppose that destructive
interference is observed for plane heights of 161 meters, 207 meters, and 253
meters directly above her home (in additional to other distances as well). WFIZ
broadcasts from a location of about 59 kilometers from
her home. Determine the frequency at which WFIZ broadcasts their radio signals.
(Assume that the reflected wave does not undergo a phase change upon reflection
off the plane.)
Answer: 6.52 x 106 Hz or 6.52 MHz
The solution strategy to this problem involves using the stated
heights to determine the wavelength of the waves which are interfering at the
antenna. Once the wavelength is determined, the frequency can be calculated
using the v = f • λ equation.
The stated heights each satisfy the criteria that the path
difference (between the reflected pathway off the plane and the direct pathway
straight to the antenna) is equal to a half number of
wavelengths. Since the transmitting station is a relatively far distance from
the antenna, the path difference is simply the heights of the plane above the
house.
So each of the heights corresponds to m•λ where
m can be 0.5, 1.5, 2.5, 3.5, ... . The difference
in heights between two adjacent plane positions is then the difference between
using (n + 0.5)•λ and (n + 1.5)•λ as the
path difference. A glance at the three positions shown (161 meters, 207 meters, and
253 meters ) reveals that the
difference in heights is 46 meters. Thus, the wavelength is 46 meters. In other
words, the path difference for the first height is 3.5•λ or 161 meters.
The path difference for the second height is 4.5•λ or 207 meters. And the
path difference for the third height is 5.5•λ or 253 meters.
Once the wavelength is determined, the frequency can be computed:
f = v / λ
f = (3 x 108 m/s) / (46 m)
f = 6.52 x 106 Hz
or 6.52 MHz
3. Anna Litical is
listening to WBBM - 780 KHz on the radio dial - when she observes destructive
interference occurs due to the reflection of radio waves off a plane that is
directly above her home. WBBM broadcasts from a location of about 40 miles from
her home. Determine the closest possible distance that the plane could be
overhead and determine the next three possible heights of the plane. (Assume
that the reflected wave does not undergo a phase change upon reflection off the
plane.)
Answers: 192 m , 577 m , 962 m , and 1346 m
An important first step is to determine the wavelength of the
radio waves using the v = f • λ where the v value is the speed of light (3
x 108m/s). Using 780 kiloHertz or 7.80 x 105 Hz
as the frequency, the wavelength is calculated as
λ = v / f = (3 x 108 m/s) / (7.80 x 105 Hz)
= 384.62 m
Destructive interference occurs when the path difference is
equal to a half-number of wavelengths. The wave traveling the
greatest distance must travel from the transmitting tower to the plane above
the house and then reflect off the plane down to the house. This wave travels
an extra distance of h compared
to the wave which travels from the transmitting tower directly to the house.
Thus, the path difference is h.
Destructive
interference will occur if the difference in distance traveled for
the direct path compared to the reflected path is some half number of
wavelengths.
|
PD = h = m • λ |
where m = 0.5, 1.5, 2.5, ... |
By substituting values of wavelength and m = 0.5, 1.5, 2.5, ... into the above equation, various possible
values for the height of the plane above the house can be determined. The
minimum height occurs when m = 0.5 is used:
h = 0.5 •
(384.62 m)
d = 192 m
The next three possible heights can be determined using the next
three half numbers - 1.5, 2.5, and 3.5.
Values for the next three heights are:
h2 =
1.5 • (384.62 m) = 577 m
h3 = 2.5 • (384.62 m) = 962 m
h4 =
3.5 • (384.62 m) = 1346 m