INTERFERENCE EFFECTS
When two or more waves travel in a medium, the resulting wave is the sum of the
displacements associated with each of the individual waves. This is known as the principle of
superposition.
If two waves are generated from different ends of a slinky spring it can be seen that waves
pass through each other. Where they meet superposition occurs. If the waves are on the
same side of the spring, displacements add together constructively to momentarily produce
a bigger wave. This is called constructive interference.
If the waves are on opposite sides of the spring, again displacements are added, but the
resultant has lower amplitude that the original waves. This is known as destructive
interference.
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The Essentials – Unit 4 Physics – Book 1
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Interference effects can be demonstrated with light, providing evidence to support the wavelike nature of light. These effects include non-reflective coatings on glass and colours from
oil films.
Optical glasses and camera lenses are frequently coated with a thin layer of clear plastic film
which is designed to reduce reflection. A small proportion of incident light always reflects
from the surface of glass/transparent plastic while most of the light passes through. By
coating the surface with a thin transparent film, a second surface can be provided from
which reflection can occur. How does creating two surfaces from which reflection occurs
actually reduce reflection?
If the film is

4
thick, the light reflecting from the second surface has to travel a total of an
extra half a wavelength compared to light reflecting from the first layer. This puts the rays
reflecting from the two surfaces half a wavelength out of phase, resulting in destructive
interference when they are superimposed. While rays actually reflect from both surfaces,
what is observed is less intense than the reflection from a single surface.
This principle is clearly demonstrated with monochromatic light (single wavelength),
however, the situation is more complicated if the incident radiation is white light which
consists of a range of wavelengths. Because the film is a set thickness, it will result in
cancellation of the wavelength that is four times the film thickness, but for wavelengths
greater or smaller, the cancellation will not be as complete, resulting in only partial
cancellation of some colours. Therefore some reflected light will be observed, and it
will be coloured, sometimes with a pink or green tinge.
Why does a film of oil on water reflect a spectrum of colours?
Other applications for principles of destructive and constructive interference of waves
includes music production, design of sporting equipment, bridge and building design to name
just a few.
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The Essentials – Unit 4 Physics – Book 1
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YOUNG’S INTERFERENCE
If light passes through a narrow slit, it diffracts, just like sound waves do.
If light behaved as we expect particles to behave, we would expect to see this:
Instead, we see this:
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Waves diffract.
The narrower the slit or the longer the wavelength, the greater the amount of diffraction.
If light behaved as we expect particles to behave, we would expect to see this:
 The School For Excellence 2011
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Instead, we see this:
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The Essentials – Unit 4 Physics – Book 1
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Further support for the wave model was provided in 1901 by Thomas Young. He passed
light from a single source through two parallel slits and produced an interference pattern of
alternating bands of constructive and destructive interference.
Interference is strictly a wave phenomenon.
The first slit, So, ensures the light passing through S1 and S2 is coming from a single source,
and is hence ‘in phase’.
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The Essentials – Unit 4 Physics – Book 1
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At any point P on the perpendicular bisector of the line joining two point sources transmitting
waves in phase, there will be constructive interference.
The reason is this: The sources are transmitting in phase. This means that as a crest is
leaving Source 1 (or S1), a crest is also leaving Source 2 (S2). Half a period later, a trough
will leave S1 and, simultaneously, a trough will leave S2. By implication, of course, the
sources are transmitting at the same frequency. Since the waves emitted by the sources
propagate through the same medium, they must propagate at the same speed. (And since
frequency and speed are the same, so must the wavelengths be the same.) Now, note that
the distance PS1 is the same as the distance PS2. Therefore, a crest from S1 will arrive at P
at the same time that a crest from S2 arrives at P. Similarly, when a trough from S1 arrives at
P, a trough from S2 arrives at P at the same time. Thus, constructive interference will always
occur at P and the amplitude of the resultant wave will be greater than that of the two
individual interfering waves.
And remember:
Large amplitude implies high intensity.
If P is moved closer to the sources along the perpendicular bisector, the intensity at point P
increases (simply because P is closer to the sources).
What would a graph of Intensity – distance along OP look like?
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If P is moved along the line RT, as in the following diagram, it will pass through a succession
of maxima and minima. Maxima are points at which constructive interference occurs, while
minima are points at which destructive interference occurs.
R
S1
PS1
S
P
S2
PS2
T
There will be a maximum at S, but if P is moved away from S, say towards T, the distances
PS1 and PS2 will no longer be equal (because PS1 becomes longer while PS2 becomes
shorter). If PS1 becomes longer than PS2 by a distance of /2, then the wave from S1 will be
half a wavelength behind the wave from S2 by the time the waves reach point P.
Wave 1
Wave 2
In this situation, destructive interference occurs – the waves almost completely cancel each
other and the intensity is thus close to zero. This is how a minimum is created.
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If P is moved even further towards T, then the difference in path lengths becomes even
greater. If the path difference becomes equal to a whole wavelength, then the wave from S1
will be a whole wavelength behind the wave from S2 by the time the waves reach point P.
Wave11
Wave
Wave 2
In this case, we again have constructive interference (so a maximum is created).
Continuing to move P even further along causes the path difference to become even greater.
When it becomes equal to 3/2 destructive interference occurs again.
Wave 11
Wave
Wave 2
If we move P even further, so that the path difference becomes 2, then we yet again have
constructive interference (so another maximum is created).
Wave 1
Wave 2
And so on . . .
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In summary:
There will be constructive interference if the Path Difference (For Example: PS1 - PS2) = 0,
, 2, 3, . . .
Path difference for a maximum:
P.D.  n  , where n is the nth maximum from the central maximum.
In this case, a crest from one source arrives at P at the same time that a crest arrives at
P from the other source. Half a period later, two crests will combine at P.
There will be destructive interference if the Path Difference (For Example: PS1 - PS2) = /2,
3/2,
5/2,
.
.
.
Path difference for a minimum:
1

P.D.   n    , where n is the nth minimum from the central maximum.
2

In this case, a crest from one source arrives at P at the same time that a trough arrives at P
from the other source.
If we now travel along the line R-T from an earlier diagram we obtain an intensity diagram as
follows:
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The maxima and minima are arranged as follows:
Min.
2nd minimum, P.D. = 
Max.
1st maximum, P.D. = 
Min.
1st minimum, P.D. = 
Max.
central maximum, P.D. = 0
Min.
1st minimum, P.D. = 
Max.
1st maximum, P.D. = 
Min.
2nd minimum, P.D. = 
Source
Source1 1
Source 22
Source
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The distance between adjacent minima (or maxima), x, depends on the wavelength
and on the geometry of the situation. Knowledge of the following relationship may
prove useful.
R
Min.
Source 1
Source 1
Max.
Min.
d
Max.
x
Min.
Max.
Source2 2
Source
Min.
T
L
x 
L
d
Where L and d are as shown in the above diagram.
This formula demonstrates the relationship between the nodal spacings and distance
between slits, and distance from the source.
i.e.
x  
x  L
x 
1
d
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QUESTION 1
Red light of wavelength 600 nm falls on two narrow slits whose separation is 1.2  104 m .
An interference pattern is formed on a screen 1.6 m away and the observed nodal pattern
has dark bands spaced 8 mm apart.
(a)
Each slit is made narrower but the separation between the centres of the two slits is
kept the same as before. How does the pattern now appear?
(b)
What would be the effect on the band pattern if the slits are brought closer together?
(c)
The slits are returned to their original width and separation. Blue light is substituted for
the red light. Compared to the original pattern, the spacing between the dark bands will
be the same, decreased or increased?
(d)
White light is now used. How will the bands appear?
(e)
The slit separation is halved compared to the original separation. What is the spacing
between the dark bands now?
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QUESTION 2
A double slit arrangement is illuminated by light that consists of two wavelengths, one of
which is known to be 450 nm . The interference pattern on a screen shows that the fifth dark
fringe for the known wavelength coincides with the fifth bright fringe for the known
wavelength. What is the unknown wavelength?
Solution
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