INTERFERENCE AND DIFFRACTION

1. Two waves (of any kind) add up together, with the net result being the simple sum of the
two waves. Consider two waves, both of the same frequency, shown below. If they start
together (i.e. are with each other) then the net amplitude is increased. This is
called . But if they start at different times (i.e. are
with each other) th
in phase
constructive interference out of phase
en the net amplitude is decreased. This is called destructive interference.
In the example above, both waves have the same frequency and amplitude, and so the
resulting amplitude is doubled (constructive) or zero (destructive). But interference
occurs for any two waves even when their amplitudes and frequencies are different.
2. Although any waves from different sources interfere, if one
wants to observe the interference of light then it is necessary
to have a source of light. Coherent means that both
coherent
waves should have a fixed phase relative to each other. Even
with lasers, it is very difficult to produce coherent light from
two separate sources. Observing interference usually requires
taking two waves from a single source, with each going along a
different path. In the figure, an incoherent light source
illuminates the first slit. This creates a uniform and coherent
ill
1 2
umination of the second screen. Then waves from the slits
S and S meet on the third screen and create a pattern of
alternating light and dark fringes.
3. Wherever there is a bright fringe, constructive interference has occured, and wherever
there is a dark fringe, destructive interference has occured. We shall now calculate
where on the third
1 2
screen the interference is constructive. Take any point on the third
screen. Light reaches this point from both S and S , but it will take different amounts
of time to get there. Hence there will be a phase difference that we can calculate. Look
at the diagram below. You can see that light from one of the slits has to travel an extra
distance equal to sin , and so the extra amount
of time it takes is ( sin ) / . There will be
constructive interference if this is equal to ,2 ,
3 , (remember that the time period is
d
d c
T T
T
θ
θ
⋅ ⋅ ⋅ inversely
related to the frequency, 1/ , and that ).
We find that sin sin ,
where 1,2,3, What about for destructive
interference? Here the waves will
T c
d nT n d n
c
n
ν λν
θ
θ λ
ν
= =
= = ⇒ =
= ⋅ ⋅ ⋅
cancel each other
if the extra amount of time is ,3 ,5 , The
2 2 2
1 condition then becomes sin ( ) .
2
T T T
d θ n λ
⋅ ⋅ ⋅
= +
4. Example: two slits with a separation of 8.5× 10-5m
create an interference pattern on a screen 2.3m away.
If the 10 bright fringe above the central is a linear
distance of 12 cm from it,
n =
1 1
what is the wavelength of
light used in the experiment?
Answer: First calculate the angle to the tenth bright
fringe using tan . Solving for gives,
tan tan 0.12
2.3
y L
y m
L m
θ θ
θ − −
=
= ⎛⎜ ⎞⎟= ⎛⎜ ⎞⎟
⎝ ⎠ ⎝ ⎠
= 3.0􀁄. From this,
5
sin 8.5 10 sin(3.0 ) 4.4 10 7 440 (nanometres).
10
d m nm
n
λ θ
−
− ⎛ × ⎞
= =⎜ ⎟ = × =
⎝ ⎠
􀁄
5. When a wave is reflected at the interface of
two media, the phase will not change if it
goes from larger refractive index to a smaller
one. But for smaller to larger, there will be
phase change of a half-wavelength. One can
show this using Maxwell's equations and
applying the boundary conditions, but this
will require some more advanced studies.
Instead let's just use this fact below.
When light falls upon a thin film of soapy water, oil, etc. it is reflected from two surfaces.
On the top surface, the reflection is with change of phase by whereas at the lower
surface t
π
here is no change of phase. This means that when waves from the two surfaces
combine at the detector (your eyes), they will interfere.
To simplify matters, suppose that you are looking at the thin film almost directly from
above. is the index of refraction for the medium. Then,
The condition for destructive interf
Here n
erence is:
The condition for constructive interference is : 2 ( 1) ( 0,1,2,...).
2
Prove it!
2 ( 0,1,2,...)
nd m m
nd m m
λ
λ
= + =
= =
6. Interference is why thin films give rise to colours.
A drop of oil floating on water spreads out until it
is just a few microns thick. It will have thick and
thin portions. Thick portions of any non-uniform
thin film appear blue because the long-wavelength
red light experiences destructive interference.
Thinner portions appear red because the shortwavelength
blue light interferes destructively.
7. The bending of light around objects
(into what would otherwise be a shadowed region)
is known as diffraction. Diffraction occurs when
light passes through very small apertur
Diffraction :
es or near
sharp edges. Diffraction is actually just interference,
with the difference being only in the source of the
interfering waves. Interference from a single slit, as
in the figure here, is called diffraction.
180o phase change
0o phase change
n > 1 d
We can get an interference pattern with a single slit provided its size is approximately
equal to the wavelength of the light (neither too small nor too large).
8. Let's work out the condition necessary for diffraction of
light from a single slit. With reference to the figure,
imagine that a wave is incident from the left. It will
cause secondary waves to be radiated from the edges
of the slit. If one looks at angle , the extra distance
that the wave emitted from the lower slit must travel is
W sin . If this is a multiple of the wave
θ
θ length , then
constructive interference will occur. So the condition
becomes sin with 1, 2, 3... So, even
from a single slit one will see a pattern of light and
dark fringes
W m m
λ
θ = λ =± ± ±
when observed from the other side.
-6
9. Light with wavelength of 511 nm forms a diffraction pattern after passing through a
single slit of width 2.2×10 m. Find the angle associated with (a) the first and (b) the
second bright fr
9
1 1
6
9
1 1
inge above the central bright fringe.
SOLUTION: For 1, sin sin (1)(511 10 13.4
2.20 10
For 2, sin sin (2)(511 10
2.20 10
m m m
W m
m m m
W
λ
θ
λ
θ
−
− −
−
−
− −
= = ⎜⎝⎛ ⎟⎠⎞= ⎜⎝⎛ ×× ⎟⎠⎞=
= = ⎛⎜⎝ ⎞⎟⎠= ××
􀁄
6 27.7
− m
⎛ ⎞
⎜ ⎟=
⎝ ⎠
􀁄
10. Diffraction puts fundamental limits on the capacity
of telescopes and microscopes to separate the objects
being observed because light from the sides of a circular
aperture inter
min
feres. One can calculate that the first dark
fringe is at 1.22 , where D is diameter of the
aperture. Two objects can be barely resolved if the
diffraction maximum of one obje
D
λ
θ =
ct lies in the diffraction
minimum of the second object. Clearly, the larger D is,
the smaller the angular diameter separation. We say
that larger apertures lead to better resolution