Unit 19
Interference and Diffraction
19.1
Superposition revisit
19.2
Young’s double-slit experiment
19.3
Phase change due to reflection
19.4
Diffraction
19.1 Superposition Revisit
The simple addition of two or more waves to give a resultant wave is referred to as
superposition. When waves are superposed, the result may be a wave of greater amplitude
(constructive interference) or of reduced amplitude (destructive interference).
1
Constructive interference occurs: l 2 − l1 = mλ , where m = 0, ± 1, ± 2, ± 3, ....
1
Destructive interference occurs: l 2 − l1 = (m − )λ , where m = 0, ± 1, ± 2, ± 3, ....
2
The source waves that we often used:
•
Monochromatic light: it consists of waves with a single frequency and hence a single
color.
•
Coherent/incoherent light: Light waves that maintain a constant phase relationship
with one another are referred to as coherent. Light waves in which the relative phases
vary randomly with time are said to be incoherent.
Example
Two friends tune their radios to the same frequency and
pick up a signal transmitted simultaneously by a pair of
antennas. The friend who is equidistant from the antennas,
at P 0 , receives a strong signal. The friend at point Q 1
receives a very weak signal. Find the wavelength of the
radio waves if d = 7.50 km, L = 14.0 km, and y = 1.88 km. Assume that Q 1 is the first point
of minimum signal as one moves away from P 0 in the y direction.
Answer
The first minimum in the y direction from the maximum at point P 0 , we know that the path
difference, l 2 – l 1 , is half a wavelength.
The path length l 1 :
2
l1 =
d

L +  − y =
2

2
2
 (7.50 km)

(14.0 km) + 
− 1.88 km  = 14.1 km
2


2
The path length l 2 :
2
l2 =
d

L +  + y =
2

2
2
 (7.50 km)

(14.0 km) + 
+ 1.88 km  = 15.1 km
2


2
λ
By the relation l2 − l1 = , it gives λ = 2.0 km . This wavelength corresponds to a frequency
2
of about 150 kHz.
2
19.2 Young’s double-slit experiment
Monochromatic light from a narrow vertical slit S falls on two other narrow slits S 1 and S 2
which are very close together and parallel to S. S 1 and S 2 act as two coherent sources (both
being derived from S) and if they (as well as S) are narrow enough, diffraction causes the
emerging beams to spread into the region beyond the slits. Superposition occurs in the shaded
area where the diffracted beams overlap. Alternate bright and dark equally-spaced vertical
bands (interference fringes) can be observed on a screen. Note that if either S 1 or S 2 is
covered the bands disappear.
An expression for the separation of two bright (or dark) fringes can be obtained as follows.
For the central bright fringe: since S1O = S 2 O , the two waves are in phase and constructive
interference occurs.
At P, distance x 1 from O, there will be a bright fringe if the path difference is a whole number
wavelength. The path difference between waves
S 2 P − S1 P = nλ ,
3
where n is an integer. We say the nth bright fringe is formed at P. If d is the distance from the
screen to the double-slit and a is the slit separation. Hence,
( S 2 P) 2 = d 2 + ( x1 + a / 2) 2 = d 2 + x1 + ax1 + a 2 / 4
2
( S1 P) 2 = d 2 + ( x1 − a / 2) 2 = d 2 + x1 − ax1 + a 2 / 4
2
Therefore, ( S 2 P) 2 − ( S1 P) 2 = 2ax1
But, ( S 2 P) 2 − ( S1 P) 2 = ( S 2 P − S1 P) ( S 2 P + S1 P) , so we have
( S 2 P − S1 P) 2d = 2ax1 or ( S 2 P − S1 P)d = ax1
For the nth bright fringe at P we have
nλd = ax1 or nλ =
ax1
.
d
(1)
For the (n+1)th bright fringe at Q, say, we have
(n + 1)λd = ax 2 or (n + 1)λ =
(2) – (1) gives λ =
ax 2
.
d
(2)
ay
, where y = x 2 − x1 .
d
Remarks:
1
, if λ and d are constant.
a
1.
y∝
2.
y ∝ d , if λ and a are constant.
3.
y ∝ λ , if a and d are constant.
In fact, as the double-slit is far from the screen, the paths for bright fringes or dark fringes are
parallel. Refer to the following figure and note that the parameters d and y are defined
differently from the above discussion. Hence, we have
4
for bright fringe: d sin θ = mλ , m = 0, ± 1, ± 2, ± 3, ....
1
for dark fringe: d sin θ = (m − )λ , m = 0, ± 1, ± 2, ± 3, ....
2
The linear distance from central fringe is y = L tan θ .
Example
Red light (λ = 750 nm) passes through a pair of slits with a separation of 6.2 × 10−5 m. Find
the angles corresponding to (a) the first bright fringe and (b) the second dark fringe above the
central bright fringe.
Answer
(a)
The relation d sin θ = mλ with m = +1, gives
−9
m) 
−1  mλ 
−1  (1)(750 × 10
o
θ sin
=
=
=
sin

 0.69 .


−5
 d 
 6.2 × 10 m 
(b)
1
The relation d sin=
θ (m − )λ , with m = +2, gives
2
1 λ
1 750 × 10−9 m 

−1 
o
=
θ sin −1  (m − )=
sin
(2
−
)×
=

 1.04 .

−5
2
d
2
6.2
×
10
m




Example
Two slits with a separation of 8.5 × 10−5 m create an interference
pattern on a screen 2.3 m away. If the tenth bright fringe above the
central fringe is a linear distance of 12 cm from it, what is the
wavelength of light used in the experiment?
Answer
The relation y = L tan θ gives
0.12 m
−1 y
=
θ tan
=
( ) tan −1 (=
) 3.0o
L
2.3 m
For bright fringe: d sin θ = mλ which gives
d
m
 8.5 × 10−5 m 
−7
o
 sin(3.0 ) =4.4 × 10 =440 nm .
10


λ = sin θ =
The light used in the experiment is blue in color.
5
Example
A two-slit experiment is performed in the air. Later, the same apparatus is immersed in water
and the experiment is repeated. When the apparatus is in water, are the interference fringes (a)
more closely spaced, (b) more widely spaced, or (c) spaced the same as when the apparatus is
in air?
Answer
The wavelength is smaller in water than in air, e.g. λw =
λa
naw
, where n aw > 1. According to
the equation d sin θ = mλ for the bright fringe, a decreased wavelength implies a decreased θ,
and hence the fringes are closely packed.
19.3 Phase change due to reflection
•
There is no phase change when light reflects from a region with a lower index of
refraction.
•
There is half-wavelength (λ/2 or π) phase change when light reflects from a region
with a higher index of refraction, or a solid surface.
6
19.3.1 Air wedge
Interference effect is between light (ray 1) reflected from the bottom surface of the top glass
plate and light (ray 2) reflected from the upper surface of the lower plate. Ray 1 experience
no phase change. Ray 2, that is the reflection from air to glass results in a half-wavelength
(λ/2 or π) phase change, thus the effective path difference of ray 2 and ray 1 is
∆l eff = d +
λ
2
+ d = 2d +
λ
2
. For constructive interference (bright fringe):
∆l eff = mλ , i.e. 2d +
λ
2
= mλ , where m = 1, 2, 3, …
For destructive interference (dark fringe):
∆l eff = m
λ
2
, i.e. 2d +
λ
2
=m
λ
2
, where m = 1, 3, 5, …
Or 2d = mλ , where m = 0, 1, 2, 3, …
When m = 0, d = 0, it corresponds to the first dark fringe located at the joint of the glass
plates.
Remarks:
The bright (or dark) fringes are equally spaced. Take a look at adjacent bright fringes:
2d m +1 +
λ
2
= (m + 1)λ ,
2d m +
λ
2
= mλ
The difference of this two equations gives 2(d m +1 − d m ) = λ , hence 2(l m +1 − l m ) sin θ = λ
where l m+1 and l m are the lengths measured from the joint of glass plates to the (m+1)th and
mth bright fringes respectively. The angle between glass plates is θ. Thus, we have
2∆l sin θ = λ , which shows that the fringes are equally spaced.
7
Example
An air wedge is formed by placing a human hair between two glass plates on one end, and
allowing them to touch on the other end. When this wedge is illuminated with red light (λ =
771 nm) it is observed to have 179 bright fringes. How thick is the hair?
Answer
Setting that the thickness of hair as t which corresponds to the 179th bright fringe, we have
from the relation 2t +
2t +
λ
2
=
mλ ,
771 × 10−9 m
=
(179)(771 × 10−9 m)
2
Thus, we obtain the thickness of hair = 68.8 ×10−6 m = 68.8 µm.
Remarks:
If a thicker hair is used, then by the relation 2d +
λ
1
(m − )λ , the
=
mλ again, 2d=
m
2
2
number of fringes increased under the same frequency of light.
Example
When light is reflected from a medium of several layers as shown in the right figure, the first
destructive interference is given by the path difference t =
λa
4naf
. Given that the film is less
dense than the glass. Explain this.
8
Answer
As the reflected rays 1 and 2 has both a phase change of π. The effect in the path difference is
cancelled. But there is a path difference due to the thickness of film. So, 2tnaf =
t=
λa
4naf
λa
2
or
. Thus, we get the result. A glass (n = 1.52) is often coated with a thin film of
magnesium fluoride (n = 1.38) to minimize reflected light. For yellow-green light
λ = 565 nm , hence, we have the thickness of thin film equals to 102 nm. For your reference,
the left figure shows no phase change in the reflected ray from the third medium.
19.4 Diffraction
The waves are initially traveling directly to the right. After passing through the gap in the
barrier, however, they spread out and travel in all possible forward directions, in accordance
with Huygen’s principle. That’s why an observer located at P (not along the line with the
incoming wave and the gap) detects the wave. In general, waves always bend, or diffract
when they pass by a barrier or through an opening (or slit).
A familiar example of diffraction is the observation that you can hear a person talking even
when that person is out of sight around a corner. The sound waves from the person bend
around a corner. It might seem, then, that light cannot be wave, since it does not band around
a corner along with the sound. In fact, the observation is due to the great difference between
the sound and light waves – of many orders. For sound: the wavelength is in the order of
meter, but for light, the wavelength is in the order of 10–7 meter.
Remarks:
9
According to Huygens, every point on a wavefront may be regarded as a source of secondary
spherical wavelets which spread out with the wave velocity. The new wavefront is the
envelope of these secondary wavelets.
19.4.1 Single-slit diffraction
As shown in figure, since the screen is distant, the waves from 1 and 1’ travel on
approximately parallel paths to reach the screen. So, the path difference for these waves is
(W / 2) sin θ . When θ = 0, constructive interference occurs and bright fringe is observed on
the screen. As θ is increased, the path difference is also increased, when (W / 2) sin θ =
λ
2
, we
observe the first dark fringe (the first minimum). Or we can write
W sin θ = λ .
The second dark fringe is obtained by considering the following figure. (W / 4) sin θ = λ / 2 ,
hence we have W sin θ = 2λ . In general, we have W sin θ = mλ , where m = ±1, ± 2, ± 3, ... .
10
For the first bright fringe, we have (W / 3) sin θ = λ / 2 ,
giving sin θ =
3λ
.
2W
Similarly, the second bright can be obtained by
(W / 5) sin θ = λ / 2 , giving sin θ =
5λ
.
2W
Remark:
The width of the central bright fringe can be estimated by first considering the first dark
fringes which has angular separation
locate at sin θ = ±
λ
W
2λ
(approximately). It is because the first dark fringes
W
, and for small θ, sin θ ≈ θ .
Example
If the width of the slit through which light passes is reduced, does the central bright fringe (a)
become wider, (b) become narrower, or (c) remain the same size?
Answer
The answer is wider due to the above discussion. In addition, we know that a narrower slit
will give more diffraction.
Example
Light with a wavelength of 510 nm forms a diffraction pattern after passing through a single
slit of width 2.2×10−6 m. Find the angle associated with (a) the first and (b) the second dark
fringe above the central bright fringe.
Answer
 mλ 
We apply the formula for the dark fringe, θ = sin −1 

W 
−9
−1  (1)(510 × 10 ) 
o
=
=
sin
θ
When m = 1,

 13 .
−6
 2.2 × 10

−9
−1  (2)(510 × 10 ) 
o
=
=
When m = 2, θ sin

 28 .
−6
 2.2 × 10

11
19.4.2 Diffraction grating
A system with a large number of slits is referred to as diffraction
grating. In some cases, it is possible to produce gratings with as
many as 40,000 slits per centimeter. The interference pattern formed
by a diffraction grating consists of a series of sharp, widely spaced
bright fringes – called principal maxima – separated by relatively
dark regions with a number of weak secondary maxima. In the limit
of a large number of slits, the principal maxima become more
sharply peaked, and the secondary maxima become insignificant.
For constructive interference in a diffraction grating
d sin θ = mλ ,
where m = ±1, ± 2, ± 3, ...
Notice that a smaller d gives a larger spread angle θ. That’s
why we observe a widely spread of light if a grating of great
number of lines per centimeter is passed with light beam.
Example
Find the slit spacing necessary for 450-nm light to have a first-order (m = 1) principal
maximum at 15o.
Answer
For constructive interference in a diffraction grating, we have
d sin θ = mλ ,
Now, plug in the given data, we have =
d
(1)(450 × 10−9 m)
mλ
=
= 1.7 × 10−6 m .
o
sin θ
sin(15 )
That means the diffraction grating has 1/1.7 × 10−6 ≈ 588000 lines per meter, or 5,880 lines
per cm.
12
Example
When 546-nm light passes through a particular diffraction
grating a second-order principal maximum is observed at an
angle of 16.0o. How many lines per centimeter does this
grating have?
Answer
For constructive interference in a diffraction grating, we have
d sin θ = mλ ,
Now, plug in the given data, we have
=
d
(2)(546 × 10−9 m)
mλ
=
= 3.96 × 10−6 m .
o
sin θ
sin(16.0 )
That means the diffraction grating has 1/ 3.96 × 10−6 =
253, 000 lines per meter, or 2,530 lines
per cm.
13