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Interference and Diffraction
- Chapter
- 14.1 Superposition of Waves 14-
- 14.2 Young’s Double-Slit Experiment 14-
- Example 14.1: Double-Slit Experiment................................................................ 14-
- 14.3 Intensity Distribution 14-
- Example 14.2: Intensity of Three-Slit Interference 14-
- 14.4 Diffraction....................................................................................................... 14-
- 14.5 Single-Slit Diffraction..................................................................................... 14-
- Example 14.3: Single-Slit Diffraction 14-
- 14.6 Intensity of Single-Slit Diffraction 14-
- 14.7 Intensity of Double-Slit Diffraction Patterns.................................................. 14-
- 14.8 Diffraction Grating 14-
- 14.9 Summary......................................................................................................... 14-
- 14.10 Appendix: Computing the Total Electric Field............................................. 14-
- 14.11 Solved Problems 14-
- 14.11.1 Double-Slit Experiment 14-
- 14.11.2 Phase Difference 14-
- 14.11.3 Constructive Interference....................................................................... 14-
- 14.11.4 Intensity in Double-Slit Interference 14-
- 14.11.5 Second-Order Bright Fringe 14-
- 14.11.6 Intensity in Double-Slit Diffraction 14-
- 14.12 Conceptual Questions 14-
- 14.13 Additional Problems 14-
- 14.13.1 Double-Slit Interference......................................................................... 14-
- 14.13.2 Interference-Diffraction Pattern............................................................. 14-
- 14.13.3 Three-Slit Interference 14-
- 14.13.4 Intensity of Double-Slit Interference 14-
- 14.13.5 Secondary Maxima 14-
- 14.13.6 Interference-Diffraction Pattern............................................................. 14-
Interference and Diffraction
14.1 Superposition of Waves
Consider a region in space where two or more waves pass through at the same time. According to the superposition principle, the net displacement is simply given by the vector or the algebraic sum of the individual displacements. Interference is the combination of two or more waves to form a composite wave, based on such principle. The idea of the superposition principle is illustrated in Figure 14.1.1.
(a) (b)
(c) (^) (d)
Figure 14.1.1 Superposition of waves. (b) Constructive interference, and (c) destructive interference.
Suppose we are given two waves,
ψ 1 ( , ) x t = ψ 10 sin( k x 1 ± ω 1 t +φ 1 ), ψ 2 ( , ) x t = ψ 20 sin( k x 2 ± ω 2 t + φ 2 ) (14.1.1)
the resulting wave is simply
ψ ( , ) x t = ψ 10 sin( k x 1 ± ω 1 t + φ 1 ) + ψ 20 sin( k x 2 ± ω 2 t + φ 2 ) (14.1.2)
The interference is constructive if the amplitude of ψ ( , ) x t is greater than the individual
ones (Figure 14.1.1b), and destructive if smaller (Figure 14.1.1c).
As an example, consider the superposition of the following two waves at t = 0 :
1 ( )^ x^^ sin^ x ,^^2 ( ) x^^ 2sin^ x 4
The resultant wave is given by
In order to form an interference pattern, the incident light must satisfy two conditions:
(i) The light sources must be coherent. This means that the plane waves from the sources must maintain a constant phase relation. For example, if two waves are completely out of phase with φ = π, this phase difference must not change with time.
(ii) The light must be monochromatic. This means that the light consists of just one
wavelength λ = 2 π/ k.
Light emitted from an incandescent lightbulb is incoherent because the light consists o waves of different wavelengths and they do not maintain a constant phase relationship. Thus, no interference pattern is observed.
Figure 14.1.3 Incoherent light source
14.2 Young’s Double-Slit Experiment
In 1801 Thomas Young carried out an experiment in which the wave nature of light was demonstrated. The schematic diagram of the double-slit experiment is shown in Figure 14.2.1.
Figure 14.2.1 Young’s double-slit experiment.
A monochromatic light source is incident on the first screen which contains a slit. The
emerging light then arrives at the second screen which has two parallel slits S
S 0
1 and S 2. which serve as the sources of coherent light. The light waves emerging from the two slits then interfere and form an interference pattern on the viewing screen. The bright bands (fringes) correspond to interference maxima, and the dark band interference minima.
Figure 14.2.2 shows the ways in which the waves could combine to interfere constructively or destructively.
Figure 14.2.2 Constructive interference (a) at P , and (b) at P 1. (c) Destructive interference at P 2.
The geometry of the double-slit interference is shown in the Figure 14.2.3.
Figure 14.2.3 Double-slit experiment
Consider light that falls on the screen at a point P a distance from the point O that
lies on the screen a perpendicular distance L from the double-slit system. The two slits are separated by a distance d. The light from slit 2 will travel an extra distance
y
δ = r 2 (^) − r 1
to the point P than the light from slit 1. This extra distance is called the path difference. From Figure 14.2.3, we have, using the law of cosines,
2 2 2 2 2 1 2 cos^2 2 sin
d d r r dr r dr
and 2 2 2 2 2 2 cos^ sin 2 2 2
d d r r dr r dr
= + ⎛⎜^ ⎞⎟^ − ⎛⎜^ + θ⎞⎟^ = + ⎛⎜^ ⎞⎟ + θ
Subtracting Eq. (142.1) from Eq. (14.2.2) yields
Figure 14.2.5 (a) Destructive interference. (b) Constructive interference.
To locate the positions of the fringes as measured vertically from the central point O , in addition to L , we shall also assume that the distance between the slits is much greater than the wavelength of the monochromatic light, d
d
λ. The conditions imply
that the angle θ is very small, so that
sin tan
y L
Substituting the above expression into the constructive and destructive interference conditions given in Eqs. (14.2.5) and (14.2.6), the positions of the bright and dark fringes are, respectively,
b
L
y m d
and
d 2
L
y m d
⎛ ⎞^ λ
Example 14.1: Double-Slit Experiment
Suppose in the double-slit arrangement, d = 0.150 mm, L = 120 cm, λ = 833nm, and
y =2.00 cm.
(a) What is the path difference δ for the rays from the two slits arriving at point P?
(b) Express this path difference in terms of λ.
(c) Does point P correspond to a maximum, a minimum, or an intermediate condition?
Solutions:
(a) The path difference is given by δ= d sin θ. When L y , θ is small and we can
make the approximation sin θ ≈ tan θ= y / L. Thus,
( )
2 1.50 10 4 m 2.00^10 m 2.50 10 m 1.20 m
y d L
− ≈ ⎛^ ⎞= × −^ × = × ⎜ ⎟ ⎝ ⎠
− 6
(b) From the answer in part (a), we have
6 7
2.50 10 m
8.33 10 m
− −
×
×
or δ = 3.00λ.
(c) Since the path difference is an integer multiple of the wavelength, the intensity at point P is a maximum.
14.3 Intensity Distribution
Consider the double-slit experiment shown in Figure 14.3.1.
Figure 14.3.1 Double-slit interference
The total instantaneous electric field E
G
at the point P on the screen is equal to the vector
sum of the two sources:. On the other hand, the Poynting flux S is
proportional to the square of the total field:
E = E 1 + E
G G G
2
2
2 2 2 2 S ∝ E = ( E 1 (^) + E 2 (^) ) = E 1 (^) + E 2 + 2 E 1 ⋅ E
G G G G
Taking the time average of S , the intensity I of the light at P may be obtained as:
2 2 I = S ∝ E 1 (^) + E 2 + 2 E 1 ⋅ E 2
G G
d sin
Assuming that both fields point in the same direction, the total electric field may be obtained by using the superposition principle discussed in Section 13.4.1:
1 2 0 [^ sin^ sin(^ )]^2 0 cos^ sin 2 2
E E E E t t E t
= + = ω + ω + φ = ⎛⎜^ ⎞⎟^ ⎛⎜^ ω + ⎞⎟
where we have used the trigonometric identity
sin sin 2sin cos 2 2
The intensity I is proportional to the time average of the square of the total electric field:
2 2 2 2 2 2 I E 4 E 0 cos 2 sin t (^) 2 2 E 0 cos 2
or
2 I I 0 cos (^2)
where I (^) 0 is the maximum intensity on the screen. Upon substituting Eq. (14.3.4), the
above expression becomes
2 0
sin cos
d I I
Figure 14.3.2 Intensity as a function of d sin θ /λ
For small angle θ , using Eq. (14.2.5) the intensity can be rewritten as
2 0 cos^
d I I L
y
Example 14.2: Intensity of Three-Slit Interference
Suppose a monochromatic coherent source of light passes through three parallel slits, each separated by a distance d from its neighbor, as shown in Figure 14.3.3.
Figure 14.3.3 Three-slit interference.
The waves have the same amplitude E 0 and angular frequency ω , but a constant phase
difference φ = 2 π d sin θ/ λ.
(a) Show that the intensity is
2 (^0 1) 2 cos^2 sin 9
I d I
⎣ ⎝^ ⎠⎦
where I (^) 0 is the maximum intensity associated with the primary maxima.
(b) What is the ratio of the intensities of the primary and secondary maxima?
Solutions:
(a) Let the three waves emerging from the slits be
E 1 (^) = E 0 (^) sin ω t , E 2 (^) = E 0 (^) sin (^) ( ω t + φ) , E 3 (^) = E 0 sin (^) ( ω t + 2 φ) (14.3.17)
Using the trigonometric identity
sin sin 2 cos sin 2 2
the sum of E 1 and E 3 is
E 1 (^) + E 3 (^) = E 0 (^) ⎡⎣ sin ω t + sin (^) ( ω t + 2 φ (^) )⎤⎦ = 2 E 0 cos φ sin( ω t +φ) (14.3.19)
The total electric field at the point P on the screen is
14.4 Diffraction
In addition to interference, waves also exhibit another property – diffraction , which is the bending of waves as they pass by some objects or through an aperture. The phenomenon of diffraction can be understood using Huygens’s principle which states that
Every unobstructed point on a wavefront will act a source of secondary spherical waves. The new wavefront is the surface tangent to all the secondary spherical waves.
Figure 14.4.1 illustrates the propagation of the wave based on Huygens’s principle.
Figure 14.4.1 Propagation of wave based on Huygens’s principle.
According to Huygens’s principle, light waves incident on two slits will spread out and exhibit an interference pattern in the region beyond (Figure 14.4.2a). The pattern is called a diffraction pattern. On the other hand, if no bending occurs and the light wave continue to travel in straight lines, then no diffraction pattern would be observed (Figure 14.4.2b).
Figure 14.4.2 (a) Spreading of light leading to a diffraction pattern. (b) Absence of diffraction pattern if the paths of the light wave are straight lines.
We shall restrict ourselves to a special case of diffraction called the Fraunhofer diffraction. In this case, all light rays that emerge from the slit are approximately parallel to each other. For a diffraction pattern to appear on the screen, a convex lens is placed between the slit and screen to provide convergence of the light rays.
14.5 Single-Slit Diffraction
In our consideration of the Young’s double-slit experiments, we have assumed the width of the slits to be so small that each slit is a point source. In this section we shall take the width of slit to be finite and see how Fraunhofer diffraction arises.
Let a source of monochromatic light be incident on a slit of finite width a , as shown in Figure 14.5.1.
Figure 14.5.1 Diffraction of light by a slit of width a.
In diffraction of Fraunhofer type, all rays passing through the slit are approximately parallel. In addition, each portion of the slit will act as a source of light waves according to Huygens’s principle. For simplicity we divide the slit into two halves. At the first minimum, each ray from the upper half will be exactly 180 out of phase with a corresponding ray form the lower half. For example, suppose there are 100 point sources, with the first 50 in the lower half, and 51 to 100 in the upper half. Source 1 and source 51 are separated by a distance and are out of phase with a path difference
a / 2 δ = λ/ 2.
Similar observation applies to source 2 and source 52, as well as any pair that are a distance a / 2apart. Thus, the condition for the first minimum is
sin 2 2
a λ
or
sin a
Applying the same reasoning to the wavefronts from four equally spaced points a
distance a / 4apart, the path difference would be δ = a sin θ/ 4, and the condition for
destructive interference is
2 sin a
The argument can be generalized to show that destructive interference will occur when
a sin θ = m λ, m = ±1, ± 2, ± 3, ... (destructive interference) (14.5.4)
Figure 14.5.2 illustrates the intensity distribution for a single-slit diffraction. Note that
θ = 0 is a maximum.
The first minimum corresponds to m = 1. If y 1 (^) = 1.00 mm, then
( )( ) ( )
4 3 1 9
8.00 10 m 1.00 10 m 1.33 m 1 600 10 m
ay L
m λ
− − −
× ×
×
(b) The width of the central maximum is (see Figure 14.5.2)
( ) 3 w 2 y 1 2 1.00 10 m 2.00 mm = = × − =
14.6 Intensity of Single-Slit Diffraction
How do we determine the intensity distribution for the pattern produced by a single-slit diffraction? To calculate this, we must find the total electric field by adding the field contributions from each point.
Let’s divide the single slit into N small zones each of width ∆ y = a / N , as shown in
Figure 14.6.1. The convex lens is used to bring parallel light rays to a focal point P on the
screen. We shall assume that ∆ y λso that all the light from a given zone is in phase.
Two adjacent zones have a relative path length δ= ∆ y sin θ. The relative phase shift∆ β
is given by the ratio
sin 2 , sin 2
y y
Figure 14.6.1 Single-slit Fraunhofer diffraction
Suppose the wavefront from the first point (counting from the top) arrives at the point P on the screen with an electric field given by
E 1 = E 10 sin ω t (14.6.2)
The electric field from point 2 adjacent to point 1 will have a phase shift∆ β , and the
field is
E 2 (^) = E 10 (^) sin( ω t + ∆β (^) ) (14.6.3)
Since each successive component has the same phase shift relative the previous one, the electric field from point N is
EN = E 10 (^) sin (^) ( ω t + ( N − 1)∆ β (^) ) (14.6.4)
The total electric field is the sum of each individual contribution:
E = E 1 (^) + E 2 (^) + " EN = E 10 (^) ⎡⎣sin ω t + sin (^) ( ω t + ∆β (^) ) + " + sin (^) ( ω t + ( N − 1)∆β)⎤⎦ (14.6.5)
Note that total phase shift between the point N and the point 1 is
N N y sin a sin
where. The expression for the total field given in Eq. (14.6.5) can be simplified
using some algebra and the trigonometric relation
N ∆ y = a
cos( α − β ) − cos( α + β) =2sin αsin β (14.6.7)
[See Appendix for alternative approaches to simplifying Eq. (14.6.5).] To use the above in Eq. (14.6.5), consider
cos( / 2) cos( / 2) 2sin sin( / 2) cos( / 2) cos( 3 / 2) 2sin( )sin( / 2) cos( 3 / 2) cos( 5 / 2) 2sin( 2 ) sin( / 2)
cos[ ( 1/ 2) ] cos[ ( 3 / 2) ] 2sin[ ( 1) ]sin( / 2)
t t t t t t t t t
t N t N t N
ω β ω β ω β ω β ω β ω β β ω β ω β ω β β
ω β ω β ω β
β
)
Adding the terms and noting that all but two terms on the left cancel leads to
( ) (
cos( / 2) cos[ ( 1/ 2) ] 2sin( / 2) sin sin sin ( 1)
t t N t t t N
The two terms on the left combine to
cos( / 2) cos[ ( 1/ 2) ] 2sin( ( 1) / 2)sin( / 2)
t t N t N N
ω β ω β ω β
= + − ∆ ∆ β
with the result that
From Eq. (14.6.15), we readily see that the condition for minimum intensity is
a sin m , m 1, 2, 3, ...
or
sin m , m 1, 2, 3, ... a
In Figure 14.6.3 the intensity is plotted as a function of the angle θ , for a = λ and
a = 2 λ. We see that as the ratio a / λ grows, the peak becomes narrower, and more light
is concentrated in the central peak. In this case, the variation of I (^) 0 with the width a is
not shown.
Figure 14.6.3 Intensity of single-slit diffraction as a function of θ for a = λand a = 2 λ.
14.7 Intensity of Double-Slit Diffraction Patterns
In the previous sections, we have seen that the intensities of the single-slit diffraction and the double-slit interference are given by:
( )
2 0
sin sin / single-slit diffraction sin /
a I I a
2 2 0 0
sin cos cos double-slit interference 2
d I I I
= ⎛^ ⎞^ = ⎛^ ⎞
Suppose we now have two slits, each having a width a , and separated by a distance d. The resulting interference pattern for the double-slit will also include a diffraction pattern due to the individual slit. The intensity of the total pattern is simply the product of the two functions:
( )
2 2 0
sin sin^ sin^ / cos sin /
d^ a I I a
π θ^ π^ θ^ λ
⎛ ⎞^ ⎡^ ⎤
The first and the second terms in the above equation are referred to as the “interference factor” and the “diffraction factor,” respectively. While the former yields the interference substructure, the latter acts as an envelope which sets limits on the number of the interference peaks (see Figure 14.7.1).
.
Figure 14.7.1 Double-slit interference with diffraction.
We have seen that the interference maxima occur when d sin θ = m λ. On the other hand,
the condition for the first diffraction minimum is a sin θ = λ. Thus, a particular
interference maximum with order number m may coincide with the first diffraction minimum. The value of m may be obtained as:
sin sin
d m a
or d m a
Since the m th fringe is not seen, the number of fringes on each side of the central fringe is m − 1. Thus, the total number of fringes in the central diffraction maximum is
N = 2( m + 1) + 1 = 2 m − 1 (14.7.3)
14.8 Diffraction Grating
A diffraction grating consists of a large number of slits each of width and separated from the next by a distance , as shown in Figure 14.8.1.
N a d
