Lecture Notes for PHYS 2911 – Optics – Sem. 1 2016
Prof. Tim Bedding (tim.bedding@sydney.edu.au)
School of Physics, Room 216, Tel: 9351 2537
The subject of Optics can be divided into three areas:
Geometrical Optics where light is described by rays which show the paths of energy transfer. Geometrical
optics provides a good understanding of the propagation of light in transparent media and the operation of
optical imaging systems such as cameras, telescopes and microscopes.
Physical Optics where the wave nature of light is taken into account. Physical optics covers polarisation,
interference and diffraction of light. The physical optics approach is necessary for understanding the limits
of resolution of optical imaging systems.
Quantum Optics where the particle nature of light is taken into account. This description, where light is
considered to consist of massless particles called photons, is needed to understand fully the interaction
of light and matter. Topics which require the quantum optics approach include the photoelectric effect,
photodetectors and lasers.
This course is on Physical Optics; it covers general principles, phenomena which require physical optics
for their explanation and applications. The basic elements of geometrical optics, covered in the Optics
module of the Physics I Experimental Physics course, will be assumed; a brief summary is provided below.
Reference books
Young & Freedman, University Physics (Junior Physics textbook)
Hecht, Eugene, Optics (in Scitech Library Reserve).
Fowles, Grant R., Introduction to Modern Optics (in Scitech Library Reserve).
1 Geometrical Optics - Summary
1.1 Refractive Index
The speed of light in a transparent medium is given by c/n where c is the speed of light in a vacuum and n
is the refractive index of the medium. The refractive index is around 1.5 for most glasses; 1.33 for water;
1.000294 for air at atmospheric pressure and 0◦ C; and exactly 1 for vacuum.
1.2 Light Rays
Light rays are lines which show the direction of energy flow. For example, light rays from a point source
are radial lines from the source; for a collimated beam the light rays are parallel lines.
1.3 Laws of Reflection and Refraction
In general, when light is incident upon an interface between two transparent media with different refractive
indices, some light is reflected and some (most) is transmitted through the interface as shown in Figure 1.
1
n1
θ1 θ’1
n2
θ2
Figure 1: Reflection and refraction at an interface between two transparent media
The Law of Reflection is
θ1′ = θ1
(1)
where the angle of incidence θ1 and the angle of reflection θ1′ are measured with respect to the normal to
the interface.
The Law of Refraction - also called Snell’s Law - describes the way in which a light ray bends when it
passes from one medium to the other:
n1 sin θ1 = n2 sin θ2
(2)
1.4 Total Internal Reflection
Total internal reflection will occur at an interface between two media when
sin θ1 ≥
n2
.
n1
(3)
This can occur only when light is incident from the side of higher refractive index (ie when n1 > n2 ). The
angle of incidence for which θ2 = π/2 is called the critical angle, θc . Thus
sin θc =
n2
n1
(4)
and the condition for total internal reflection is θ1 > θc .
1.5 Thin Lens Formula
1
1
1
+ =
s s′
f
(5)
where s is the distance of the object from the lens, s′ is the distance of the image from the lens, and f is the
focal length of the lens. For a converging lens f > 0; for a diverging lens f < 0. For a real image s′ > 0;
for a virtual image s′ < 0. The lateral magnification of the lens is given by
m=−
s′
s
(6)
If follows that a collimated beam (parallel rays, as if from an object at infinity) will be brought to a focus
in the focal plane. This is true whether or not the rays are parallel to the optical axis of the lens.
2
2 Electromagnetic radiation
We know the amplitude of the electric field from a stationary charge falls off with distance squared
(Coulomb’s Law). The same is true for the magnetic field from a constant current (Biot-Savart Law).
In the case of an accelerating charge (or a changing current), Maxwell’s equations predict additional components of the electric and magnetic fields which fall off much more slowly, as the first power of distance.
This implies that electric currents in one place can affect other charges far away. We now recognise light
to be electric and magnetic influences extending over vast distances, generated by incredibly rapid oscillations of the electrons in atoms. The atoms in a distant star create electromagnetic waves which propagate
through space and cause electrons in your eye to oscillate.
Any configuration of electric and magnetic fields must satisfy Maxwell’s equations. In integral form, as
studied in Junior Physics, they are:
I
1
E · dA =
Qenc
(7)
ǫ0
I
B · dA = 0
(8)
I
dΦB
E · dl = −
(9)
dt
I
dΦE
(10)
B · dl = µ0 Ienc + µo ǫ0
dt
The third of these equations (Faraday’s Law) indicates that a changing magnetic field produces an electric field. The fourth equation (Ampere’s Law with Maxwell’s additional term) indicates that a changing
electric field produces a magnetic field. Thus, even if we are far from any charges and currents, we can
have electric and magnetic fields that are continually changing and self-sustaining. As you saw in Junior
physics, such a configuration can exist in the form of a moving “slab” of electric and magnetic fields that
are perpendicular to each other and to their direction of propagation. It can be shown that this moving
√
pattern of crossed fields satisfies Maxwell’s equations provided the pattern moves at speed 1/ ǫo µ0 . As
Maxwell realised, this quantity (which contains the constants in the force laws from electrostatics and
magnetostatics) is equal to the measured speed of light, to within experimental uncertainty. He was led to
conclude that light is a propagating pattern of mutually perpendicular electric and magnetic fields.
How does such a pattern get produced in the first place? As mentioned above, the fields are produced by
accelerating charges (see Fig. 2). Consider the electric field at large distances from an accelerating charge
that is moving nonrelativistically. The electric field vector lies in the plane perpendicular to the line of
sight. Its component in some direction x in this plane depends on the component of the acceleration of the
charge in that same direction:
r
−1 q 1
t
−
.
(11)
a
Ex (t) =
x
4πǫ0 r c2
c
We see that Ex varies inversely as r and is proportional to the component of the acceleration of the charge
in the direction x, evaluated at an earlier time t − r/c. Note that the magnetic field has magnitude 1/c
times the electric field, and is perpendicular to both E and the line of sight. All of this can be deduced from
Maxwell’s equations, and this is covered in Senior Physics.
Equation 11 implies that the field moves as a wave outward from the source at speed c. An interesting
case: the charge oscillates with a displacement proportional to cos ωt. Then the acceleration, and hence the
radiation component of the electric field, also oscillate as cos ωt.
2.1 Energy of radiation
The energy content of a wave is proportional to the square of the amplitude. For example, placing a charge
in an oscillating electric field produces forces proportional to field strength, hence acceleration and velocity
3
Figure 2: Electric field from a single positive charge. Left: the charge is stationary. Right: the charge
underwent a brief acceleration a short time ago. The “kink” in the field lines propagates outwards at
speed c.
are also proportional to field strength. So the kinetic energy developed in the charge is proportional to the
square of the field.
The energy that the source can deliver from radiation therefore falls off as the square of the distance. This
makes sense, since it implies that the total energy integrated over an sphere around the source is independent
of the radius of the sphere.
The power per unit area is called the irradiance of light (in the past, it was called intensity).
2.2 Sinusoidal waves
Suppose we have an oscillator whose displacement is proportional to cos ωt. The electric field at some
(large) distance r is then
E=
Ar
cos ω t − rc .
r
(12)
Note that the argument of the cosine function (called the phase) must be dimensionless (it is measured in
radians). The quantity Ar is the amplitude of the electric field at distance r from the source.
First let us fix the position r and watch the field as a function of time. It oscillates at the angular frequency ω.
This quantity ω = 2πf is the rate of change of phase with time (radians per second). In practice, with light
we cannot see these fast oscillations. Instead, at a fixed point we measure the irradiance, which will be
proportional to E 2 .
Now take a snapshot at a fixed time. That is, fix t and look at the wave as a function of distance r. We notice
that as a function of r, the field is also oscillatory. The wavelength of this spatial variation is λ = c/f . We
define the wave number to be k = 2π/λ, which is the rate of change of phase with distance (radians per
metre). With this definition, and noting that c = ω/k, we can write Equation 12 as
E=
Ar
cos (ωt − kr).
r
(13)
Provided we are far from the source charge, the factor 1/r varies very slowly with r and can therefore be
considered as a constant and absorbed into the amplitude. This approximation means that we are treating
the wave as plane rather than a spherical. With this assumption, we have
E = A cos (ωt − kr).
(14)
4
3 Interference between radiation sources
3.1 Calculation for two sources
Suppose we have two sources which have the same frequency and are radiating in phase. We want to find
the resultant field at some distant point P at some particular angle (Fig. 4). We have to add two cosines
with the same frequency but with different phases.
The resultant field at P will be:
Etot
=
E1 + E2
(15)
=
A1 cos(ωt + φ1 ) + A2 cos(ωt + φ2 )
(16)
Since we assume the two sources are exactly in phase, we can see from the previous section that
φ1 = −kr1
and φ2 = −kr2 ,
(17)
where r1 and r2 are the distances from each of the two sources to point P.
Our aim is to calculate Etot , although note that we are actually interested in the amplitude of the result
(which gives the irradiance).
The algebraic method For simplicity, set A1 = A2 = A. We use the rule1 :
X−Y
cos X + cos Y = 2 cos( X+Y
2 ) cos( 2 ).
(18)
This gives
Etot
=
=
=
A cos(ωt + φ1 ) + A cos(ωt + φ2 )
2
2
cos ωt + φ1 +φ
2A cos φ1 −φ
2
2
2
1
cos ωt − k r1 +r
2A cos k r2 −r
2
2
(19)
(20)
(21)
In the final step we have used Equation 17. This is in the same form as Equation 14 – an oscillatory
plane wave with the same frequency as the two original signals,
but with a new amplitude and phase. The
1
amplitude of the resulting wave is Atot = 2A cos k r2 −r
.
2
Phasor addition method Any cosine function of ωt can be considered the horizontal projection of a
rotating phasor. To add two cosine functions, we add their phasors and measure the horizontal component
of the result. Both phasors are rotating with the same angular speed ω, so the angle between them stays
constant. Applying the cosine rule to the triangle (Fig. 3), it can be shown (do it!) that
A2tot
=
=
A21 + A22 + 2A1 A2 cos(φ2 − φ1 )
1
,
A21 + A22 + 2A1 A2 cos k r2 −r
2
(22)
(23)
which agrees with Method 1 for the special case A1 = A2 = A (prove it).
The complex method We write a complex number for each of the phasors. The real parts of the complex
numbers are the physical quantities (projection onto the horizontal axis).
Etot
1 To
=
A1 ei(ωt+φ1 ) + A2 ei(ωt+φ2 )
(24)
=
(A1 eiφ1 + A2 eiφ2 )eiωt
(25)
prove this rule, evaluate cos(a + b) + cos(a − b) and then set X = a + b and Y = a − b.
5
=
2
A
A1
+
ω t +φ1
ot
At
φ−φ
2
1
ω t +φ2
Figure 3: Addition of two phasors.
To get the amplitude, we multiply by the complex conjugate:
A2tot
=
(A1 eiφ1 + A2 eiφ2 )eiωt (A1 e−iφ1 + A2 e−iφ2 )e−iωt
=
=
A21 + A22 + A1 A2 (ei(φ1 −φ2 ) + ei(φ2 −φ1 ) )
A21 + A22 + 2A1 A2 cos(φ2 − φ1 )
1
,
A21 + A22 + 2A1 A2 cos k r2 −r
2
=
as before.
(26)
(27)
Thus the sum of the fields from the two sources produces an irradiance which is the sum of the two
individually, plus an extra term. It is called the interference term, and it may be positive or negative.
3.2 Application to two sources
To apply Equation 26 to the case of two oscillators, we simply need to find the phase difference. Consider
the case of two oscillators of equal amplitude, separated by distance d. As before, we assume the two
sources are in phase.
to point P
θ
d
d sin θ
Figure 4: Two oscillators with equal amplitude.
What is the irradiance at some distant point P? The difference in distance from P to the two oscillators is
|r2 − r1 | = d sin θ (Fig. 4). This introduces a phase difference equal to the number of wavelengths in
d sin θ, multiplied by 2π. Equivalently, the phase difference equals the distance difference multiplied by k,
since k is the rate of change of phase with distance. The phase difference at P is therefore
φ2 − φ1 = 2πd sin θ/λ
(28)
This allows us to calculate the intensity in any direction, as shown by the following two examples:
Example 1 Suppose the two oscillators are separated by half a wavelength in the North-South direction,
as shown in Fig. 5. We assume they are in phase. Then for points along the E–W direction, both oscillators
6
contribute equally and their electric fields add in phase. The resulting electric field is twice that of a single
oscillator and the intensity is 4I0 . At points along the N–S direction, the extra λ/2 path difference means
the two sources are exactly out of phase, so the resulting electric field is zero. Of course, this is only true
at distances large enough to neglect the difference between the 1/r factors.
I=0
I=2
I=2
λ/2
I=4
I=4
I=2
I=2
I=0
Figure 5: Two oscillators in phase and with equal amplitude, separated by half a wavelength.
Does all this agree with the above calculations? If we set d = λ/2 then the phase difference as a function
of angle in Equation 28 becomes φ2 − φ1 = π sin θ. Putting this into Equation 26 gives A2tot = 2A2 (1 +
cos(π sin θ)). If we evaluate this for θ = 0 and π/2 we get 4A2 and zero, respectively. One final example:
the angle at which the intensity is 2I0 is θ = π/6 (i.e., 30◦ ).
Example 2 Now suppose the separation is many wavelengths. In this case, the phase difference (Equation 28) is sin θ times a large number, which goes from zero to a large positive number, back through zero
and to a large negative number. That is, the phase goes through many multiples of 2π as θ goes around
once. When we put this inside the cosine function (Equation 26), the resulting intensity goes through many
peaks and zeroes.
3.3 Interference between N sources
Now consider N equally spaced oscillators, all having the same amplitude but different in phase from one
another, either because they are driven at different phases or because we are looking at them from an angle.
The total electric field at P is
Etot
=
=
E1 + E2 + . . . + EN
A [cos ωt + cos(ωt + φ) + cos(ωt + 2φ) + . . . + cos(ωt + (N − 1)φ)] ,
(29)
where φ is the phase difference between each adjacent pair of oscillators: φ = α + 2πd sin θ/λ.
Phasor addition method We can add the terms in Equation 29 geometrically (Fig. 6). All the phasors
have length A. The first has zero phase, the next has phase φ, the next has phase 2φ, and so on. All the
vertices lie on a circle.
The radius of the circle satisfies both
sin(φ/2) =
A/2
r
(30)
7
Nφ/2
φ
φ
A
Figure 6: Phasor addition of N = 7 sources.
and also
Atot /2
.
r
(31)
sin(N φ/2)
.
sin(φ/2)
(32)
sin(N φ/2) =
Eliminating r gives:
Atot = A
The complex method We write a complex number for each of the phasors. The real parts of the complex
numbers are the physical quantities (projection onto the horizontal axis).
i
h
(33)
Etot = A eiωt + ei(ωt+φ) + ei(ωt+2φ) + . . . + ei(ωt+(N −1)φ)
i
h
(34)
= A eiωt 1 + eiφ + e2iφ + . . . + e(N −1)iφ
eiN φ − 1
eiφ − 1
eiN φ/2 eiN φ/2 − e−iN φ/2
eiφ/2 eiφ/2 − e−iφ/2
=
A eiωt
=
A eiωt
=
A eiωt ei(N −1)φ/2
sin(N φ/2)
.
sin(φ/2)
(35)
(36)
(37)
Taking the amplitude gives the same result as Equation 32.
Result
The irradiance is
I = I0
sin2 (N φ/2)
.
sin2 (φ/2)
(38)
This allows us to calculate the intensity in any direction.
8
Practice problems (Set 1):
1. Red plane waves from a helium-neon laser (632.8 nm) in air pass through two parallel slits in an
opaque screen. A fringe pattern forms on a distant wall, and we see the fourth bright band 0.8◦
above the central axis (counting the central bright fringe as zero). (a) Calculate the separation of the
slits. (b) What is the orientation of the slits?
2. Show that for N = 2, the irradiance calculated in Equation 38 agrees with the two-source case
(Equation 26).
3. Use the complex representation to find the resultant E = E1 + E2 , where
E1 = A cos(kx + ωt)
and
E2 = −A cos(kx − ωt).
Describe the resultant wave.
4. Show that that the sum of N sinusoids with identical periods but different phases is also a sinusoid.
Hint: you need to consider adding terms like Aj cos(ωt + φj ), and it is only necessary to show for
two sinusoids (why?). Try using complex notation!
Describe briefly the implication of this result for the electric radiation field from a large number of
oscillating electrons that is viewed through a monochromatic filter.
5. Consider the sum of N vectors
EN =
N
X
ej ,
j=1
where all the ej have unit amplitude but random directions. Show that the expected value for |EN |2 ,
which we write as h|EN |2 i, is N . Hint: use mathematical induction and apply the cosine rule to a
vector diagram.
Describe briefly the implication of this result for the electric radiation field from a large number of
oscillating electrons that is viewed through a monochromatic filter.
9
4 Introduction to Diffraction
4.1 Huygens-Fresnel principle
Every unobstructed point of a wavefront, at a given instant in time, serves as a source of secondary wavelets
of the same frequency as the primary wave. The amplitude of the optical field at any point beyond is the
superposition of all these wavelets (considering their amplitudes and relative phases).
Note that this statement would predict secondary wavelets in all directions, even backwards. In practice
this is not observed. Instead, the secondary wave is concentrated in the forwards direction. We can account
for this by saying that the amplitude of the secondary wavelet in direction θ is reduced by an amount K(θ),
which is the so-called obliquity factor. More detailed calculations show that K(θ) = 12 (1 + cos θ). Note
that this factor is unity in the forward direction (θ = 0) and zero in the backwards direction (θ = π). In
most practical situations the obliquity factor can be approximated by unity.2
What do we mean by a wavefront? The best definition is a surface of constant phase. We can also talk
about rays, especially in geometrical optics, and these are lines that are perpendicular to the wavefront and
indicate the direction of energy flow.
Diffraction by an aperture Our task is to find out what happens when we shine a light at an opaque
mask with some holes in it. Most people would say that light shines through the holes and produces and
effect on the other side. Huygens’ principle says that we should pretend there are sources distributed
uniformly across the open holes and that the phases of these sources are the same as they would be if the
opaque material were absent. Of course there are no sources at the holes – indeed, this is the only place
where there are certainly no sources. For an explanation of why Huygen’s principle gives (almost) the right
answer, see Hecht (section 10.1.1) or Feynman (section 31-6).
4.2 Calculating diffraction using Huygens-Fresnel principle
Consider light from a point source S which passes through an aperture (Fig. 7). We want to calculate the
resulting field at P , on the other side of the aperture.
Q
r’
r
x
P
O
S
Figure 7: Diffraction through an aperture.
To apply the H-F principle, we need to know the field at each point in the aperture. Let Q be a point in the
aperture with position vector x (where the origin is point O).
Suppose the source at S is monochromatic with angular frequency ω, so that it oscillates as cos ωt. We
discussed this in Section 2.2. The amplitude of the field at Q goes as 1/r′ (recall that amplitude falls off
2 An
exception is in calculating the Fresnel pattern from a circular aperture.
10
linearly with distance). The phase of the field at Q is delayed by the wave number (k = 2π/λ) times the
path length (r′ = the distance from S to Q). The field at Q is therefore
EQ = E(x) =
A i(ωt−k r′ (x))
e
.
r′ (x)
(39)
Note that r′ is a function of x, the position in the aperture plane.
Now we can calculate the field at P using the H-F principle. It will be the sum of contributions from all
points like Q in the aperture:
EP
=
ZZ
aperture
=
ZZ
aperture
E(x) −ik r(x)
e
dx
r(x)
′
A
ei(ωt−k r(x)−k r (x)) dx
r′ (x)r(x)
(40)
(41)
In fact, this is not completely correct: we have left out the obliquity factor, and also a constant factor ik/2π.
4.3 Limiting cases
If the source is at a large distance from the aperture, the incident wave will be a plane wave and r′ will be
constant over the aperture, so the terms involving it can be taken outside the integral:
E = A′
ZZ
aperture
e−ik r(x)
dx.
r(x)
(42)
Note that the rapid time dependence (eiωt ) has also been taken outside the integral and absorbed into A′ ,
which is therefore complex.
A neat way to make the source appear to be at infinity is to insert a lens between the source and the aperture.
This is shown in Fig. 8, where wavefronts are shown as dashed lines and rays as solid lines. Since light
slows down in glass, the central parts of the wavefront are delayed more, which alters the shape of the
wavefront. If the source is placed at the focal point of the lens then the effect is to change the spherical
wavefront into a plane wavefront. The aperture can then be placed at any point to the right of the lens, even
right next to it.
source
aperture
Figure 8: Using a lens to convert a spherical wavefront into a plane wavefront.
11
An even more limited case occurs if the point P is also at a large distance. In this case, r is also approximately constant over the aperture (but not when it is inside the cosine function!):
E = A′′
ZZ
e−ik r(x) dx
(43)
aperture
Again, we have absorbed the constant term into A′′ .
Equation 43 is valid if the distance from the aperture to the screen is large, in which case we speak of
Fraunhofer diffraction. In practice, this can be achieved by placing a lens after the diffracting aperture and
observing the pattern at the focal plane of the lens.
If the screen is not far from the aperture then we are in the so-called Fresnel regime, and we must go back
to Equation 42. Calculating Fresnel diffraction patterns is not part of this course; it is difficult analytically
but fairly straightforward with a computer.
Figure 9: A succession of diffraction patterns at increasing distance from a single slit: Fresnel at the bottom
(nearby), going to Fraunhofer at the top (far away). From Hecht (1998).
12
5 Fraunhofer Diffraction
Suppose the coordinates of P are (X, Y, Z), and the coordinates of a point Q on the aperture are (x, y, z),
where z = 0 (see Fig. 10). If the distance QP is r then
r2
= (X − x)2 + (Y − y)2 + (Z − z)2
= X 2 + Y 2 + Z 2 + x2 + y 2 − 2xX − 2yY
x2 + y 2
xX + yY
2
= L 1+
−2
L2
L2
xX + yY
≃ L2 1 − 2
L2
(44)
(45)
(46)
(47)
Here, L = (X 2 + Y 2 + Z 2 )1/2 is the distance OP . Hence
r
xX + yY
L 1−
L2
≃
(48)
and so Equation 43 becomes
E=A
ZZ
eik(xX+yY )/L dx dy.
(49)
aperture
We can rewrite this as
E=A
ZZ
ei(ux+vy) dx dy,
(50)
aperture
where we have introduced new variables u and v to specify the position of P on the screen in terms of the
angles involved (see Fig. 10):
u =
v =
kX/L = k sin θx = 2π sin θx /λ
kY /L = k sin θy = 2π sin θy /λ.
(51)
(52)
y
x
P
r
Q
X Y
L
θy
z
O
Figure 10: Fraunhofer diffraction.
13
Relationship to Fourier transforms
If we write f (x, y) for the aperture function, so that f = 0 where the mask is opaque and f = 1 where it
is transparent, then our equation for Fraunhofer diffraction (Equation 50) becomes
E(u, v) = A
Z
∞
−∞
Z
∞
f (x, y) ei(ux+vy) dx dy.
(53)
−∞
This is the definition of the two-dimensional Fourier transform. In other word, the electric field E(u, v) at
the screen is the Fourier transform of the aperture function f (x, y).
5.1 Condition for Fraunhofer diffraction
The condition for Fraunhofer diffraction is that phase difference kr varies linearly with x. That is, for a
fixed observation point P , the phase difference between contributions from various parts of the aperture
varies linearly with position on the aperture. This is reflected in Equation 48. In deriving this equation for
r, we have neglected second order terms like x2 /L and y 2 /L. For these terms to be ignored, they must be
much less than a wavelength. The condition for Fraunhofer approximation to be valid is therefore
D2 /L ≪ λ,
(54)
where D is the size of the aperture.
5.2 Single slit
If we have a long slit of width b then
f (x, y) =
1
0
if |x| ≤ b/2
if |x| > b/2.
(55)
We assume the slit is infinitely long in the y direction, so we only need to work out things in the x direction.
The integral in Equation 53 then is:
E(u) = A
Z
b/2
Z
= A
eiux dx
(56)
−b/2
b/2
cos(ux) dx + i
−b/2
sin(ux)
= A
u
Z
!
b/2
sin(ux) dx
−b/2
x=b/2
+0
(57)
(58)
x=−b/2
sin(bu/2)
bu/2
sin
β
= A′
β
= A′
(59)
(60)
where
β = bu/2 = bk sin θ/2 = πb sin θ/λ.
(61)
Note that the constants A, A′ , etc., are in general complex (see Equation 43). In particular, they include the
rapid time variation eiωt .
14
For more complicated examples, it is best to stay in complex notation as long as possible. Therefore, let us
repeat the derivation of Equation 60 with this advice in mind:
E(u) = A
Z
b/2
eiux dx
(62)
−b/2
A iux x=b/2
e
x=−b/2
iu
A iub/2
=
e
− e−iub/2
iu
sin(bu/2)
= A′
,
bu/2
=
(63)
(64)
(65)
as before. Here, we have used the identity
sin θ =
eiθ − e−iθ
.
2i
(66)
The function sin β/β is called the sinc function. Note that sinc(0) = 1 (see Fig. 11).
1
sin(x)/x
(sin(x)/x)**2.0
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-25
-20
-15
-10
-5
0
5
10
15
20
25
Figure 11: Fraunhofer diffraction pattern from a single slit.
The irradiance is
I(θ) = I(0)
sin β
β
2
.
(67)
Zeroes in irradiance occur when sin β = 0, except for β = 0 where the sinc function has a maximum.
Hence, zeroes occur when
β
sin θ
= ±π, ±2π, ±3π, . . .
λ
λ
λ
= ± , ±2 , ±3 , . . .
b
b
b
(68)
(69)
Maxima in irradiance occur approximately half way between the zeroes. By differentiating the sinc function, it can be shown that they occur when tan β = β:
β
sin θ
= 0, ±1.43π, ±2.46π, ±3.47π, . . .
λ
λ
λ
= 0, ±1.43 , ±2.46 , ±3.47 , . . .
b
b
b
15
(70)
(71)
5.3 Circular aperture
We want to evaluate Equation 50 for the case of a circular aperture with radius a. We convert to polar
coordinates by setting
x
y
= r cos φ
= r sin φ
X
Y
= R cos Φ
= R sin Φ
so that dx dy = r dr dφ. The integral then becomes
Z a Z 2π
rR
E(R, Φ) = A
eik L (cos φ cos Φ+sin φ sin Φ) r dr dφ
= A
r=0 φ=0
Z a Z 2π
r=0
eik
rR
L
cos(φ−Φ)
r dr dφ
(73)
φ=0
By symmetry this is the same for all values of Φ, so we may as well evaluate at Φ = 0.
Z a Z 2π
rR
E(R) = A
eik L cos φ r dr dφ
r=0
(72)
(74)
φ=0
J1 (kaR/L)
kaR/L
J
1 (ka sin θ)
= A′
ka sin θ
= A′
(75)
(76)
Figure 12: The function J1 (z).
The function J1 (z) is the first-order Bessel function. As shown in Fig. 12, it resembles a sine wave. Note
that as z approaches zero, J1 (z)/z approaches 12 . Contrast with sin z/z, which is 1 at the origin. Hence,
the irradiance is
2
2J1 (ka sin θ)
(77)
I(θ) = I(0)
ka sin θ
The irradiance is very similar to the sinc function (see Fig. 13) and is known as the Airy pattern. Zeroes in
irradiance occur when J1 (ka sin θ) = 0, which happens when
ka sin θ
=
sin θ
=
±3.83, ±7.02, ±10.2, . . .
λ
λ
λ
±1.22 , ±2.23 , ±3.24 , . . .
D
D
D
where D = 2a is the diameter of the aperture. Compare these values with Equation 69.
16
(78)
(79)
Figure 13: Fraunhofer diffraction pattern from a circular aperture. The solid line is 2J1 (z)/z and the
dashed line is the square of that function.
5.4 Spatial resolution and Rayleigh’s criterion
The spatial resolution of an optical system is a measure of how well it can reveal fine detail. The ultimate limit on resolution is diffraction at the most restrictive aperture in the system. This can be seen by
considering the image of two distant objects (for example, a double star) produced by a converging lens.
Since the objects are effectively at infinity, their images will be in the focal plane of the lens. If the angular
separation of the object points is very small, their individual diffraction patterns will overlap and the two
points will not be resolved in the image.
Rayleigh’s criterion is one useful (but arbitrary) way of deciding whether two point sources are resolved. It
states that the minimum angular separation which can be resolved is that which causes the first minimum
of the diffraction pattern of one source to coincide with the central maximum of the diffraction pattern of
the other. Thus, for a circular aperture the angular resolution θmin is given by
sin(θmin ) = 1.22
λ
D
(80)
where D is the diameter of the aperture of the lens. This equation gives the limit of angular resolution
of any imaging system (camera, telescope, microscope, etc.). For any imaging system, diffraction-limited
resolution may not be achieved for other reasons, such as aberrations in the optical system or fluctuations
in the medium through which the light passes (e.g., the Earth’s atmosphere).
Figure 14: Two diffraction patterns which are just resolved according to Rayleigh’s criterion.
17
Figure 15: Fraunhofer diffraction pattern from a double slit with a/b = 3.5. The dashed line is the singleslit sinc envelope.
5.5 Double slit
Suppose each slit has width b and the slit separation is a. The irradiance can be obtained from the integral
in Equation 53, which is the sum of two integrals: one where the integration range is from x = −b/2 to
+b/2; the other from x = a − b/2 to a + b/2. The result is
I(θ) = I(0)
sin β
β
2
cos2 α
(81)
where
α =
β
=
πa
sin θ
λ
πb
bu/2 = 21 bk sin θ =
sin θ.
λ
au/2 = 12 ak sin θ =
(82)
(83)
Note that the result is equivalent to the product of Young’s interference pattern for two slits separated by
a and the diffraction pattern of a single slit of width b. Thus we see interference fringes which have their
intensity modulated by the single slit diffraction pattern. The Fraunhofer diffraction pattern is shown in
Fig. 15 for the case where the slit separation is 3.5 times larger than the slit width. Note that the first
minimum of the diffraction envelope occurs midway between interference maxima of order 3 and 4, that is
at “order” 3.5. Note also that maxima of the double-slit fingres occur when a sin θ is an integral multiple
of λ.
5.6 Multiple slits (diffraction gratings)
The irradiance from an array of N slits is
2 2
sin N α
sin β
I(θ) = I(0)
β
N sin α
(84)
where α and β have the same definitions as before. This result is the product of an N -source pattern
(Equation 38) with a single-slit pattern (Fig. 16). As more slits are added the maxima become sharper
(Fig. 17). The positions of the maxima occur when
a sin θ = mλ,
(85)
where m is an integer. If N is large then the pattern of slits is known as a diffraction grating and Equation 85
is called the grating equation.
18
Figure 16: The components that make up the Fraunhofer pattern from mulitple slits. From top to bottom:
2
2
2
Nα
Nα
(a) sin2 N α, (b) sin2 α, (c) sin
for N = 2 and N = 6, and (d) sinβ 2 β sin
for N = 2 and a = 4b.
sin2 α
sin2 α
From Hecht (1998).
19
Figure 17: Fraunhofer diffraction patterns from multiple slits. The slit widths are all the same (note the
scale on the vertical axes) and a/b = 3.5.
20
5.6.1 Angular dispersion of a diffraction grating
Differentiating Equation 85 with respect to θ gives the angular dispersion of the grating:
dθ
m
=
.
dλ
a cos θ
(86)
This equation gives the angular dispersion in radians per metre, which can then be converted into more
useful units such as degrees per nanometre.
5.6.2 Spectral resolving power of a grating
We wish to know whether some optical system can distinguish two closely-spaced spectral lines. If the
smallest wavelength separation that can be resolved is ∆λ then the resolving power of the system is defined
as
def
R =
λ
∆λ
(87)
To use Rayleigh’s criterion, we need to determine the angle between the principle maximum and the adjacent zero. Referring to Fig. 18, suppose the mth order peak for a given wavelength is at an angle θ. The
path difference between two adjacent slits in direction θ is a sin θ, so the path difference between the first
and N th slits is (N − 1)a sin θ (which we approximate as N a sin θ, assuming N is very large). To find
the nearby angle at which the irradiance drops to zero, we look at a slightly different angle, θ + ∆θ. There
will now be an extra path difference and the irradiance will be zero when this is one wavelength. Thus, the
zero occurs when the path difference between the first and N th slits is N a sin θ + λ. At this angle, the path
difference between two adjacent slits will be a sin θ + λ/N
Suppose the peak of the mth order at λ1 occurs at angle θ1 , and the peak at λ2 occurs at angle θ2 (also
in order m). Then we know that a sin θ1 = mλ1 and a sin θ2 = mλ2 . We now use Rayleigh’s criterion
by arranging that the peak at λ1 coincides with the zero at λ2 . From the above, it follows that a sin θ2 =
a sin θ1 + λ1 /N . Combining these gives m(λ2 − λ1 ) = λ1 /N , which means the resolving power is
R = mN.
(88)
This simple formula shows that the resolving power depends only on the total number of slits and the order.
For gratings used in spectrographs, R is typically somewhere between a few hundred and a few hundred
thousand.
5.6.3 The grating equation at non-normal incidence
If light strikes the grating at an incident angle θi , the maxima occur when
a(sin θ − sin θi ) = mλ,
(89)
where m is an integer. The same equation applies to a reflection grating, provided positive angles are
defined correctly (Fig. 19).
5.6.4 Blazed reflection gratings
When a grating is used to disperse light, it is often desirable to direct as much light as possible into one
order. With a reflection grating this can be achieved by tilting the individual elements (facets), producing
a blazed grating (Fig. 20). The multiple-slit pattern is unchanged but the single-element envelope shifts
according to simple reflection, with its maximum occurring at an angle θ where θi − θb = θ + θb . By
choosing θb (the blaze angle) and θi (the incident angle), it is possible to make θ (the direction of the blaze
peak, which is the single-element envelope) coincide with a non-zero order.
21
a
a sin θ
θ
(N-1)a sin θ
Figure 18: A diffraction grating with N slits. Maxima occur when a sin θ = mλ.
m=1
m=0
m=-1
θi
θ
θ
θi
m=-1
m=0
m=1
Figure 19: Diffraction gratings with non-normal incidence. Left: transmission grating. Right: reflection
grating. In both cases, all angles are positive as drawn.
θb
et
e fac
th
al to
θi
θ
θb
norm
normal to the grating
blaze peak
m=-1
m=0
m=1
Figure 20: A blazed reflection grating.
22
Practice problems (Set 2):
1. The Fourier tranform (FT) of a one-dimensional function f (x) is defined to be
F (u) =
Z
∞
f (x) e−iux dx.
(90)
−∞
(Note that there are slight variations on this√definition, such as including a factor 2π in the exponent
or multiplying by a normalisation factor 1/ 2π.)
The FT of a two-dimensional function f (x, y) is defined to be
F (u, v) =
Z
∞
−∞
Z
∞
f (x, y) e−i(ux+vy) dx dy.
(91)
−∞
(a) Suppose that f is separable, i.e., it can written as the product of functions of x and y alone:
f (x, y) = g(x)h(y). Show that the FT of f is the product of the one-dimensional FTs of g and h.
(b) Use this result to deduce the equation for the Fraunhofer pattern from a rectangular aperture.
2. In Section 5.2 of the notes, the Fraunhofer pattern for a single slit (Eq. 43) is derived for slit whose
edges are at x = −b/2 and x = +b/2. Show that the same intensity pattern is obtained for a slit
which is not centred at the origin (i.e., the slit edges are at x = a − b/2 and x = a + b/2).
This illustrates a general property of Fourier transforms (FTs) known as the Shift Theorem: if a
function is shifted then its FT is multiplied by a phase factor (and therefore the squared modulus of
the FT is unchanged).
3. For the one-dimensional case, show that the squared modulus of the FT of a real function is even.
Make sure you identify the step that relies on the function being real.
What does this imply for the Fraunhofer pattern from an asymmetric aperture, such as a parallel pair
of slits with different widths?
4. Derive the Fraunhofer pattern for an array of N slits and show that it gives the familiar result for
double slits when N = 2.
Hints: Choose coordinates so that the slits are centred at x = 0, a, 2a, . . . and stay in complex
notation for as long as possible.
2
2
5. Consider the Gaussian function: f (x) = e−x /(2σ ) . It can be shown that the Fourier transform of
√
2 2
f (x) is another Gaussian: F (u) = 2πσe−u σ /2 .
Sketch graphs of both functions, indicating the relationships between their heights and widths. Suggestion: show the width of each at the point where the function falls to e−1/2 of the peak. Discuss
how this illustrates the reciprocal relationship between the scales of a function and its Fourier transform. How could this transform pair be illustrated using Fraunhofer diffraction?
6. Let f (x) be the function that is 1 for all values of x. The Fourier transform of f (x) is called the
Dirac delta function, δ(u). Decide what this function looks like and draw a sketch (hint: start with
the single-slit Fourier transform). How can you interpret this result in terms of optics?
7. Babinet’s theorem concerns the interference patterns of two complementary screens (meaning that
the opaque parts of one screen correspond exactly to the transparent parts of the other, and vice
versa). The theorem says that the interference patterns of two complementary screens are identical,
except for a small region near the centre. Prove the one-dimensional version of Babinet’s theorem
in the Fraunhofer region by showing that the Fraunhofer pattern of a screen with aperture function
f (x) is identical to that of the complementary screen for all points not on the axis. Hint: the solution
to the previous question may be useful.
23
8. The ultimate limit for the angular resolution of a telescope is set by diffraction. In movies, people’s faces can sometimes be recognised from spy satellite images. Estimate the minimum diameter
needed for a space telescope in order to achieve this. Assume that face recognition requires about 10
resolution elements (pixels) across a 15 cm human head, and that the satellite is orbiting at an altitude
of 200 km. In terms of diffraction theory, was the movie realistic? What other considerations would
limit the resolving power of a spy satellite?
9. A reflection grating is required that can resolve wavelengths as close as 0.025 Å in second order for
the spectral region around 370 nm (note that the Ångström is a unit of length that is often used in
spectroscopy, where 1 Å equals 10−10 m). The grating is to be installed in an instrument where light
from the entrance slit is incident normally on the grating. If the manufacturer provides rulings over
a 12-cm grating width, determine
(a) the minimum number of grooves/cm required
(b) the optimum blaze angle for work in this region
(c) the angle of diffraction where irradiance is maximum (show both blaze angle and diffraction
angle on a sketch)
(d) the dispersion in nanometers per degree
24
6 Interference from multiple reflections
So far we have dealt with inference between different parts of a wavefront after it passed through an
aperture such as one consisting of two slits. Now we turn to interference which arises when light is divided
by partial reflection and then recombined. You should review the basic laws of geometrical optics given in
Section 1. Also keep in mind that interference between two beams will depend on the optical path length
of each, which is the physical path length times the refractive index: OP L = nd.
6.1 Reflection
Suppose light is incident on an interface between two materials having refractive indices n1 and n2 , respectively. It can be shown (using electromagnetic theory) that, if the incidence is normal, the ratio of the
reflected to the incident electric field amplitudes will be
r=
n1 − n2
n1 + n2
(92)
If n1 < n2 (incidence onto a medium of higher refractive index), the wave undergoes a π phase change
(and r is negative).
The ratio of the reflected to the incident irradiance is r2 , which is called the reflectance. We also define
the transmittance, t2 . Assuming no absorption losses, we have r2 + t2 = 1, but this only holds at normal
incidence. For other angles, the cross-sectional areas of the incident and transmitted beams will not be
equal (because of refraction).
Unit
amplitude
rt 2
−r
t2
r 3t 2
r 2t 2
r 5t 2
r 4t 2
r 6t 2
Figure 21: Multiple internal reflections in a transparent plate, showing the amplitudes of each wave.
6.2 Thin film interference – fringes of equal inclination
Figure 22 shows fringes from a thin film with parallel faces. It can be shown (do it!) that the difference
in OPL between the two beams is 2n2 d cos θ2 . Also note that one or both of the beams may undergo a π
phase change on reflection, depending on the relative refractive indices n1 , n2 and n3 . For example, for a
glass or soap film in air we have n2 > n1 = n3 and one beam will undergo a π phase change (the one that
reflects off the top surface. In this case we therefore expect minima when 2n2 d cos θ2 = mλ and maxima
when 2n2 d cos θ2 = (m + 12 )λ.
The fringes are called fringes of equal inclination. As shown in Fig. 22, they can be seen a the focal plane
of a lens (which may be your eye!) and are therefore said to be localised at infinity.
25
extended source
n1
θ2
d
n2
n3
Figure 22: Fringes of equal inclination from a thin film.
extended source
d
Figure 23: Fringes of equal thickness from a thin film.
26
Figure 24: Michelson interferometer. Drawings copyright HyperPhysics (C.R. Nave, 2006)
6.3 Thin film interference – fringes of equal thickness
Figure 22 shows fringes from a wedge. The rays reflected from the top and bottom surfaces diverge and
can be brought together to interfere by a converging lens. The rays are brought together where the lens
produces an image of the thin film. Thus to see these fringes by eye it is necessary to be focus on the
thin film – hence we say that the fringes are localised at the film. We assume the wedge angle is small
(nearly parallel surfaces) and that the light is incident close to normally (cos θ2 ≃ 1). Noting again that
the phase changes upon reflection at the top and bottom interfaces differ by π, the condition for minima is
2n2 d = mλ and for maxima is 2n2 d = (m + 12 )λ. The fringes are contours of equal thickness.
6.4 Michelson interferometer
Figure 24 shows a Michelson interferometer, which is an important and versatile instrument. Light from
the source is split into two beams by a partially reflecting beamsplitter set at 45◦ . Each beam travels to a
mirror and is reflected back to the beamsplitter, where each beam is again split into two equal beams. The
intensity pattern at the output depends on the path difference between the two arms. Note that moving the
adjustable mirror backwards or forwards by a distance d changes the path length in that arm by 2d.
27
6.5 Fabry-Perot interferometer
As shown in Fig. 25, this device produces interference from multiple reflections between two flat plates.
θ
θ
source
screen
d
Figure 25: Fabry-Perot interferometer showing the formation of circular fringes from multiple reflections.
The optical path difference between two adjacent rays is 2n2 d cos θ2 , where d is the separation of the
plates. The gaps is usually air, so we can set n2 = 1. We also drop the subscript on θ2 . The phase
difference between two adjacent rays is k times the path difference:
δ = 2kd cos θ =
4π
d cos θ
λ
(93)
Referring to Fig. 21, the amplitude of the transmitted wave is a geometric series:
A = t2 + t2 r2 eiδ + t2 r4 ei2δ + . . .
= t2 (1 + r2 eiδ + r4 ei2δ + . . .)
t2
.
=
1 − r2 eiδ
The irradiance of the transmitted light is
I
= AA∗
t2
t2
=
2
iδ
1 − r e 1 − r2 e−iδ
t4
=
4
1 + r − 2r2 cos δ
1
=
1 + F sin2 δ2
(94)
where
F =
4r2
.
(1 − r2 )2
(95)
The function in Equation 94 is called the Airy function and consists of a series of peaks, as shown in
Fig. 26. When δ/2 = mπ the Airy function is equal to unity for all values of F and therefore r. When r
28
Figure 26: Airy function.
approaches one, the transmitted irradiance is very small except within the sharp spikes centred about the
points δ/2 = mπ.
√
It can be shown that the full width at half maximum (FWHM) of these peaks is 4/ F and that the resolving
power of the Fabry-Perot is
R = mF
(96)
√
π F
F=
2
(97)
where
is called the finesse.
Compare this with a diffraction grating, whose resolving power is R = mN , where N is the total
number of grooves in the grating. In the case of the Fabry-Perot, m is very large because d is many
wavelengths. Recall that the order number, m, is the number of wavelengths in the path difference:
m = (optical path difference)/λ. For the on-axis rays (cos θ = 1), we have m = 2d/λ, which will be
many thousands.
29
Practice problems (Set 3):
1. Referring to the left-hand figure, show that the optical path difference between the two reflected rays
is 2n2 d cos θ2 .
θ1
θ1
D
θ1
A
n1
C
d
θ2
n2
0.050 mm
B
n3
200 mm
2. The right-hand figure shows a thin wedge-shaped air film between two optically flat plates of glass,
each of length 200 mm. At one end the plates are in contact; at the other end they are separated by
a thin wire of diameter 0.050 mm. When illuminated from above by 550 nm monochromatic light,
fringes of equal thickness are observed.
(a) What is the fringe spacing?
(b) Is the fringe at the apex bright or dark? Justify your answer.
3. Oil (n = 1.25) leaking from a damaged tanker creates a large oil slick on the harbour (n = 1.33). In
order to determine the thickness of the slick, a plane is flown when the sun is overhead. The sunlight
reflected from directly below the plane is found to have intensity maxima at 455 nm and 610 nm, and
at no wavelengths in between. What is the thickness of the oil slick?
4. On television, it is common to see a suspect being watched by police through a one-way mirror.
From what you have learned about reflection and tranmission of light, discuss why a one-way mirror
is not realistic. Suggest how you would actually create the situation desired by the police (we see
him, but he cannot see us very well).
5. The intensity transmitted by a Fabry-Perot interferometer is given by the Airy function (Equation 68),
where δ is the phase difference between two adjacent paths (Equation 67). Here we calculate the
spectral resolving power of the Fabry-Perot. We cannot apply the Rayleigh criterion because the
intensity does not drop to zero. Instead, we use the criterion that two closely spaced wavelengths are
just resolved when their peaks are separated by one FWHM (full width at half maximum).
(a) Draw a diagram illustrating this criterion.
√
(b) Show that the FWHM of the Airy function is approximately 4/ F .
(c) Suppose that two close wavelengths λ and λ + ∆λ are just resolved. Argue that the resolving
power λ/∆λ is equal to −δ/∆δ, and hence show that the resolving power is R = mF , where
F is defined in Equation 71.
(d) If each plate has a reflectance r2 of 85%, calculate the minimum separation of the plates that
is required if we are to resolve the two components of the 656.3 nm hydrogen line, which are
separated by 0.0136 nm.
30