lecture notes for phys 2911 – optics - school of...
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Lecture Notes for PHYS 2911 – Optics
Prof. Tim Bedding ([email protected])
The subject ofOpticscan be divided into three areas:
Geometrical Opticswhere light is described byrayswhich show the paths of energy transfer. Geometricaloptics provides a good understanding of the propagation of light in transparent media and the operation ofoptical imaging systems such as cameras, telescopes and microscopes.
Physical Opticswhere the wave nature of light is taken into account. Physical optics coverspolarisation,interferenceanddiffractionof light. The physical optics approach is necessary for understanding the limitsof resolution of optical imaging systems.
Quantum Opticswhere the particle nature of light is taken into account. This description, where light isconsidered to consist of massless particles calledphotons, is needed to understand fully the interactionof light and matter. Topics which require the quantum opticsapproach include thephotoelectric effect,photodetectorsandlasers.
This course is onPhysical Optics; it covers general principles, phenomena which require physical opticsfor their explanation and applications. The basic elementsof geometrical optics, covered in the Opticsmodule of the Physics IExperimental Physicscourse, 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
1 Geometrical Optics - Summary
1.1 Refractive Index
The speed of light in a transparent medium is given byc/n wherec is the speed of light in a vacuum andnis 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 and0◦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 sourceare 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 betweentwo transparent media with different refractiveindices, some light is reflected and some (most) is transmitted through the interface as shown in Figure 1.
1
θ1
θ2
2n
θ1n1
’
Figure 1: Reflection and refraction at an interface between two transparent media
TheLaw of Reflectionis
θ′1 = θ1 (1)
where the angle of incidenceθ1 and the angle of reflectionθ′1 are measured with respect to the normal tothe interface.
The Law of Refraction- also calledSnell’s Law- describes the way in which a light ray bends when itpasses 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 betweentwo media when
sin θ1 ≥ n2
n1. (3)
This can occur only when light is incident from the side of higher refractive index (ie whenn1 > n2). Theangle of incidence for whichθ2 = π/2 is called the critical angle,θc. Thus
sin θc =n2
n1(4)
and the condition for total internal reflection isθ1 > θc.
1.5 Thin Lens Formula
1
s+
1
s′=
1
f(5)
wheres is the distance of the object from the lens,s′ is the distance of the image from the lens, andf is thefocal length of the lens. For a converging lensf > 0; for a diverging lensf < 0. For a real images′ > 0;for a virtual images′ < 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 focusin 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 aconstant current (Biot-Savart Law).In the case of anacceleratingcharge (or a changing current), Maxwell’s equations predict additional com-ponents 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 lightto be electric and magnetic influences extending over vast distances, generated by incredibly rapid oscilla-tions of the electrons in atoms. The atoms in a distant star create electromagnetic waves which propagatethrough space and cause electrons in your eye to oscillate.
Any configuration of electric and magnetic fields must satisfy Maxwell’s equations. In integral form, asstudied in Junior Physics, they are:
∮
E · dA =1
ǫ0Qenc (7)
∮
B · dA = 0 (8)∮
E · dl = −dΦB
dt(9)
∮
B · dl = µ0Ienc + µoǫ0dΦE
dt(10)
The third of these equations (Faraday’s Law) indicates thata changing magnetic field produces an elec-tric field. The fourth equation (Ampere’s Law with Maxwell’sadditional term) indicates that a changingelectric field produces a magnetic field. Thus, even if we are far from any charges and currents, we canhave electric and magnetic fields that are continually changing and self-sustaining. As you saw in Juniorphysics, such a configuration can exist in the form of a moving“slab” of electric and magnetic fields thatare perpendicular to each other and to their direction of propagation. It can be shown that this movingpattern of crossed fields satisfies Maxwell’s equations provided the pattern moves at speed1/
√ǫoµ0. As
Maxwell realised, this quantity (which contains the constants in the force laws from electrostatics andmagnetostatics) is equal to the measured speed of light, to within experimental uncertainty. He was led toconclude 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 byaccelerating charges (see Fig. 2). Consider the electric field at large distances from an accelerating chargethat is moving nonrelativistically. The electric field vector lies in the plane perpendicular to the line ofsight. Its component in some directionx in this plane depends on the component of the acceleration ofthecharge in that same direction:
Ex(r, t) =−1
4πǫ0
q
r
1
c2ax
(
t − r
c
)
. (11)
We see thatEx varies inversely asr and is proportional to the component of the acceleration of the chargein the directionx, evaluated at an earlier timet − r/c. Note that the magnetic field has magnitude1/ctimes the electric field, and is perpendicular to bothE and the line of sight. All of this can be deduced fromMaxwell’s equations, and this is covered in Senior Physics.
Equation 11 implies that the field moves as a wave outward fromthe source at speedc. An interestingcase: the charge oscillates with a displacement proportional tocosωt. Then the acceleration, and hence theradiation component of the electric field, also oscillate ascosωt.
2.1 Energy of radiation
The energy content of a wave is proportional to the square of the amplitude. For example, placing a chargein an oscillating electric field produces forces proportional to field strength, hence acceleration and velocity
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Figure 2: Electric field from a single positive charge. Left:the charge is stationary. Right: the chargeunderwent a brief acceleration a short time ago. The “kink” in the field lines propagates outwards atspeedc.
are also proportional to field strength. So the kinetic energy developed in the charge is proportional to thesquareof the field.
The energy that the source can deliver from radiation therefore falls off as the square of the distance. Thismakes sense, since it implies that the total energy integrated over an sphere around the source is independentof the radius of the sphere.
The power per unit area is called theirradianceof light (in the past, it was calledintensity).
2.2 Sinusoidal waves
Suppose we have an oscillator whose displacement is proportional to cosωt. The electric field at some(large) distancer is then
E =Ar
rcos(
ω(
t − rc
))
. (12)
Note that the argument of the cosine function (called the phase) must be dimensionless (it is measured inradians). The quantityAr is the amplitude of the electric field at distancer from the source.
First let us fix the positionr and watch the field as a function of time. It oscillates at the angular frequencyω.This quantityω = 2πf is the rate of change of phase with time (radians per second).In practice, with lightwe cannot see these fast oscillations. Instead, at a fixed point we measure the irradiance, which will beproportional toE2.
Now take a snapshot at a fixed time. That is, fixt and look at the wave as a function of distancer. We noticethat as a function ofr, the field is also oscillatory. The wavelength of this spatial variation isλ = c/f . Wedefine thewave numberto bek = 2π/λ, which is the rate of change of phase with distance (radians permetre). With this definition, and noting thatc = ω/k, we can write Equation 12 as
E =Ar
rcos (ωt − kr). (13)
Provided we are far from the source charge, the factor1/r varies very slowly withr and can therefore beconsidered as a constant and absorbed into the amplitude. This approximation means that we are treatingthe wave as plane rather than a spherical. With this assumption, we have
E = A cos (ωt − kr). (14)
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3 Interference between radiation sources
3.1 Calculation for two sources
Suppose we have two sources which have the same frequency andare radiating in phase. We want to findthe resultant field at some distant point P at some particularangle (Fig. 4). We have to add two cosineswith 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)
wherer1 andr2 are the distances from each of the two sources to point P.
Our aim is to calculateEtot, although note that we are actually interested in the amplitude of the result(which gives the irradiance).
The algebraic method For simplicity, setA1 = A2 = A. We use the rule1:
cosX + cosY = 2 cos(X+Y2 ) cos(X−Y
2 ). (18)
This gives
Etot = A cos(ωt + φ1) + A cos(ωt + φ2) (19)
= 2A cos(
φ1−φ2
2
)
cos(
ωt + φ1+φ2
2
)
(20)
= 2A cos(
k r2−r1
2
)
cos(
ωt − k r1+r2
2
)
(21)
In the final step we have used Equation 17. This is in the same form as Equation 14 – an oscillatoryplane wave with the same frequency as the two original signals, but with a new amplitude and phase. Theamplitude of the resulting wave isAtot = 2A cos
(
k r2−r1
2
)
.
Phasor addition method Any cosine function ofωt can be considered the horizontal projection of arotating phasor. To add two cosine functions, we add their phasors and measure the horizontal componentof the result. Both phasors are rotating with the same angular speedω, so the angle between them staysconstant. Applying the cosine rule to the triangle (Fig. 3),it can be shown (do it!) that
A2tot = A2
1 + A22 + 2A1A2 cos(φ2 − φ1) (22)
= A21 + A2
2 + 2A1A2 cos(
k r2−r1
2
)
, (23)
which agrees with Method 1 for the special caseA1 = A2 = A (prove it).
The complex method We write a complex number for each of the phasors. The real parts of the complexnumbers are the physical quantities (projection onto the horizontal axis).
Etot = A1ei(ωt+φ1) + A2e
i(ωt+φ2) (24)
= (A1eiφ1 + A2e
iφ2)eiωt (25)
1To prove this rule, evaluatecos(a + b) + cos(a − b) and then setX = a + b andY = a − b.
5
ω +φ t 1
A1+ =
t 2ω +φ
φ − φ12
A tot
A 2
Figure 3: Addition of two phasors.
To get the amplitude, we multiply by the complex conjugate:
A2tot = (A1e
iφ1 + A2eiφ2)eiωt(A1e
−iφ1 + A2e−iφ2)e−iωt
= A21 + A2
2 + A1A2(ei(φ1−φ2) + ei(φ2−φ1))
= A21 + A2
2 + 2A1A2 cos(φ2 − φ1) (26)
= A21 + A2
2 + 2A1A2 cos(
k r2−r1
2
)
, (27)
as before.
Thus the sum of the fields from the two sources produces an irradiance which is the sum of the twoindividually, plus an extra term. It is called theinterferenceterm, and it may be positive or negative.
3.2 Application to two sources
To apply Equation 26 to the case of two oscillators, we simplyneed to find the phase difference. Considerthe case of two oscillators of equal amplitude, separated bydistanced. As before, we assume the twosources are in phase.
d θ
to point P
θd sin
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 aphasedifference equal to the number of wavelengths ind sin θ, multiplied by2π. Equivalently, the phase difference equals the distance difference multiplied byk,sincek 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 pointsalong the E–W direction, both oscillators
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contribute equally and their electric fields add in phase. The resulting electric field is twice that of a singleoscillator and the intensity is4I0. At points along the N–S direction, the extraλ/2 path difference meansthe two sources are exactly out of phase, so the resulting electric field is zero. Of course, this is only trueat distances large enough to neglect the difference betweenthe1/r factors.
λ/2
I=0
I=2
I=4I=4
I=0
I=2I=2
I=2
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 setd = λ/2 then the phase difference as a functionof angle in Equation 28 becomesφ2 − φ1 = π sin θ. Putting this into Equation 26 givesA2
tot = 2A2(1 +cos(π sin θ)). If we evaluate this forθ = 0 andπ/2 we get4A2 and zero, respectively. One final example:the angle at which the intensity is2I0 is θ = π/6 (i.e.,30◦).
Example 2 Now suppose the separation is many wavelengths. In this case, the phase difference (Equa-tion 28) issin θ times a large number, which goes from zero to a large positivenumber, back through zeroand to a large negative number. That is, the phase goes through many multiples of2π asθ goes aroundonce. When we put this inside the cosine function (Equation 26), the resulting intensity goes through manypeaks and zeroes.
3.3 Interference betweenN sources
Now consider N equally spaced oscillators, all having the same amplitude but different in phase from oneanother, either because they are driven at different phasesor because we are looking at them from an angle.The total electric field at P is
Etot = E1 + E2 + . . . + EN
= A [cosωt + cos(ωt + φ) + cos(ωt + 2φ) + . . . + cos(ωt + (N − 1)φ)] , (29)
whereφ 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 phasorshave lengthA. The first has zero phase, the next has phaseφ, the next has phase2φ, and so on. All thevertices lie on a circle.
The radius of the circle satisfies both
sin(φ/2) =A/2
r(30)
7
φ
φA
N /2φ
Figure 6: Phasor addition ofN = 7 sources.
and also
sin(Nφ/2) =Atot/2
r. (31)
Eliminatingr gives:
Atot = Asin(Nφ/2)
sin(φ/2). (32)
The complex method We write a complex number for each of the phasors. The real parts of the complexnumbers are the physical quantities (projection onto the horizontal axis).
Etot = A[
eiωt + ei(ωt+φ) + ei(ωt+2φ) + . . . + ei(ωt+(N−1)φ)]
(33)
= Aeiωt[
1 + eiφ + e2iφ + . . . + e(N−1)iφ]
(34)
= Aeiωt eiNφ − 1
eiφ − 1(35)
= Aeiωt eiNφ/2(
eiNφ/2 − e−iNφ/2)
eiφ/2(
eiφ/2 − e−iφ/2) (36)
= Aeiωt ei(N−1)φ/2 sin(Nφ/2)
sin(φ/2). (37)
Taking the amplitude gives the same result as Equation 32.
Result The irradiance is
I = I0sin2(Nφ/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 (632.8nm) in air pass through two parallel slits in anopaque 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 theslits. (b) What is the orientation of the slits?
2. Show that forN = 2, the irradiance calculated in Equation 38 agrees with the two-source case(Equation 26).
3. Use the complex representation to find the resultantE = E1 + E2, where
E1 = A cos(kx + ωt)
and
E2 = −A cos(kx − ωt).
Describe the resultant wave.
4. Show that that the sum ofN sinusoids with identical periods but different phases is also a sinusoid.Hint: you need to consider adding terms likeAj cos(ωt + φj), and it is only necessary to show fortwo sinusoids (why?). Try using complex notation!
Describe briefly the implication of this result for the electric radiation field from a large number ofoscillating electrons that is viewed through a monochromatic filter.
5. Consider the sum ofN vectors
EN =
N∑
j=1
ej ,
where all theej have unit amplitude but random directions. Show that the expected value for|EN |2,which we write as〈|EN |2〉, is N . Hint: use mathematical induction and apply the cosine ruleto avector diagram.
Describe briefly the implication of this result for the electric radiation field from a large number ofoscillating electrons that is viewed through a monochromatic filter.
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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 waveletsof the same frequency as the primary wave. The amplitude of the optical field at any point beyond is thesuperposition of all these wavelets (considering their amplitudes and relative phases).
Note that this statement would predict secondary wavelets in all directions, even backwards. In practicethis is not observed. Instead, the secondary wave is concentrated in the forwards direction. We can accountfor this by saying that the amplitude of the secondary wavelet in directionθ is reduced by an amountK(θ),which is the so-called obliquity factor. More detailed calculations show thatK(θ) = 1
2 (1 + cos θ). Notethat this factor is unity in the forward direction (θ = 0) and zero in the backwards direction (θ = π). Inmost practical situations the obliquity factor can be approximated by unity.2
What do we mean by awavefront? The best definition isa surface of constant phase. We can also talkaboutrays, especially in geometrical optics, and these are lines thatare perpendicular to the wavefront andindicate the direction of energy flow.
Diffraction by an aperture Our task is to find out what happens when we shine a light at an opaquemask with some holes in it. Most people would say that light shines through the holes and produces andeffect on the other side. Huygens’ principle says that we should pretend there are sources distributeduniformly across the open holes and that the phases of these sources are the same as they would be if theopaque material were absent. Of course there areno sources at the holes – indeed, this is the only placewhere there are certainly no sources. For an explanation of why Huygen’s principle gives (almost) the rightanswer, see Hecht (section 10.1.1) or Feynman (section 31-6).
4.2 Calculating diffraction using Huygens-Fresnel principle
Consider light from a point sourceS which passes through an aperture (Fig. 7). We want to calculate theresulting field atP , on the other side of the aperture.
r’
Q
O
S
xP
r
Figure 7: Diffraction through an aperture.
To apply the H-F principle, we need to know the field at each point in the aperture. LetQ be a point in theaperture with position vectorx (where the origin is pointO).
Suppose the source atS is monochromatic with angular frequencyω, so that it oscillates ascosωt. Wediscussed this in Section 2.2. The amplitude of the field atQ goes as1/r′ (recall that amplitude falls off
2An exception is in calculating the Fresnel pattern from a circular aperture.
10
linearly with distance). The phase of the field atQ is delayed by the wave number (k = 2π/λ) times thepath length (r′ = the distance fromS to Q). The field atQ is therefore
EQ = E(x) =A
r′(x)ei(ωt−k r′(x)). (39)
Note thatr′ is a function ofx, the position in the aperture plane.
Now we can calculate the field atP using the H-F principle. It will be the sum of contributions from allpoints likeQ in the aperture:
EP =
∫ ∫
aperture
E(x)
r(x)e−ik r(x) dx (40)
=
∫ ∫
aperture
A
r′(x)r(x)ei(ωt−k r(x)−k r′(x)) dx (41)
In fact, this is not completely correct: we have left out the obliquity factor, and also a constant factorik/2π.
4.3 Limiting cases
If the source is at a large distance from the aperture, the incident wave will be a plane wave andr′ will beconstant over the aperture, so the terms involving it can be taken outside the integral:
E = A′
∫ ∫
aperture
e−ik r(x)
r(x)dx. (42)
Note that the rapid time dependence (eiωt) has also been taken outside the integral and absorbed intoA′,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, wherewavefrontsare shown as dashed lines andrays as solid lines. Since lightslows down in glass, the central parts of the wavefront are delayed more, which alters the shape of thewavefront. If the source is placed at the focal point of the lens then the effect is to change the sphericalwavefront into a plane wavefront. The aperture can then be placed at any point to the right of the lens, evenright next to it.
source
aperture
Figure 8: Using a lens to convert a spherical wavefront into aplane wavefront.
11
An even more limited case occurs if the pointP is also at a large distance. In this case,r is also approxi-mately constant over the aperture (but not when it is inside the cosine function!):
E = A′′
∫ ∫
aperture
e−ik r(x) dx (43)
Again, we have absorbed the constant term intoA′′.
Equation 43 is valid if the distance from the aperture to the screen is large, in which case we speak ofFraunhofer diffraction. In practice, this can be achieved by placing a lensafter the diffracting aperture andobserving the pattern at the focal plane of the lens.
If the screen isnot far from the aperture then we are in the so-called Fresnel regime, and we must go backto Equation 42. Calculating Fresnel diffraction patterns is not part of this course; it is difficult analyticallybut 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 ofP are(X, Y, Z), and the coordinates of a pointQ on the aperture are(x, y, z),wherez = 0 (see Fig. 10). If the distanceQP is r then
r2 = (X − x)2 + (Y − y)2 + (Z − z)2 (44)
= X2 + Y 2 + Z2 + x2 + y2 − 2xX − 2yY (45)
= L2
[
1 +x2 + y2
L2− 2
xX + yY
L2
]
(46)
≃ L2
[
1 − 2xX + yY
L2
]
(47)
Here,L = (X2 + Y 2 + Z2)1/2 is the distanceOP . Hence
r ≃ L
[
1 − xX + yY
L2
]
(48)
and so Equation 43 becomes
E = A
∫ ∫
aperture
eik(xX+yY )/L dx dy. (49)
We can rewrite this as
E = A
∫ ∫
aperture
ei(ux+vy) dx dy, (50)
where we have introduced new variablesu andv to specify the position ofP on the screen in terms of theanglesinvolved (see Fig. 10):
u = kX/L = k sin θx = 2π sin θx/λ (51)
v = kY/L = k sin θy = 2π sin θy/λ. (52)
y
y
z
x
rP
Q
O
Y
θ
XL
Figure 10: Fraunhofer diffraction.
13
Relationship to Fourier transforms
If we write f(x, y) for the aperture function, so thatf = 0 where the mask is opaque andf = 1 where itis transparent, then our equation for Fraunhofer diffraction (Equation 50) becomes
E(u, v) = A
∫
∞
−∞
∫
∞
−∞
f(x, y) ei(ux+vy) dx dy. (53)
This is the definition of the two-dimensional Fourier transform. In other words, the electric fieldE(u, v)at the screen is the Fourier transform of the aperture functionf(x, y).
5.1 Condition for Fraunhofer diffraction
The condition for Fraunhofer diffraction is that phase differencekr varies linearly withx. That is, for afixed observation pointP , the phase difference between contributions from various parts of the aperturevaries linearly with position on the aperture. This is reflected in Equation 48. In deriving this equation forr, we have neglected second order terms likex2/L andy2/L. For these terms to be ignored, they must bemuch less than a wavelength. The condition for Fraunhofer approximation to be valid is therefore
D2/L ≪ λ, (54)
whereD is the size of the aperture.
5.2 Single slit
If we have a long slit of widthb then
f(x, y) =
{
1 if |x| ≤ b/20 if |x| > b/2.
(55)
We assume the slit is infinitely long in they direction, so we only need to work out things in thex direction.
The integral in Equation 53 then is:
E(u) = A
∫ b/2
−b/2
eiux dx (56)
= A
(
∫ b/2
−b/2
cos(ux) dx + i
∫ b/2
−b/2
sin(ux) dx
)
(57)
= A
[
sin(ux)
u
]x=b/2
x=−b/2
+ 0 (58)
= A′sin(bu/2)
bu/2(59)
= A′sinβ
β(60)
where
β = bu/2 = bk sin θ/2 = πb sin θ/λ. (61)
Note that the constantsA, A′, etc., are in general complex (see Equation 43). In particular, they include therapid time variationeiωt.
14
For more complicated examples, it is best to stay in complex notation as long as possible. Therefore, let usrepeat the derivation of Equation 60 with this advice in mind:
E(u) = A
∫ b/2
−b/2
eiux dx (62)
=A
iu
[
eiux]x=b/2
x=−b/2(63)
=A
iu
(
eiub/2 − e−iub/2)
(64)
= A′sin(bu/2)
bu/2, (65)
as before. Here, we have used the identity
sin θ =eiθ − e−iθ
2i. (66)
The functionsin β/β is called the sinc function. Note thatsinc(0) = 1 (see Fig. 11).
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-25 -20 -15 -10 -5 0 5 10 15 20 25
sin(x)/x(sin(x)/x)**2.0
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 whensin β = 0, except forβ = 0 where the sinc function has a maximum.Hence, zeroes occur when
β = ±π, ±2π, ±3π, . . . (68)
sin θ = ±λ
b,±2
λ
b,±3
λ
b, . . . (69)
Maxima in irradiance occur approximately half way between the zeroes. By differentiating the sinc func-tion, it can be shown that they occur whentanβ = β:
β = 0,±1.43π, ±2.46π, ±3.47π, . . . (70)
sin θ = 0,±1.43λ
b,±2.46
λ
b,±3.47
λ
b, . . . (71)
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5.3 Circular aperture
We want to evaluate Equation 50 for the case of a circular aperture with radiusa. We convert to polarcoordinates by setting
x = r cosφ
y = r sin φ
X = R cosΦ
Y = R sin Φ
so thatdx dy = r dr dφ. The integral then becomes
E(R, Φ) = A
∫ a
r=0
∫ 2π
φ=0
eik rR
L(cos φ cosΦ+sin φ sin Φ)r dr dφ (72)
= A
∫ a
r=0
∫ 2π
φ=0
eik rR
Lcos(φ−Φ)r dr dφ (73)
By symmetry this is the same for all values ofΦ, so we may as well evaluate atΦ = 0.
E(R) = A
∫ a
r=0
∫ 2π
φ=0
eik rR
Lcos φr dr dφ (74)
= A′J1(kaR/L)
kaR/L(75)
= A′J1(ka sin θ)
ka sin θ(76)
Figure 12: The functionJ1(z).
The functionJ1(z) is the first-order Bessel function. As shown in Fig. 12, it resembles a sine wave. Notethat asz approaches zero,J1(z)/z approaches12 . Contrast withsin z/z, which is 1 at the origin. Hence,the irradiance is
I(θ) = I(0)
(
2J1(ka sin θ)
ka sin θ
)2
(77)
The irradiance is very similar to the sinc function (see Fig.13) and is known as the Airy pattern. Zeroes inirradiance occur whenJ1(ka sin θ) = 0, which happens when
ka sin θ = ±3.83, ±7.02, ±10.2, . . . (78)
sin θ = ±1.22λ
D,±2.23
λ
D,±3.24
λ
D, . . . (79)
whereD = 2a is the diameter of the aperture. Compare these values with Equation 69.
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Figure 13: Fraunhofer diffraction pattern from a circular aperture. The solid line is2J1(z)/z and thedashed 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 ulti-mate limit on resolution is diffraction at the most restrictive aperture in the system. This can be seen byconsidering the image of two distant objects (for example, adouble 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 angularseparation of the object points is very small, their individual diffraction patterns will overlap and the twopoints will not be resolved in the image.
Rayleigh’s criterion is one useful (but arbitrary) way of deciding whether two point sources are resolved. Itstates that the minimum angular separation which can be resolved is that which causes the first minimumof the diffraction pattern of one source to coincide with thecentral maximum of the diffraction pattern ofthe other. Thus, for a circular aperture the angular resolution θmin is given by
sin(θmin) = 1.22λ
D(80)
whereD is the diameter of the aperture of the lens. This equation gives the limit of angular resolutionof any imaging system (camera, telescope, microscope, etc.). For any imaging system,diffraction-limitedresolution may not be achieved for other reasons, such as aberrations in the optical system or fluctuationsin the medium through which the light passes (e.g., the Earth’s atmosphere).
Figure 14: Two diffraction patterns which are just resolvedaccording to Rayleigh’s criterion.
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Figure 15: Fraunhofer diffraction pattern from a double slit with a/b = 3.5. The dashed line is the single-slit sinc envelope.
5.5 Double slit
Suppose each slit has widthb and the slit separation isa. The irradiance can be obtained from the integralin Equation 53, which is the sum of two integrals: one where the integration range is fromx = −b/2 to+b/2; the other fromx = a − b/2 to a + b/2. The result is
I(θ) = I(0)
(
sin β
β
)2
cos2 α (81)
where
α = au/2 = 12ak sin θ =
πa
λsin θ (82)
β = bu/2 = 12bk sin θ =
πb
λsin θ. (83)
Note that the result is equivalent to the product of Young’s interference pattern for two slits separated bya and the diffraction pattern of a single slit of widthb. Thus we see interference fringes which have theirintensity modulated by the single slit diffraction pattern. The Fraunhofer diffraction pattern is shown inFig. 15 for the case where the slit separation is 3.5 times larger than the slit width. Note that the firstminimum of the diffraction envelope occurs midway between interference maxima of order 3 and 4, that isat “order” 3.5. Note also that maxima of the double-slit fringes occur whena sin θ is an integral multipleof λ.
5.6 Multiple slits (diffraction gratings)
The irradiance from an array ofN slits is
I(θ) = I(0)
(
sin β
β
)2(sin Nα
N sin α
)2
(84)
whereα andβ 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)
wherem is an integer. IfN is large then the pattern of slits is known as a diffraction grating and Equation 85is called the grating equation.
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Figure 16: The components that make up the Fraunhofer pattern from mulitple slits. From top to bottom:(a) sin2 Nα, (b) sin2 α, (c) sin2 Nα
sin2 αfor N = 2 andN = 6, and (d)sin2 β
β2
sin2 Nαsin2 α
for N = 2 anda = 4b.From Hecht (1998).
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Figure 17: Fraunhofer diffraction patterns from multiple slits. The slit widths are all the same (note thescale on the vertical axes) anda/b = 3.5.
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5.6.1 Angular dispersion of a diffraction grating
Differentiating Equation 85 with respect toθ gives theangular dispersionof the grating:
dθ
dλ=
m
a cos θ. (86)
This equation gives the angular dispersion in radians per metre, which can then be converted into moreuseful 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 thesmallest wavelength separation that can be resolved is∆λ then theresolving powerof the system is definedas
Rdef=
λ
∆λ(87)
To use Rayleigh’s criterion, we need to determine the angle between the principle maximum and the adja-cent zero. Referring to Fig. 18, suppose themth order peak for a given wavelength is at an angleθ. Thepath difference between two adjacent slits in directionθ is a sin θ, so the path difference between the firstandN th slits is(N − 1)a sin θ (which we approximate asNa sin θ, assumingN is very large). To findthe nearby angle at which the irradiance drops to zero, we look at a slightly different angle,θ + ∆θ. Therewill now be an extra path difference and the irradiance will be zero when this is one wavelength. Thus, thezero occurs when the path difference between the first andN th slits isNa sin θ +λ. At this angle, the pathdifference between twoadjacentslits will bea sin θ + λ/N
Suppose the peak of themth order atλ1 occurs at angleθ1, and the peak atλ2 occurs at angleθ2 (alsoin orderm). Then we know thata sin θ1 = mλ1 anda sin θ2 = mλ2. We now use Rayleigh’s criterionby arranging that the peak atλ1 coincides with the zero atλ2. From the above, it follows thata sin θ2 =a sin θ1 + λ1/N . Combining these givesm(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 hundredthousand.
5.6.3 The grating equation at non-normal incidence
If light strikes the grating at an incident angleθi, the maxima occur when
a(sin θ − sin θi) = mλ, (89)
wherem is an integer. The same equation applies to a reflection grating, provided positive angles aredefined 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 oneorder. With a reflection grating this can be achieved by tilting the individual elements (facets), producinga blazedgrating (Fig. 20). The multiple-slit pattern is unchanged but the single-element envelope shiftsaccording to simple reflection, with its maximum occurring at an angleθ whereθi − θb = θ + θb. Bychoosingθb (theblazeangle) andθi (the incident angle), it is possible to makeθ (the direction of the blazepeak, which is the single-element envelope) coincide with anon-zero order.
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θ
θ(N-1)a sin
θa sin a
Figure 18: A diffraction grating withN slits. Maxima occur whena sin θ = mλ.
θθ
i
m=1m=0
m=-1
θ
m=-1
m=1m=0
θi
Figure 19: Diffraction gratings with non-normal incidence. Left: transmission grating. Right: reflectiongrating. In both cases, all angles are positive as drawn.
normal to the grating
θb
m=0m=-1
m=1
θbθiθ
blaze peak
normal to the facet
Figure 20: A blazed reflection grating.
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Practice problems (Set 2):
1. The Fourier tranform (FT) of a one-dimensional functionf(x) is defined to be
F (u) =
∫
∞
−∞
f(x) e−iux dx. (90)
(Note that there are slight variations on this definition, such as including a factor2π in the exponentor multiplying by a normalisation factor1/
√2π.)
The FT of a two-dimensional functionf(x, y) is defined to be
F (u, v) =
∫
∞
−∞
∫
∞
−∞
f(x, y) e−i(ux+vy) dx dy. (91)
(a) Suppose thatf is separable, i.e., it can written as the product of functions of x andy alone:f(x, y) = g(x)h(y). Show that the FT off is the product of the one-dimensional FTs ofg andh.(b) Use this result to deduce the equation for the Fraunhoferpattern 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 whoseedges are atx = −b/2 andx = +b/2. Show that the same intensity pattern is obtained for a slitwhich is not centred at the origin (i.e., the slit edges are atx = a − b/2 andx = a + b/2).
This illustrates a general property of Fourier transforms (FTs) known as the Shift Theorem: if afunction is shifted then its FT is multiplied by a phase factor (and therefore the squared modulus ofthe 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 pairof slits with different widths?
4. Derive the Fraunhofer pattern for an array ofN slits and show that it gives the familiar result fordouble slits whenN = 2.
Hints: Choose coordinates so that the slits are centred atx = 0, a, 2a, . . . and stay in complexnotation for as long as possible.
5. Consider the Gaussian function:f(x) = e−x2/(2σ2). It can be shown that the Fourier transform off(x) is another Gaussian:F (u) =
√2πσe−u2σ2/2.
Sketch graphs of both functions, indicating the relationships between their heights and widths. Sug-gestion: show the width of each at the point where the function falls toe−1/2 of the peak. Discusshow this illustrates the reciprocal relationship between the scales of a function and its Fourier trans-form. How could this transform pair be illustrated using Fraunhofer diffraction?
6. Let f(x) be the function that is1 for all values ofx. The Fourier transform off(x) is called theDirac delta function, δ(u). Decide what this function looks like and draw a sketch (hint: start withthe single-slit Fourier transform). How can you interpret this result in terms of optics?
7. Babinet’s theorem concerns the interference patterns oftwo complementary screens (meaning thatthe opaque parts of one screen correspond exactly to the transparent parts of the other, and viceversa). 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 theoremin the Fraunhofer region by showing that the Fraunhofer pattern of a screen with aperture functionf(x) is identical to that of the complementary screen for all points not on the axis. Hint: the solutionto the previous question may be useful.
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8. The ultimate limit for the angular resolution of a telescope is set by diffraction. In movies, peo-ple’s faces can sometimes be recognised from spy satellite images. Estimate the minimum diameterneeded for a space telescope in order to achieve this. Assumethat face recognition requires about 10resolution elements (pixels) across a 15 cm human head, and that the satellite is orbiting at an altitudeof 200 km. In terms of diffraction theory, was the movie realistic? What other considerations wouldlimit the resolving power of a spy satellite?
9. A reflection grating is required that can resolve wavelengths as close as 0.025A in second order forthe spectral region around 370 nm (note that theAngstrom is a unit of length that is often used inspectroscopy, where 1A equals10−10 m). The grating is to be installed in an instrument where lightfrom the entrance slit is incident normally on the grating. If the manufacturer provides rulings overa 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 diffractionangle on a sketch)
(d) the dispersion in degrees per nanometre
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6 Interference from multiple reflections
So far we have dealt with inference between different parts of a wavefront after it passed through anaperture such as one consisting of two slits. Now we turn to interference which arises when light is dividedby partial reflection and then recombined. You should reviewthe basic laws of geometrical optics given inSection 1. Also keep in mind that interference between two beams will depend on theoptical path lengthof each, which is the physical path length times the refractive index:OPL = nd.
6.1 Reflection
Suppose light is incident on an interface between two materials having refractive indicesn1 andn2, re-spectively. It can be shown (using electromagnetic theory)that, if the incidence is normal, the ratio of thereflected 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 aπ phase change(andr is negative).
The ratio of the reflected to the incidentirradiance is r2, which is called thereflectance. We also definethe transmittance, t2. Assuming no absorption losses, we haver2 + t2 = 1, but this only holds at normalincidence. For other angles, the cross-sectional areas of the incident and transmitted beams will not beequal (because of refraction).
232rt 25−r r t r t
2t r t 2 r t 22 4
Unitamplitude
2r t 6
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 differencein OPL between the two beams is2n2d cos θ2. Also note that one or both of the beams may undergo aπphase change on reflection, depending on the relative refractive indicesn1, n2 andn3. For example, for aglass or soap film in air we haven2 > n1 = n3 and one beam will undergo aπ phase change (the one thatreflects off the top surface. In this case we therefore expectminima when2n2d cos θ2 = mλ and maximawhen2n2d cos θ2 = (m + 1
2 )λ.
The fringes are calledfringes of equal inclination. As shown in Fig. 22, they can be seen a the focal planeof a lens (which may be your eye!) and are therefore said to be localised at infinity.
25
θ2
n2
extended source
dd
n1
n3
Figure 22: Fringes of equal inclination from a thin film.
d
extended source
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 andcan be brought together to interfere by a converging lens. The rays are brought together where the lensproduces an image of the thin film. Thus to see these fringes byeye it is necessary to be focus on thethin 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 thatthe phase changes upon reflection at the top and bottom interfaces differ byπ, the condition for minima is2n2d = mλ and for maxima is2n2d = (m + 1
2 )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 fromthe source is split into two beams by a partially reflecting beamsplitter set at 45◦. Each beam travels to amirror and is reflected back to the beamsplitter, where each beam is again split into two equal beams. Theintensity pattern at the output depends on the path difference between the two arms. Note that moving theadjustable mirror backwards or forwards by a distanced changes the path length in that arm by2d.
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6.5 Fabry-Perot interferometer
As shown in Fig. 25, this device produces interference from multiple reflections between two flat plates.
source
θ
screen
θ
d
Figure 25: Fabry-Perot interferometer showing the formation of circular fringes from multiple reflections.
The optical path difference between two adjacent rays is2n2d cos θ2, whered is the separation of theplates. The gaps is usually air, so we can setn2 = 1. We also drop the subscript onθ2. The phasedifference between two adjacent rays isk times the path difference:
δ = 2kd cos θ =4π
λd cos θ (93)
Referring to Fig. 21, the amplitude of the transmitted wave is a geometric series:
A = t2 + t2r2eiδ + t2r4ei2δ + . . .
= t2(1 + r2eiδ + r4ei2δ + . . .)
=t2
1 − r2eiδ.
The irradiance of the transmitted light is
I = AA∗
=t2
1 − r2eiδ
t2
1 − r2e−iδ
=t4
1 + r4 − 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 inFig. 26. Whenδ/2 = mπ the Airy function is equal to unity for all values ofF and thereforer. Whenr
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Figure 26: Airy function.
approaches one, the transmitted irradiance is very small except within the sharp spikes centred about thepointsδ/2 = mπ.
It can be shown that the full width at half maximum (FWHM) of these peaks is4/√
F and that the resolvingpower of the Fabry-Perot is
R = mF (96)
where
F =π√
F
2(97)
is called thefinesse.
Compare this with a diffraction grating, whose resolving power is R = mN , whereN is the totalnumber of grooves in the grating. In the case of the Fabry-Perot, m is very large becaused is manywavelengths. 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 havem = 2d/λ, which will bemany thousands.
29
Practice problems (Set 3):
1. Referring to the left-hand figure, show that the optical path difference between the two reflected raysis 2n2d cos θ2.
n2
θ1
θ2
θ1
θ1
dd
n1
n3B
C
D
A
200 mm
0.050 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 bya thin wire of diameter 0.050mm. When illuminated from aboveby 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 = 1.65) leaking from a damaged tanker creates a large oil slick on the harbour (n = 1.33). Inorder to determine the thickness of the slick, a plane is flownwhen the sun is overhead. The sunlightreflected from directly below the plane is found to have intensity maxima (correction: should beintensityminima) at 450 nm and 600 nm, and at no wavelengths in between. What isthe thickness ofthe 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 mirroris not realistic. Suggest how you would actually create the situation desired by the police (we seehim, but he cannot see us very well).
5. The intensity transmitted by a Fabry-Perot interferometer is given by the Airy function (Equation 68),whereδ is the phase difference between two adjacent paths (Equation 67). Here we calculate thespectral resolving power of the Fabry-Perot. We cannot apply the Rayleigh criterion because theintensity does not drop to zero. Instead, we use the criterion that two closely spaced wavelengths arejust 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 approximately4/√
F .
(c) Suppose that two close wavelengthsλ andλ + ∆λ are just resolved. Argue that the resolvingpowerλ/∆λ is equal to−δ/∆δ, and hence show that the resolving power isR = mF , whereF is defined in Equation 71.
(d) If each plate has a reflectancer2 of 85%, calculate the minimum separation of the plates thatis required if we are to resolve the two components of the 656.3nm hydrogen line, which areseparated by 0.0136nm.
30