Wave Optics

Question 1

Maxwell’s eqs (+ free space)

Answer 1

∮ₛ E.dA = Qₑₙ꜀/∈₀
∮ₛ B.dA = 0
∮꜀ E.dL = -d/dt ∫ₛB.dA
∮꜀ B.dL = µ₀∫ₛ(j+∈₀ ∂E/∂t).dA
———————————-
∇.E = ρ/ϵ₀
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀(J+ϵ₀ ∂E/∂t)
———–Free Space————————–
∇.E = 0
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀ϵ₀ ∂E/∂t

Question 2

Derive the wave equation in free space from Maxwell’s laws

Answer 2

∇ × (∇ × E) = −∂/∂t ∇×B
∇(∇.E) − ∇²E = −µ₀ϵ₀ ∂/∂t (∂/∂t E)
∇²E = µ₀ϵ₀ ∂²E/∂t²

OR

∇ × (∇ × B) = µ₀ϵ₀ ∂/∂t ∇×E
∇(∇.B) − ∇²B = -µ₀ϵ₀ ∂/∂t (∂/∂t B)
∇²B = µ₀ϵ₀ ∂²B/∂t²

Question 3

Standard wave equation

Answer 3

∇²E = 1/c² ∂²E/∂t²

Question 4

Plane wave

Answer 4

E = E₀ exp(i(k.r-ωt))

B = B₀ exp(i(k.r-ωt))

Question 5

Wavevector

Answer 5

k = (kₓ, kᵧ, kᶻ) = (2π/λₓ, 2π/λᵧ, 2π/λᶻ)

Question 6

What is a wavefront

Answer 6

Surfaces or loci of contiguous points on a wave, with the same phase.

Question 7

Show that k is perpendicular to B (and E in free space)

Answer 7

B = (Bₓ, Bᵧ, Bᶻ) = exp(−iωt) (B₀ₓexp(ik.r), B₀ᵧexp(ik.r), B₀ᶻexp(ik.r),

∇.B = ∂Bₓ/∂x + ∂Bᵧ/∂y + ∂Bᶻ/∂z = 0
ikₓBₓ + ikᵧBᵧ + ikᶻBᶻ = 0
k.B = 0

In free space:
E = (Eₓ, Eᵧ, Eᶻ) = exp(−iωt) (E₀ₓexp(ik.r), E₀ᵧexp(ik.r), E₀ᶻexp(ik.r),

∇.E = ∂Eₓ/∂x + ∂Eᵧ/∂y + ∂Eᶻ/∂z = 0
ikₓEₓ + ikᵧEᵧ + ikᶻEᶻ = 0
k.E = 0

Question 8

Derive relation between k̂, E, and B

Answer 8

Using ∇ x E = -∂B/∂t
and ∇ x E = i(kₓ,kᵧ,kᶻ) x E

-∂B/∂t = iωB
k x E = ωB
k̂ x E = cB

Question 9

Poynting vector

Answer 9

Energy flux
S = 1/µ₀ E x B

Question 10

Irradiance/ intensity of a plane wave

Answer 10

E = E₀ cos(k.r − ωt), B = B₀ cos(k.r − ωt)
S = (E₀ × B₀)/µ₀ cos²(k.r − ωt)

I = ⟨|S|⟩ =|E₀ × B₀|/2µ₀ = E₀²/2µ₀c = cϵ₀E₀²/2 = √µ₀’E₀²/2√ϵ₀’
(using E₀ = cB₀ and c²=1/µ₀ϵ₀)

Define η₀ = √µ₀/ϵ₀’

I = E₀²/2η₀

Question 11

Radiation pressure from perfect reflection

Answer 11

2⟨|S|⟩/c = ϵ₀E₀²

Question 12

Find solution to 1D wave equation

Answer 12

∂²ψ/∂x² - 1/v² ∂²ψ/∂t² = 0
factorise
(∂/∂x + 1/v ∂/∂t)(∂/∂x - 1/v ∂/∂t) ψ = 0
requires ψ= f(x-vt) + g(x+vt)
(right)+(left) travelling waves
Waves are sum of any function that preserves shape whilst travelling

Question 13

Derive Helmholtz equation

Answer 13

∇²E = 1/v² ∂²E∂t²
General solution: E = Eᵣ(r)exp(±ikvt)
Sub in, compare sides
Eᵣ = -k²exp(±ikvt)

get Helmholtz eqn
∇²Eᵣ = -k²Eᵣ

note that kv = ω
can solve helmholtz for spacial part, then multiphy by exp(±ikvt) for full solution

Question 14

Plane wave solution to 3D wave equation

Answer 14

∂²E/∂x² + ∂²E/∂y² + ∂²E/∂z² = 1/v² ∂²E/∂t²

solutions
E = E₀ exp(±i(k.r ± ωt))
Eᵣ = Eᵣ₀ exp(±ik.r)

Can be found using k.r = kₓx+kᵧy+kᶻz and ω = vk, where k² = k²x + k²y + k²z

Question 15

Spherically symmetric solution to 3D wave equation

Answer 15

Spherical symmetry, so Eᵣ scalar, can drop subscript.

Spherical polar:
∇²E = −k²E =1/r² ∂/∂r (r² ∂E/∂r)
⇒ r² ∂²E/∂r² + 2r ∂E/∂r + r²k²E = 0

Bessel equation, its solutions are Bessel functions.

Simplest solution found by defining ϕ(r) such that E(r) ≡ ϕ(r)/r, then substitution into Helmholtz eqn.
Get solution A/r exp(i(kr±ωt)).
Amplitude diminishes with r.

Question 16

Cylindrically symmetric solution to 3D wave equation

Answer 16

Cylindrical polar:
∇²E = - k²E = 1/r ∂/∂r (r ∂E/∂r)

Bessel eqn with no easy analytic solutions.

A/√r’ exp(i(kr±ωt)) is a solution IF r is large

Question 17

Huygens-Fresnel principle

Answer 17

All points on a wavefront can be considered as point sources for production of secondary spherical wavelets. At a later time the new position of the wavefront will be a surface tangent to the secondary wavelets. The wavelets have destructively interfered in all but one direction to form this new wavefront. The amplitude of the optical field at any point beyond is the superposition of all of these wavelets.

Question 18

Obliquity factor

Answer 18

Factor that prevents plane waves moving backwards.
½(1+cos θ)
Maximum transmission in direction of travel (θ=0°), none in backwards direction (θ=180°)

Question 19

Optical path

Answer 19

If light traverses a distance l through a medium of refractive index n, the optical path is nl, or more generally
∫n(l) dl
if the refractive index is variable.

Question 20

Fermat’s principle

Answer 20

A light ray going from A to B traverses a path that is stationary with respect to variations in path.

If a particular path has a “minimum or maximum or saddle point” optical path than all other paths immediately surrounding it, light will travel along this path.
In other words, a stationary point, or locally nonvarying region of optical path.

Question 21

Four regimes of optics?

Answer 21

• Geometrical optics: interference and diffraction effects are ignored. Light propagation may be understood in terms of rays, which are perpendicular to wavefronts and along which the light is considered to propagate. It is valid when all the components considered are very large compared to the wavelength.
• Physical optics: the regime where the apparatus used is comparable to the wavelength of light, and under which circumstance interference and diffraction are noticeable.
• Electromagnetism: a full solution of Maxwell’s equations, without approximations made in physical optics.
• Quantum optics: the regime where the discrete nature of photons is important, particularly in interaction of light with matter, and the study of laser emission.

Question 22

Refractive index

Answer 22

n = c/v

Question 23

Derive law of reflection (Fermat)

Answer 23

Light propagates along a path between A and C, reflecting at B.
Path length:
l = √x² + y²’ + √(h − y)² + x²’
Fermat: path length must be
stationary with respect to the variable y
dl/dy = ½(x²+y²)⁻¹’² · 2y+½((h−y)²+x²)⁻¹’² · 2(h−y) · −1 = 0

y/(x² + y²)¹’² = h−y/(x² + (h − y)²)¹’²

SOHCAHTOA, equivalent to sinθᵢ = sinθᵣ, so θᵢ = θᵣ

Question 24

Derive law of refraction (Snell’s law)

Answer 24

Ray refracts from medium of refractive index n₁ to a medium n₂. Wavelength p in n₁, meet interface at angle θᵢ. Wavelength q in n₂, meet interface at angle θᵣ.
Wavefronts perpendicular to the incoming and outgoing rays.
Time taken for light to travel distance p in medium n₁ must be the same as to travel q in medium n₂, as wavefronts must continue to meet at the interface (wave separation d).
p = d sin θi and q = d sin θr.

d sinθᵢ/(c/n₁) = d sinθᵣ/(c/n₂)

n₁ sinθᵢ = n₂ sinθᵣ

Question 25

Derive law of refraction (Fermat)

Answer 25

Minimise time taken to travel from A to B w.r.t. x (location of crossing)
d/dx (√a²+x²’/v₁ + √b²+(l−x)²’/v₂) = 0

1/v₁ · 2x(a²+x²)⁻¹’²/2 + 1/v² · ½(b²+(l−x)²)⁻¹’² · 2(l−x) · −1 = 0

Substituting for angles using trig, and v=c/n
sinθᵢ/v₁ = sinθᵣ/v₂
n₁sinθᵢ = n₂sinθᵣ

Question 26

Critical angle

Answer 26

If light travelling in an optically dense medium is incident on a boundary with a less dense medium, there will be some incidence angle θ꜀ for which the refraction angle θᵣ = 90°. This angle θ꜀ is known as the critical angle. Light incident at θ > θ꜀ will not pass into the less dense medium, and instead will be totally internally reflected back into the denser medium.

Question 27

Paraxial approximation

Answer 27

Often used in geometric optics. All processes take place close to the optic axis of the system. Consequently, all angles are small, so that θ ≃ sin θ ≃ tan θ.

Question 28

Reflection at a spherical surface

Answer 28

Light is going from P to P′, reflecting from a spherical mirror at Z. Two angles θ are equal
by the law of reflection, and the radius of curvature of the mirror is R. (refer to fig. 1).
y is vertical distance from Z to axis.

αᵥ = αᵣ + θ
αᵣ = αᵤ + θ
⇒ αᵥ + αᵤ = 2αᵣ
use paraxial approx:
⇒ tanαᵥ + tan αᵤ = 2tanαᵣ
⇒y/v + y/u = 2y/R

1/v + 1/u = 2/R

Question 29

Refraction at a spherical surface

Answer 29

Light is going from A to B, refracting at a spherical surface to a denser medium at Z. Radius of surface is R. (refer to fig. 2)
h is vertical distance from Z to axis.

αᵣ = αᵥ + θᵣ
θᵢ = αᵤ + αᵣ
Paraxial approx:
h/R = h/v + sinθᵣ
sinθᵢ = h/u + h/R
Combine with Snell’s law
nᵢ/u + nᵣ/v = (nᵣ − nᵢ) / R

Question 30

Derive equation for focal length

Answer 30

Lens of refractive index n, assumption that thickness of lens &laquo_space;other distances.
An object is at distance u₁ in front of it.
Image formed by first surface
(refractive index 1 to n):
1/u₁ + n/v₁ = (n − 1)/R₁
Second surface sees an object at −v₁:
n/−v₁ + 1/v₂ = (1 − n)/R₂
Adding:
1/u₁ + 1/v₂ = (n − 1) (1/R₁ − 1/R₂)
Assume incoming waves parallel (u₁ → ∞), v₂ is the focal length.
1/f = (n − 1) (1/R₁ − 1/R₂)

If u and v are object and image distances
1/f = 1/u + 1/v

Question 31

5 lens conventions and definitions

Answer 31

• Focal length, f, is distance from lens at which an image is formed, if object is at infinity.
• Object distance, u, is positive if the object is to the left (in front of) the lens, and image distance, v, is positive if the image is to the right (behind) the lens.
• A converging lens, with positive f, causes incident parallel rays to converge. A diverging lens, with negative f, causes incident parallel rays to diverge.
• Radii of curvature are positive if convex, as seen from the front (incoming ray). Known as the Gaussian convention.
• An image is real if it can be formed on a screen (rays will intersect at the image). It is virtual if it cannot (rays will intersect at the image only if extrapolated from other rays).

Question 32

If stuck on geometric problem

Answer 32

• PATH LENGTHS
• binomial expansion?

Question 33

Binomial expansion

Answer 33

For small x,
(1+x)^a ~ 1 + ax + 1/2 a(a-1)x^2 …

Question 34

Drawing ray diagrams

Answer 34

• Lens and optic axis
• Object in front of the lens.
• Ray from object through the centre of the lens: this ray is not deviated.
• Ray from the object parallel to the optic axis. Cnverging lens, refract through far focal point. Diverging lens, refract away from optic axis along line that, projected back, originates near focal point.
• If lines cross, the image is real and at the position and height implied by lines.
• If lines do not cross, extrapolate one or both backwards to find position and size of virtual image.

Question 35

Magnification

Answer 35

Ratio of the size of an image to the size of the object creating it. (-v/u)

Question 36

Gravitaitonal lens (Fermat)

Answer 36

Light going through a gravitational potential suffers a delay (Shapiro delay). Light tries to find a compromise minimum path between the direct path (short geometric path but lots of Shapiro delay) and a very bent path (long geometric path but less Shapiro delay). Path chosen is the minimum in between these two.

With no lensing, concentric circles show contours of constant optical path, with minimum at centre.

With lensing, contours distort, leading to a maximum, minimum, and saddle point(s?). Multiple images form at these stationary points.

Question 37

Linear polarization

Answer 37

x̂E₀ₓ cos(kz − ωt + ϕ) + ŷE₀ᵧ cos(kz − ωt + ϕ)
wave travelling in z direction

Electric field remains in same plane as it propagates, plane is at angle tan⁻¹(E₀ᵧ/E₀ₓ) to x axis

Max. amplitude is √E₀ₓ² + E₀ᵧ² ‘

No phase shift between x̂ and ŷ components.

Question 38

Circular polarization

Answer 38

x̂E₀ₓ cos(kz − ωt + ϕ) + ŷE₀ᵧ cos(kz − ωt + (ϕ ± π/2))
OR
x̂E₀ₓ cos(kz − ωt + ϕ) ± ŷE₀ᵧ sin(kz − ωt + (ϕ))
wave travelling in z direction

Electric field vector is same size, but rotates around direction of propagation.

±90° phase shift between x̂ and ŷ components.

Right-hand circular (RHC) if rotates clockwise viewed
back down the wave from the observer.

Adding LHC and RHC waves of same amplitude gives linear polarization.

LHC - +π/2
RHC - -π/2

Question 39

Elliptical polarization

Answer 39

Elliptical is the general case.
x̂E₀ₓ cos(kz − ωt + ϕ₁) + ŷE₀ᵧ cos(kz − ωt + ϕ₂)

Special case of linear polarization if ϕ₁ = ϕ₂

Special case of circular polarization if E₀ₓ = E₀ᵧ and ϕ₂ − ϕ₁ = ±π/2.

Question 40

Unpolarized light

Answer 40

Fragments of polarized light, can be resolved into two perpendicular components:
* component polarized perpendicular (in and out) to the plane is s-polarized light
* component polarized parallel (side to size) to the plane is p-polarized light

Question 41

Polarization in dielectric media

Answer 41

• Incident light’s E-vector excites oscillations of dipoles in dielectric along the direction of the E-vector (not direction of propagation).
• These dipoles do not radiate along their length but do in other directions.
• Can resolve amplitudes of E-vectors along different directions as required.

Question 42

Polarized light by reflection: Brewster’s angle

Answer 42

• Incoming light is unpolarized, can be treated as fragments of polarized light with two components (shown as dots in/out of paper and arrows in plane of paper).
• These excite dipoles in denser medium
• As well as refracted ray, weak reflected ray is formed.
• Production of the reflected ray by oscillating dipoles in the dense medium. If the angle between the reflected and refracted rays is 90°, there is no reflected ray in one of the polarizations, because dipoles do not radiate along their length.
• Since θᵢ + θᵣ = π/2, then sinθᵣ = cosθᵢ
  Combine with Snell’s law, nᵢ sinθᵢ = nᵣ sinθᵣ.
• For this totally polarised reflected ray, tanθᵢ = nᵣ/nᵢ.
• This value of θᵢ, θ_B, is the Brewster angle.

Question 43

What are the Fresnel coefficients?

Answer 43

When an unpolarised ray is both reflected and refracted, the Fresnel coefficients are the coefficients of the reflected and transmitted (refracted ray) in terms of the incident amplitude.

Question 44

Derive the Fresnel coefficients: set up figure

Answer 44

• 2 figures, one for each component of polarization, incident light, amplitude 1.
• s-polarization, a refracted, b reflected, additional line continuing incident ray
• p-polarization, A refracted, B reflected. Additional red RA triangle continuing incident ray, RA, to A. Additional blue RA triangle, B, RA, RA, to A. One angle is θᵢ, one is θᵢ - θᵣ, one is 90 - θᵢ - θᵣ.

Question 45

Derive the Fresnel coefficients: first argument

Answer 45

Argument: all of the oscillation corresponding to a produces b, but only the (blue) component of oscillation of A polarized perpendicular to the path of B produces B. This is because dipoles do not radiate along their length.

Resolve A to get component perpendicular to B (line parallel to B direction gives perpendicular oscillations)

b/a = B/Acos(θᵢ+θᵣ)

Question 46

Derive the Fresnel coefficients: second argument

b/a = B/Acos(θᵢ+θᵣ)

Answer 46

Argument: The charges in the glass must conspire to cancel the beam marked in red by producing a net field of amplitude −1. All of a produces this, but only one component of A, Acos(θᵢ − θᵣ) (line thats a continuation of incident)

a = Acos(θᵢ−θᵣ)

combine with first argument, b/a = B/Acos(θᵢ+θᵣ)

B/b = cos(θᵢ+θᵣ)/cos(θᵢ−θᵣ)

Question 47

Derive the Fresnel coefficients: third argument

Answer 47

Argument: conservation of energy

|A|² + |B|² = 1
|a|² + |b|² = 1

1-|B|² / 1-|b|² = |A|²/|a|²

Argument 2:
a = Acos(θᵢ−θᵣ)
⇒ |A|²/|a|² = 1/cos²(θᵢ−θᵣ)

B/b = cos(θᵢ+θᵣ)/cos(θᵢ−θᵣ)
⇒ |B|² = |b|²cos²(θᵢ+θᵣ)/cos²(θᵢ−θᵣ)
⇒|b|² = |B|²cos²(θᵢ−θᵣ)/cos²(θᵢ+θᵣ)

Sub into energy eqn:
|B|² = tan²(θᵢ−θᵣ)/tan²(θᵢ+θᵣ)
|b|² = sin²(θᵢ−θᵣ)/sin²(θᵢ + θᵣ)

Question 48

Derive the Fresnel coefficients: find coefficients

Answer 48

Coefficients are reflected/incident
Incident = 1, so:

fₛ = b = sin(θᵢ−θᵣ)/sin(θᵢ + θᵣ)
fₚ = B = tan(θᵢ−θᵣ)/tan(θᵢ+θᵣ)

Can now use snells law to find any case. (Brewsters, θᵢ+θᵣ = π/2, B=0)

Question 49

Polarization by scattering

Answer 49

• Wave has a polarization state that can be resolved into two orthogonal states.
• Components produce up/down oscillations and left/right oscillations.
• Enters dipole medium
• No re-radiation from up/down component in up/down direction, instead radiation to the sides and in original direction of propagation
• Similarly, no re-radiation from left/right component in left/right direction, instead radiation to the top and bottom and in original direction of propagation
• Putting these together, scattering will give highly polarized scattered light at 90°
  to the original propagation direction, and less polarized in other directions.

See fig. 3

Question 50

Polarization by dichroism

Answer 50

Selective absorbtion of one component.

A material, such as a polaroid, has long aligned molecules which absorb any component in their direction, leaving polarized light orthogonal to them.

Transmitted wave has amplitude E₀cosθ, where θ is angle between initial polarization vector and the pass axis.

Question 51

Malus’s law

Answer 51

Intensity transmitted by a dichroic when plane-polarized light is incident on it is
I(θ) = I₀ cos²θ, where θ is the angle between the initial E-vector and the pass axis.

Question 52

What is an anisotropic/birefringent crystal

Answer 52

• A crystal in which the refractive index n depends on the direction of oscillation of the E-vector of a wave.
• The velocity of the wave, and the relative permittivity ϵ also depend on the direction of oscillation.
• It is easier to polarize an anisotropic crystal in one direction relative to another.

e.g. calcite, quartz

Question 53

Three types of anisotropic/birefringent crystal

Answer 53

• Uniaxial: two of the components of ϵ are equal and the other differs.
  Because ϵ ∝ n², the refractive index (and hence velocity) differs in one direction from the others.
• Biaxial: all three components of ϵ are different.
• Negative uniaxial: the unique axis has a lower refractive index (higher velocity, i.e. a “fast axis”)

Question 54

Ordinary and extraordinary refractive indices

Answer 54

For nₒ,
k²/nₒ² = ω²/c²
(normal)

For nₑ,
kₓ²/nₑ² + kᵧ²/nₑ² + kᶻ²/nₒ² = ω²/c²
(holy shit dude!)

nₑ used for light oscillating in direction of optic axis.

Question 55

Refractive index in a arbritrary direction in an anisotropic crystal

Answer 55

Use k in spherical polar:
kₓ =|k|sinθ cosϕ
kᵧ =|k|sinθ sinϕ
kᶻ =|k|cosθ

1/n² = sin²θ/nₑ² + cos²θ/nₒ²

Question 56

Effect of light travelling in a crystal

Answer 56

• Propagation along optic axis: all components of E-vector oscillate perpendicular to axis, so travel at the same speed.
• No odd effects.
• Propagation perpendicular to optic axis: one component of polarization advances faster than the other.
• One component undergoes an increasing phase shift relative to the other.
• Modifies state of polarization

Question 57

Double wavefronts in a crystal

Answer 57

• Unpolarized source
• For propagation along the optic axis, all E-vectors propagate with the same speed.
• For other directions, any snippet polarized perpendicular to the OA travels at a speed c/nₒ, and snippets polarized parallel to the OA travel at c/nₑ.

Question 58

Double refraction

Answer 58

• Birefringent crystal.
• Optic axis at arbitrary angle, which With Love Of God we can take to be in the plane of the paper.
• o-ray E-vector oscillates ⊥ to OA - wave proceeds without deviation, wavefronts circular.
• (e-ray) has E-vector one component || to OA, and one ⊥ to OA. The former expands faster and so wavefronts are elliptical. Direction of propagation (direction of S) no longer perpendicular to wavefront.
• both cases, ∇ · D = 0 and k · Dₑ = 0.
• e-case ∇ · E≠ 0 implies k · Eₑ ≠ 0
• e-case, snells law not obeyed
• include D, E, S, k vectors in diagram

Question 59

What is a waveplate

Answer 59

Plate made from birefringent material, cut so optic axis is parallel to surface. Used to modify state of input polarization via phase shift (different components propagate at different speeds)

Question 60

Types of waveplate

Answer 60

Quater-wave plate:
produces phase shift of π/2, or λ/4
Half-wave plate:
produces phase shift of π, or λ/2

e.g. linear polarization at 45°, QWP, circular polarization output.

Question 61

Thickness of waveplates

Answer 61

QWP, relative phase shift between o and e of λ/4
HWP, relative phase shift between o and e of λ/2

Shift in optical path given as
|dnₑ-dnₒ|

Thinnest slab width is
d=λ / m|nₑ-nₒ| where m is 4 or 2

Question 62

Quartz wedge

Answer 62

Wedge with OA along long side of wedge. Light entering from below will have one E-vector oscillation direction along, and one perpendicular to, the optic axis. Distance travelled through wedge varies along it as wedge gets thicker. This causes a change in path. Relative delay between components will be λ/4, λ/2, 3λ/4, λ at equally spaced points.
Polarization state changes along wedge.

e.g. Monochromatic, linearly polarized light in at 45°. Linear at 0, λ/2, λ. Circular at λ/4, 3λ/4.
-45° analyzing polarizer above wedge will show dark bands at origin and λ, bright band at λ/2.
Non-monochromatic will give series of coloured bands.

Question 63

Glan polarizer

Answer 63

Two 45-45-90 pieces of calcite, with an interface layer between them (Fig. 4).
At the interface, two components of unpolarized light (|| and ⊥ to OA) have different refractive indices. Angles of incidence on interface arranged so one component hits interface above critical angle (total internal reflection) and the other below (refracts).

Only works for limited range of input angles.
Glan-air polarizers have air in the interface; Glan-Thompson polarizers have a layer of epoxy resin.

Question 64

Wollaston prism

Answer 64

Two prism-shaped pieces of birefringent crystal glued together (Fig. 5), but with different optic axes.
Unpolarized light enters.
At the interface in centre, the optic axis changes: e-ray becomes the o-ray, and refracts towards the normal (because nₒ > nₑ); o-ray becomes the e-ray and refracts away from the normal.
The two components of polarization are split. There is a further change in direction of both rays at the outer edge
of the device as they refract into air.

Question 65

What is the Faraday effect

Answer 65

Rotation of the plane of polarization of light passing through a magnetic field

Question 66

Faraday effect equation

Answer 66

θ = V Bd

d is distance travelled through B-field
B is strength of B-field
V is Verdet’s constant

When propagation parallel to B-field:
+ve Verdet constant corresponds to L-rotation (anticlockwise)
When propagation anti-parallel to B-field:
+ve Verdet constant corresponds to R-rotation (clockwise)

If a ray of light is passed through a material and reflected back through it, the rotation doubles.

Question 67

Faraday isolator

Answer 67

Utilises Faraday effect surrounded by two polaroids with pass angles 0° and 45°.

Forwards, Faraday effect rotates wave so it can pass through 45°. Backwarsds, rotates to 90° so can t pass through 0° polaroid. Light can only pass in one direction.

Question 68

Interference vs Diffraction

Answer 68

Interference: the combination of a finite number of waves (e.g. two combining waves from Young’s slits)
Diffraction: the combination of an infinite number of waves (e.g. different parts of a wide aperture).
Both: the resultant field is the sum of the combining fields, and the intensity is proportional to
Eₜₒₜ * · Eₜₒₜ

Question 69

Intensity of two linearly polarized e-field waves

Answer 69

E = E₁ exp(i(k₁.r − ω₁t + ϵ₁)) + E₂ exp(i(k₂.r − ω₂t + ϵ₂))

I = Eₜₒₜ * · Eₜₒₜ = E * · E
=|E₁|² + |E₂|² + 2E₁·E₂cos[(k₁ − k₂)·r − (ω₁ − ω₂)t + (ϵ₁ − ϵ₂)]

Question 70

What does the existence of the interference term show

Answer 70

I =|E₁|² + |E₂|² + 2E₁·E₂cos[(k₁ − k₂)·r − (ω₁ − ω₂)t + (ϵ₁ − ϵ₂)]

• ⟨E₁·E₂⟩ ≠ 0. Interfering waves must not be orthogonal.
• ω₁≃ω₂. We observe the intensity pattern averaged over a time interval. Unless angular frequencies equal, a term involving cos(ω₁−ω₂)t quickly averages to zero. Need ∆ω < 1/T, where T is the time over which we observe.
• Averaged over time, so ϵ₁−ϵ₂ is constant, i.e. the waves have a constant phase relation.
• If these conditions are fulfilled (coherence), add waves by adding amplitudes and forming E * · E. No coherence, interference term averages to zero, add intensities.

Question 71

Requirements for coherence

Answer 71

• Waves not orthogonal
• Angular frequencies roughly equal
• Constant phase relation

Question 72

Simplifications for adding waves

Answer 72

E = E₁ exp(i(k₁.r − ωt + ϵ₁)) + E₂ exp(i(k₂.r − ωt + ϵ₂))

Assume equal frequencies and write all exp superscripts as a simple phase.

E * · E = (A₁exp(−iϵ₁) + A₂exp(−iϵ₂))(A₁exp(iϵ₁) + A₂exp(iϵ₂))

Treat each wave as a phasor. Length is amplitude (A₁, A₂). Angle to real axis is phase (ϵ₁, ϵ₂). Waves add like vectors.

Cosine rule derives
|A₁|² + |A₂|² + 2A₁A₂ cos(ϵ₁ − ϵ₂)

Question 73

Temporal coherence

Answer 73

The measure of the average correlation of the phase of the light wave along the propagation direction.

Temporal coherence time:
t꜀ = 1/∆ν
If t > t꜀ interference will not be observed between a wave and its delayed version.
Corresponding coherence length:
ct꜀

Question 74

Three things that prevent temporal coherence

Answer 74

Doppler broadening:
Emitting atoms moving, so emitted waves from atoms with different line-of-sight velocities have a range of frequencies (Gaussian profile)
Collision broadening:
Atoms collide typically every ∼10 ns, interrupting the wavetrain and giving a range of frequencies (Lorentzian profile)
Natural (Heisenberg) broadening:
Transitions of electrons on timescale ∆t results in frequency broadening according to uncertainty principle, so ∆ν ∼ 1/∆t.

Question 75

Spacial coherence

Answer 75

Correlation of the phase of a light wave transverse to the propagation direction (how uniform is the wavefront with time).

Question 76

Illustrate temporal and spacial coherence

Answer 76

∞ temporal, ∞ spacial
Parallel lines with constant distance between them

∞ temporal, finite spacial
Sections with different, but constant, widths between them

finite temporal, ∞ spacial
Parallel lines with differing gaps between them. Can be achieved with pinhole

finite temporal, finite spacial
big wiggly mess

(see fig. 6)

Question 77

Relate frequency bandwith and wavelength bandwidth

Answer 77

ν = c/λ
∆ν = (∂ν/∂λ)∆λ = (−c/λ²)∆λ

Question 78

Division of wavefront interference

Answer 78

Two parts of a wavefront are isolated, and allowed to interfere with each other.

Examples: Young’s slits, multiple slits, Lloyd’s mirror.

Question 79

Division of amplitude interference

Answer 79

A single part of a wavefront is split, the two parts are allowed to travel along different paths before being combined and allowed to interfere with each other.

Examples: Michelson interferometery, Fabry-Perot etalon.

Question 80

Important principles of interference

Answer 80

1. If a wave is expressed in the complex form, a phase shift ϕ is implemented by multiplying by exp(iϕ)
2. If a wave travels a distance d, this induces a phase shift kd, where k is the wavenumber, and hence multiplies the wave by exp(ikd)

Question 81

Derive intensity pattern of Young’s slit

Answer 81

Add two waves. One travels a distance dsinθ further than the other, so has a relative phase shift of kdsinθ.

Combined wave is
E₁exp(i(k₁.r₁ − ω₁t + ϵ₁))+E₂exp(i(k₂.r₂ − ω₂t + ϵ₂)).

Assuming ω₁=ω2, k₁=k₂, ϵ₁−ϵ₂=kdsinθ, amplitudes equal, distance large so lines parallel so k·(r₁-r₂)=kdsinθ:
Eₜₒₜ = Eexp(i(kx−ωt)) + Eexp(i(kx−ωt+kdsinθ))

Ignored the r-dependence of amplitude, because only interested in variation of interference pattern with y on distant screen, variations in r won’t make much difference.

I = Eₜₒₜ * ·Eₜₒₜ = |E|² +|E|² + 2|E|²cos(kd sin θ) = 2|E|²(1+cos(kdsinθ)).

Double-angle formula:
I(θ) = 4|E|²cos²(kdsinθ/2)

Assuming all angles small, sinθ ≃ tanθ = y/s (s is distance from slit to screen)
I(y) = 4|E|²cos²(kdy/2s)=4|E|²cos²(πdy/λs)

I(y) is a set of bright fringes which occur when kdy/2s = nπ, aka when y = nλs/d, aka when
dsinθ = nλ, where n is an integer.

Can do this with phasors, ϵ₁ → 0 and ϵ₂ → kdsinθ

Question 82

Visibility of a set of fringes

Answer 82

V = Iₘₐₓ − Iₘᵢₙ / Iₘₐₓ + Iₘᵢₙ
where Iₘₐₓ and Iₘᵢₙ are the maximum and minimum brightnesses of the intensity pattern

Question 83

What happens to phase upon reflection?

Answer 83

There is a phase change of π on reflection from an optically denser surface.

Question 84

Lloyd’s mirror

Answer 84

Indistinguishable from Young’s double slits, source travels directly to screen, also reflects in horizontal mirror. Phase change of π upon reflection, so intensity pattern becomes:

I(y) = 4|E|²sin²(kdy/2s)=4|E|²sin²(πdy/λs)

Question 85

3 slits

Answer 85

Do by phasors

δ ≡ dsinθ ≃ dy/s

Each phasor is at an angle kδ to the previous. By symmetry, Eₜₒₜ phasor is parallel to the second phasor.

By geometry, length of Eₜₒₜ is |E|+2|E|coskδ

Intensity is:
|E|²(1+2coskδ)² = |E|²(4cos²(kδ/2)−1)² = |E|²(4cos²(πdy/λs)−1)

Question 86

n slits

Answer 86

δ ≡ dsinθ ≃ dy/s.
N slits
Eₜₒₜ = E₀exp(i(kx−ωt)) (1 + exp(ikδ) + exp(2ikδ) + …. + exp((N−1)ikδ))

Using Σ1 + x + x² + …. + xᴺ⁻¹ = xᴺ−1 / x−1 :
= E₀exp(i(kx−ωt)) exp(Nikδ)−1/exp(ikδ−1)
= E₀exp(i(kx−ωt)) exp(½Nikδ)sin(Nkδ/2)/exp(½ikδ)sin(kδ/2)

Eₜₒₜ * ·Eₜₒₜ = E₀² sin²(Nkδ/2)/sin²(kδ/2)= E₀² sin²(Nπyd/λs)sin²(πyd/λs)

Question 87

n-slit intensity pattern

Answer 87

Main peaks still separated by λs/d, main peak sharper with n-2 subsidiary minima.

Question 88

Setup and path difference of Michelson interferometer

Answer 88

See fig. 7

Division of amplitude interferometer. Light split by a semisilvered surface at beamsplitter; components reflected from the two mirrors and interfere where they recombine.

Compensator plate same thickness as beamsplitter; ensures both beams pass through 3 thicknesses of glass. Important to get achromaticity (behaviour same for all wavelength)

Path difference between beams
∆ = 2(d₁ − d₂)

Question 89

Central intensity pattern for Michelson interferometer

Answer 89

Fig. 7, solid line

Path difference:
∆ = 2(d₁ − d₂)

Effectively Eₜₒₜ = E + Eexp(ik∆)
Maths same as Young’s slits, with π phase shift from extra mirror.

I = Eₜₒₜ * · Eₜₒₜ = 4|E|²sin²(π∆/λ)

As mirror moved, intensity centre of fringe pattern varies from 0 to 4|E|² and back. Mirror motion of λ/2 needed for intensity to track from one fringe to the next (extra length traversed twice).

Question 90

Fringe pattern for Michelson interferometer

Answer 90

Fig. 8

Path delay now ∆cosθ.
Corresponding phase delay is k∆cosθ.
Dark fringe, k∆cosθ = 2mπ, integer m.
Central fringe has order m₀, so ∆=m₀λ.
Consider another dark fringe of order m₀−p for which θ = θₚ (pth ring).

∆ cos θₚ = (m₀ − p)λ = ∆ − pλ
Small θₚ, cosθₚ ≃ 1−θₚ2/2
∆θₚ2/2 = pλ₀, θₚ = √2pλ/∆ ‘

Radius of each successive fringe goes as the square root of the fringe number, when mirror is moved the fringe pattern expands (or contracts) so that the central intensity varies with the mirror motion.

Question 91

Michelson spectroscopy: two spectral lines

Answer 91

Spectral lines of wavelengths λ₁ and λ₂.

Each wavelength produces its own fringe pattern. Fringe patterns add.
As ∆ increased, fringes begin distinct, become less distinct, then more distinct.
Can quantify visibility of fringes, which varies between 1 and 0.

Define ∆ₘᵢₙ as difference in ∆ between having very high-contrast, V = 1 fringes (~ ‘in phase’) and washed out, V = 0 fringes (~ ‘antiphase’)
∆ₘᵢₙ represents 0.5 wavelengths more of λ₁ than λ₂.
∆ₘᵢₙ/λ₁ = ∆ₘᵢₙ/λ₂ + ½

If we say λ = 0.5(λ₁+λ₂) and ∆λ = λ₂−λ₁
∆ₘᵢₙ = λ₁λ₂/2∆λ ≃ λ²/2∆λ

Relation between I(λ) of source and V(∆) such that when I(λ) is two delta functions, V(∆) is a sinusoid.

Question 92

Michelson spectroscopy: general case

Answer 92

I(f) as input. Each individual frequency gives fringe pattern
dI = 4I(f)cos²(πf∆/c)df = 2I(f)(1+cos(2πf∆/c))df
where I(f)df is intensity in a small strip of frequency f → f + df.

Fringe pattern, integrating, is
const + 2 ∫₀∞(I(f)cos(2πf∆/c)df

Question 93

Thin film interference

Answer 93

Fig. 9
Two waves, one reflected off the air-glass interface, one returning to the interface after reflection off the bottom surface.
Difference in optical path is 2dn/cosθᵣ − 2dtanθᵣsinθᵢ
(First term is distance backwards and forwards through material of refractive index n, second is extra path in air)
Using Snell, sinθᵢ = nsinθᵣ Substituting, get 2d(n−nsin²θᵣ / cosθᵣ)
Optical path difference = 2dncosθᵣ

Interference maximum/minimum for 2dncosθr = mλ,
where m is an integer.
Minimum if bottom surface n < film n, maximum if bottom surface n > film n. Minimum/maximum occurs at 2dncosθr = (m−½)λ

Interference condition gives maxima at different angles for different wavelengths and thicknesses - hence the coloured bands seen in soap.

Question 94

Fabry-Perot etalon:
Basic principles

Answer 94

Division of amplitude interferometer with many interfering waves, resulting
in sharp interference fringes
Two exactly parallel, highly
reflective plates. Small portion of light transmitted at each reflection → many transmitted beams.
Transmitted beams are interfered: lens projects interference pattern to screen.
Each reflection has intensity reflection coefficient ρ, transmission coefficient σ. Each successive beam is ρ weaker than the last (amplitude)

Each wave has phase shift δ=(2π/λ)·2dcosθ with respect to the last. Derivation is exactly the same as for thin films.

Condition for maxima: δ=2mπ

Question 95

Fabry-Perot etalon:
fringe intensity distribution

Answer 95

Each wave has an extra phase shift δ w.r.t. the previous, and an extra amplitude factor ρ.
Eₜₒₜ = E₀exp(i(kz−ωt))(σ+σρe(iδ)+σρ²exp(2iδ)+….)
Using (1−x)⁻¹=1+x+x²+x³….
Eₜₒₜ = E₀σexp(i(kz−ωt)) 1/1−ρexp(iδ)
I = Eₜₒₜ * · Eₜₒₜ
= E₀²σ² 1/1−ρexp(iδ) 1/1−ρexp(−iδ)
=E₀²σ²/1+ρ²−2ρcosδ
=E₀²σ²/1+ρ²−2ρ(1−2sin²(δ/2))

I(δ) = k²E₀²σ² / (1+(4ρ/(1−ρ)²)sin²(δ/2))

δ = (2π/λ)2dcosθ
k = 1/(1−ρ)

Question 96

Free spectral range, resolution, and finesse of an etalon

Answer 96

Free spectral range: (∆ν or ∆λ)
The spacing between individual principal interference peaks.

Resolution: (δν or δλ)
The width of each principal interference peak (Half peak width).

Finesse:
Ratio of the two, represents how many individual spectral lines can be resolved without confusion.

Question 97

derive δλ from δν

Answer 97

δν is at half peak, meaning
I(δ) = k²E₀²σ² / (1+(4ρ/(1−ρ)²)sin²(δ₀.₅/2)) = 1/2

Therefore
4ρ/(1−ρ)²)sin²(δ₀.₅/2) = 1
(sinx≃x)
2δ₀.₅ = 2(1−ρ)/√ρ’

This quantity is also given by
2π/λ+(δλ/2) 2dcosθ − 2π/λ−(δλ/2) 2dcosθ

Around peak, 2dcosθ=mλ
Sub in, binomial expand
=2πmδλ/λ

2(1−ρ)/√ρ’ = 2πmδλ/λ
δλ = λ(1−ρ) / πm√ρ’

This is the minimum separation in wavelength of two lines that can just be resolved by the FP etalon.

Question 98

Free spectral range, ∆λ

Answer 98

2d/m − 2d/(m+1) ≃ 2d/m² = λ/m for large m.

This is the wavelength interval between orders, determines the range of wavelength over which spectral lines can be unambiguously studied (without one end of the range of wavelengths overlapping the other end in the next order).

Question 99

Finesse

Answer 99

∆λ/δλ = λ/mδλ = πρ¹’² / 1−ρ

Question 100

Fabry-Perot etalot: what is seen

Answer 100

δ ≡ (2π/λ)2dcosθ = 2mπ for an
interference peak.
Centre of pattern (θ = 0 → cos θ = 1) has fringe of order m₀.
m₀ = 2d/λ.
Fringe of order m₀ − p, (p fringes away from central)
(cos θ ≃ −θ²/2)
2π/λ 2d(1− θₚ²/2) = 2π(m₀−p)

Sub m0 = 2d/λ and rearrange
θₚ = √pλ/d ‘

Single spectral line, series of concentric circles, angular separation from centre goes as √p’.
High finesse (small δλ), fringes sharp, low finesse, fringes fuzzy

Multiple spectral lines, multiple concentric rings per Free Spectral Range.

Question 101

Edser-Butler fringes

Answer 101

Fringes formed with white light.
Constant θ, white light focused on etalon. Only certain wavelengths, corresponding to interference condition, are passed. If split using a prism and focussed on to a detector, get coloured fringes - known as Edser-Butler fringes.

Question 102

Fourier series

Answer 102

The orthogonal basis functions sin(nπx/L) and cos(nπx/L) can be used to make any function in the range x = −L to x = L (or periodic function with period 2L) using the series:

Trig form:
f(x) = a₀ + Σaₙ sin(nπx/L) + Σbₙ cos(nπx/L)
where
a₀ = 1/2L ∫±ᴸ f(x) dx
aₙ =1/L ∫±ᴸ f(x) sin(nπx/L) dx
bₙ =1/L ∫±ᴸ f(x) cos(nπx/L) dx

Complex form:
f(x) = Σ±∞ aₙexp(inπx/L)
where
aₙ = 1/2L ∫±ᴸ f(x) exp(−inπx/L) dx

Two functions f(x), g(x) are orthogonal if ∫±ᴸ f * (x) g(x) dx = 0

Question 103

Fourier transforms

Answer 103

Non-periodic function, let L → ∞, w gets smaller so wL finite and constant. Frequency becomes infinitesimal, sum becomes integral.

f(x) = 1/2π ∫±∞ F(ξ) exp(iξx) dξ
F(ξ) = ∫±∞ f(x) exp(−iξx) dx

(1/2π can be on 2nd eq, or 1/√2π’ on both, product must be 1/2π)

Question 104

Common Fourier transforms

Answer 104

• Top hat ←→ Sinc (sinx/x)
• Gaussian ←→ Gaussian
• Constant ←→ Delta function
• 2 delta functions ←→ Cos
• Small and spiky ←→ Broad and smooth
• Widen ←→ Shrink

Remember that FT of a function tells you about the range of frequencies present. Something small and spiky, lots of high frequencies.

Question 105

Fraunhofer diffraction (by integration)

Answer 105

Far-field, screen far from aperture.

Phase at screen varies along aperture, at height x, path difference xsinθ, phase difference kxsinθ.

Represent waves as complex exponentials E₀exp(i(kz−ωt))exp(−ikxsinθ)
Can find Eₜₒₜ by integrating over aperture. (lets say from -a/2 to a/2)

(E0/a)exp(i(kz−ωt)) ∫±a/2 exp(−ikxsinθ) dx

Get E₀exp(i(kz−ωt)) sinc(πasinθ/λ)
I = E₀²sinc²(πasinθ/λ)

Question 106

Fraunhofer diffraction (by phasors)

Answer 106

Each phasor represents wave from a small part of the source,
and is an angle k dx sinθ ≡ 2π dx sinθ /λ from the previous one. The last has an angle 2πasinθ /λ from the first, where a is height of aperture.
They form an arc of a circle, with the central angle equal to kasinθ.
Make kite (fig. 12)
Eₜₒₜ = 2rsin(πasinθ /λ)

Length of the arc
E₀ = r(2πasinθ)/λ

Eₜₒₜ = E₀ sin(πasinθ /λ) / (aπsinθ /λ)
I = E₀²sinc²(πasinθ /λ)

Question 107

Phase shift in fraunhofer diffraction

Answer 107

If part of the aperture has a ϕ phase shift, treat it separately and multiply the integrand by exp(iϕ)

Question 108

Convolution

Answer 108

Convolution of two functions f(x) and g(x) is defined as
h(x) = ∫±∞ f(x′)g(x−x′)dx′

Question 109

Convolution theorem

Answer 109

If the function h(x) is the convolution of two functions f(x) and g(x), then the Fourier transform of h(x) is the product of the Fourier transforms of f(x) and g(x).

Question 110

Prove the convolution theorem

Answer 110

h(x) = ∫±∞ f(x′)g(x−x′)dx′
⇒ H(ξ) = ∫±∞ ∫±∞ f(x′)g(x−x′)dx′exp(−iξx)dx
= ∫±∞ ∫±∞ f(x′)g(x′′)dx′exp(−iξ(x′+x′′))dx′′ where (x′′=x−x′)
= ∫±∞ f(x′)exp(−iξx′)dx′ ∫±∞ g(x′′)exp(−iξx′′)dx′′
= F(ξ)G(ξ).

Question 111

Apply convolution to Fraunhofer diffraction

Answer 111

If an aperture can be made up as the convolution of two other apertures, then the diffraction pattern is the product of the individual diffraction patterns.

Example 1:
Two wide slits, d apart and of width a, can be thought of as the convolution of two narrow slits and one wide slit.
I(θ) = I₀sinc²(πasinθ /λ)cos²(πdsinθ /λ)
‘Missing orders’ when sinc goes to zero.

Example 2:
A triangular aperture can be written as the convolution of two top-hat functions, the diffraction intensity pattern has a sinc⁴ dependence on sin θ.

Question 112

Transmission grating of slits with width a

Answer 112

N narrow slits,
I = E₀² sin²(Nϕ/2)/sin²(ϕ/2)
where ϕ = 2πdsinθ /λ is phase difference between successive slits (separation d).
Each slit has finite width a. Convolution:
I = E₀² sinc²(πasinθ /λ) sin²(Nπdsinθ /λ)/sin²(πdsinθ /λ)

Missing orders when m = d/a

Larger N gives sharper grating maxima (width ∝ 1/N).

Maxima spacing inversely proportional to line spacing

Amplitude of maxima determined by the slit width (falls off faster with order if slits wider)

Question 113

Diffraction grating as a spectrometer

Answer 113

d sin θ = mλ
Blue light lower angle than red.

Dispersion = dθ/dλ
dcosθ dθ = m dλ
⇒ dθ/dλ = m/dcosθ

If θ small, dispersion ~m/d

Free spectral range:
Red starts to overlap blue if
λʳᵉᵈ m = λᵇˡᵘᵉ (m+1)
Δλ = λʳᵉᵈ - λᵇˡᵘᵉ = λᵇˡᵘᵉ/m
Quite large as m quite small.

Resolution:
Two lines are just resolved if one’s peak coincides with the other’s zero closest to its peak.
Get zeros if sin(Nπdsinθ /λ) = 0, i.e. Nπdsinθ /λ = nπ.
First zero for wavelength λ at dsinθ = mλ + λ/N
λ zero coincides with peak for wavelength λ + δλ
m(λ + δλ) = mλ + λ/N
so δλ = λ/Nm.

Question 114

Comparison of FP etalon and diffraction grating

Answer 114

……………………………………Diffraction grating | Fabry-Perot etalon
Resolving power λ/δλ | mN ∼ 10⁵ …………| πm√ρ’/(1−ρ) ∼ 10⁶
Free spectral range ∆λ | λ/m ∼ 100nm | λ/m ∼ 0.01nm
Finesse ∆λ/δλ …………..| N ∼ 10⁴ …………….| π√ρ’/(1−ρ) ∼ 100

Question 115

Reflection gratings

Answer 115

Fig. 13

d of order λ, so use Huygens principle
Path difference
∆α − ∆β = d(sinα − sinβ)
Diffraction maxima when d(sinα − sinβ) = mλ (Grating Equation)

Question 116

Fraunhofer diffraction with rectangular aperture

Answer 116

Fig. 14

|r|² = (X−x)² + (Y−y)² + D²
|R|² = X² + Y² + D²

Subtracting, and ignore terms in x² and y²
|r|²−|R|² = −2(xX+yY )

|r|+|R|≃2|R|, so |r|²−|R|²≃2|R|(|r|−|R|)

Phase difference
k(|R|−|r|) ≃ ux + vy
where u = kX/|R| ≃ kX/D, v ≃ kY/D.

Aperture −a/2 < x < a/2 and −b/2 < y < b/2, diffraction pattern
(E₀/ab)exp(i(kz−ωt)) ∫±a/2 exp(−iux) dx ∫±b/2 exp(−ivy) dy

Intensity diffraction pattern
I₀sinc²(ua/2)sinc²(vb/2)

Question 117

Fraunhofer diffraction with circular aperture

Answer 117

Use polar coordinates. Parametrise distances in aperture plane using x = pcosα and y = psinα.

Parametrise distances across observer plane using u = kqcosβ, v = kqsinβ.

With A for a normalising area,
E(q, β) = (E₀/A) ∫∫ f(x,y) exp(−i(ux+vy)) dxdy
becomes
(E₀/A) ∫d/2 ∫2π exp(ik(qpcosβcosα+qpsinβsinα) p dpdα
= (E₀/A) ∫d/2 ∫2π exp(ik(qpcos(α−β))) p dpdα

The square of this integral is a an Airy function.
A uniformly illuminated circular aperture produces a circularly-symmetric Airy function intensity diffraction pattern. The function has a radius to the first dark fringe of 1.22λ/d and a width of ∼λ/d, where d is the diameter of the circular aperture, and this
width is in radians. You can use λ/d when asked for the
resolution.

Question 118

Resolution of circular aperture

Answer 118

~λ/d !!!!!!!

Question 119

Derive Huygens-Fresnel integral

Answer 119

Fig.15

Spherical wavefront from source, S
At point Q,
E(Q) ∝ e(iks)/s (1/s dependence as found for a spherical solution to Helmholtz equation)
Wave recieved at P
dE = K(α)ϵ′ exp(iks)/s exp(ikr)/r dA
ϵ′ related to field strength.
Obliquity factor K(α) ≡ ½(1+cosα)
dA elemental area of total area

Sum of all waves at P
E(P) = ϵ′ e(iks)/s ∫ K(α) exp(ikr)/r dA

Question 120

Fraunhofer diffraction as a special case of Hugens-Fresnel

Answer 120

r and R are the distances to the observation point P(X, Y) from the point Q(x, y) on the aperture and the centre of the aperture, respectively.

Assume source is far away, so wavefront is a plane, not spherical.

By Pythagoras, r² = (X−x)² + (Y−y)² + Z²

Ignore terms in x² and y² (aperture is small).
r² ≃ R² − 2(xX + yY)
r ≃ R(1 − 2(xX + yY)/R²)¹’²
≃ R(1 − (xX + yY)/R²)

Assume obliquity factor constant. HF integral becomes
E(P) ≃ ϵ′ exp(iks₀)/s₀R ∫ exp(ikR)exp(−ik(xX+yY)/R) dA.

Ignore R in denominator, because it doesn’t vary appreciably over diffraction pattern (R ≫ X, Y ).

Combine all the constants, think about one dimension, and write sin θ ≃ tan θ = X/R, obtaining
C ∫ exp(−ikxsinθ) dx

Question 121

Phase shift of Fresnel diffraction

Answer 121

Phase across aperture not linear.

On-axis, distance z from aperture. Wave originating y above centre of the aperture. Corresponds to a distance δ, given by
δ = √z²+y² ‘ − z = z(1 + y²/z²)¹’² − z ≃ y²/2z (binomial)

Corresponding phase shift is kδ = πy²/λz

Question 122

Rayleigh distance

Answer 122

The distance at which non-linear effects become important. It is usually defined as 2y²/λ, where y is the size of the aperture. Far-field diffraction theory valid if we are at a much greater distance than this from the aperture.

Question 123

Fresnel circular aperture

Answer 123

Point on aperture, h from the centre, receiving a spherical wave from source and re-radiating towards on-axis observer.

Total path length compared to the straight-through path
= (s₀² + h²)¹’² + (R² + h²)¹’² − s₀ − R
≃ s₀(1 + h²/2s₀²) + R(1 + h²/2R²) − s₀ − R
≃ h²/2z
(using binomial expansion where R, s₀ ≫ h)

Have defined z ≡ (1/R +1/s₀)⁻¹
→ z = sR/(s₀ + R), so z = R if s₀ → ∞ and z = s₀ if R → ∞.

Circular aperture allows through a set of concentric circle parts of wavefront. Divide the area of wavefronts into half-period zones, from which waves reinforce each other when added at P.

Consider central zone. Reinforcement requires the path from all waves to observer to be
within λ/2 of length from centre of aperture to observer. Boundary of this zone (HPZ1) is when h = √zλ’

Next zone requires path lengths from h = √zλ’ to √2zλ’

To this approximation, all zones have same area.

Phasors within first HPZ will add until the phase rotates by 180° w.r.t. waves from the centre, creating semi-circle phasor diagram.

Adding second HPZ, phasors continue circle (almost) all the way back down again. Letting through waves from first two HPZs, the intensity on-axis will be
small.

If Kₘ ≃ Kₘ+1 and rₘ ≃ rₘ+1, (rₘ is distance from observer to mth zone), E₂ goes all the way back to zero.

Deeper level of approximation, not quite. Areas slightly bigger with m but both obliquity and amplitude fall off with m → E₂ doesn’t return to zero.

Can construct zone plates which block out alternate zones → powerful focusing effect, very bright points on axis.

Question 124

Assumptions made by Fresnel and Fraunhofer

Answer 124

Assumption | Fresnel ….| Fresnel………….|Fraunhofer
……………………| spherical | straight-edge |
λ ≪ R ………….| yes ………..| yes ……………….| yes
s, r const in |
denominator |no…………..| yes……………….| yes
K(α) constant |no…………..| yes……………….| yes
x, y ≪ X, Y ……|no ………….|no……………….| yes
phases linear |
across |
aperture……….|no ………….|no……………….| yes
s = s0 in |
phase term |
(plane wave)…|no ………….|no……………….| yes

Question 125

Fresnel straight edges and slits

Answer 125

On-axis, fig. 16
HF integral:
E(P) = ϵ′ exp(iks)/s ∫ K(α) exp(ikr)/r dA

Assumptions: K(α) is constant, s and r const. in denominator (but not in phase term)
E(P) ≃ ϵ′/s₀R ∫ exp(ik(r+s)) dA

Geometry
s² = s₀² + x² + y²
r² = R² + x² + y²

r+s = R(1+ x²+y²/R²)¹’² + s₀(1+ x²+y²/s₀²)¹’²

Binomial, but keeping terms in x² and y²
r + s = s₀ + R + (x²+y²) (s₀+R/2s₀R)

define z ≡ sR/(s₀+R)
E(P) = ϵ′/s₀R exp(ik(s₀+R)) ∫[ˣ²ₓ₁] exp(iπx²/λz) dx ∫[ʸ²ᵧ₁] exp(iπy²/λz) dy

define u = x√2/λz ‘, v = y√2/λz ‘
E(P) = ϵ′λz/s₀R exp(ik(s₀+R)) ∫[ᵘ²ᵤ₁] exp(iu²π/2) du ∫[ᵛ²ᵥ₁] exp(iv²π/2) dv

Can express integrals as real and imaginary parts in Argand diagram. These are known as the Fresnel integrals (w is either u or v):
C(w) = ∫ʷ cos(πw′²/2)dw’
S(w) = ∫ʷ sin(πw′²/2)dw’

Question 126

Cornu spiral

Answer 126

Phasor diagram of
C(w) = ∫ʷ cos(πw′²/2)dw’
S(w) = ∫ʷ sin(πw′²/2)dw’

Unobscured aperture, w=0, centre of spiral. w=√1’, phase shift π/2, first vertical. Each quater turn around spiral, corresponds to w=√n’ and phase shift nπ/2.

Open aperture with a slit infinitely large in one direction, calculate w₁ and w₂ (w₁ = x₁√(2/λz)’) (z is dist. to observer) . Add points to spiral, length of line directly connecting them gives amplitude seen by observer as a result of an open aperture.

Completely unobstructed amplitude given by the distance between the two eyes of the spiral, √2’

No slit, just blockage from w₁ to w₂, form two vectors from the spiral eyes (w = ±∞) points corresponding to w₁ and w₂, then add the two vectorially.

Rectangular aperture, perform above operations for both dimensions separately, then multiply resulting amplitudes.

Question 127

What’s important in quantum optics

Answer 127

Interaction with matter, photon nature of light is important.
LIGHT IS PARTICLES

Question 128

Photon frequency

Answer 128

Transition between two energy levels 1 and 2,

hν = E2 − E1

Question 129

Spontaneous emission

Answer 129

Transition from energy level 2 to 1, with electron densities N₂ and N₁

Electron moves from 2 to 1, emitting photon of energy hν.

dN₂/dt = −N₂A₂₁
⇒ N₂(t) = N₂(0)exp(−A₂₁t)

A₂₁ is the Einstein A coefficient

Question 130

Stimulated emission

Answer 130

Transition from energy level 2 to 1, with electron densities N₂ and N₁

Photon of energy hν effects the EM field, an electron moves from 2 to 1, emitting a second photon also of energy hν.

dN₂/dt = −N₂B₂₁ρ

ρ = I/c is energy density in radiation field (at transition freq.)
B₂₁ is the Einstein B coefficient for stimulated emission

Question 131

Absorbtion

Answer 131

Transition from energy level 1 to 2, with electron densities N₁ and N₂

Photon of energy hν absorbed, an electron moves from 1 to 2.

dN₁/dt = −N₁B₁₂ρ

ρ = I/c is energy density in radiation field (at transition freq.)
B₁₂ is the Einstein B coefficient for absorbtion.

Question 132

Three processes of quantum optics

Answer 132

dN₂/dt = −N₂B₂₁ρ − A₂₁N₂ + N₁B₁₂ρ

Question 133

Emission and absorption in thermal equilibrium

Answer 133

dN₂/dt = −N₂B₂₁ρ − A₂₁N₂ + N₁B₁₂ρ

dN₂/dt = 0

ρ = 8πhν³/c³ 1/exp(hν/kT)−1
(blackbody radiation)

N₂/N₁ = exp(−hν/kT)
(Boltzmann)

Sub and compare coeffs
B₁₂ = B₂₁
A₂₁/B₂₁ = 8πhν³/c³

Question 134

Optical gain

Answer 134

Length dz, gain in intensity of light dI:
dI = s(N₂B₂₁ − N₁B₁₂) I/c hνdz

B₁₂ = B₂₁, so
I(z) = I(0)exp(γz)
where
γ = s(N₂ − N₁)B₂₁ hν/c ≡ (N₂ − N₁)σ

σ ≡ sB₂₁hν/c is the stimulated emission cross section
γ is the gain coeffcient

System has optical gain if γ > 0 (γ > α, accounting for losses), which happens if N₂ > N₁ (population inversion).

Question 135

Stimulated emission cross section

Answer 135

σ ≡ sB₂₁hν/c

Question 136

Gain coefficient

Answer 136

γ = s(N₂ − N₁)B₂₁ hν/c ≡ (N₂ − N₁)σ

Question 137

Adding pump rate for optical gain

Answer 137

dN₂/dt = −N₂B₂₁ρ − A₂₁N₂ + N₁B₁₂ρ + R

Question 138

Derive threshold gain coefficient

Answer 138

• Medium in optical cavity with loss coefficient α and gain coefficient γ. Cavity length L is bounded by two mirrors of reflectance R₁ and R₂.
• Original intensity I₀, after one passage through cavity intensity becomes I₀exp[(γ−α)L].
• Reflection from mirror 2, I₀exp[(γ−α)L]R₂
• Back through medium and reflection off mirror 1, I₀exp[2(γ−α)L]R₁R₂.
• Threshold operation, there is just a net gain, hence exp[2(γₜₕ−α)L]R₁R₂ = 1.
• Rearrange, γₜₕ = α − 1/2L ln(R₁R₂)
• Lasers tend to operate near threshold. Pumped more, N₂ − N₁ increases, provokes more
  stimulated emission, N₂ − N₁ decreases.

Question 139

Energy-levels and intensity for practical lasers

Answer 139

Practical laser systems typically 4 levels. Pumping operates from ground state to top level (pump band), followed by transition to the upper laser level. Laser action takes place between the two middle levels.

Assume lifetime of transitions 3→2 and 1→0 is short compared to laser transition, to keep N₂ ≫ N₁

Rate equation:
(N = N₂−N₁ ≃ N₂, so dN/dt = dN₂/dt−dN₁/dt = 2dN₂/dt)
½ dN/dt = R − N(B₂₁ρ + A₂₁)

= 0 in steady-state conditions, and N = Nₜₕ.

ρ = (R/Nₜₕ − A₂₁) 1/B₂₁ = 1/NₜₕB₂₁ (R − Nₜₕ/τ₁₂)
where τ₁₂ is the lifetime of state 2.

I = ρc and σ = sB₂₁hν/c, so
I = shν/Nₜₕσ (R − Nₜₕ/τ₁₂)

Rₜₕ ≡ Nₜₕ/τ₁₂ is the threshold pumping rate
We define slope efficiency,
η ≡ shν/Nₜₕσ, so
I(R) = η(R − Rₜₕ)
[for R > Rₜₕ]