WRITTEN EXAMINATION - 23.7.19

SoSe 19 Institute of High-Frequency and Quantum Electronics / Laboratory of Nano-Optics

Photonic Devices

WRITTEN EXAMINATION - 23.7.19

Please answer the following questions as a text or with simple diagrams. The emphasis is to show that you have understood the topic. Please be concise and do not spend too much time.

Typically, 3-5 relevant sentences should be sufficient to answer the question. Quality and not quantity of the answer is relevant! Formulas can be used if they are helpful for the explanation.

Depending on the quality of the answer 0-3 points per question are given.

Questions

1 - Electromagnetic waves

1. Explain the difference between a plane wave and a standing wave (2P)

A plane wave travels with a planar wavefront (surface of constant phase) perpendicular to

the propagation direction , e.g. , with .

A standing wave results from the superposition of two identical counter-propagating waves of identical frequency, e.g.

. Thus, there is no propagation of energy and the wave becomes an electromagnetic oscillation with nodes and anti-nodes, e.g.

. The nodes occurs for and the anti-

nodes for .

2. When does an electromagnetic wave is linearly polarized? (2P)

An electromagnetic wave is linearly polarized, when the electric field is confined to a single plane along the direction of propagation (say z-axis). This happens when there is no phase difference between the x- and y-component of the electric field, e.g.

. The state of a linearly polarized wave can also be represented by the Jones vector , where is an angle the electric field makes with the x-axis, e.g. [1,0] is linearly polarized electromagnetic wave along x-axis.

2 - Thin-film optics

1. Explain the working principle of an anti-reflection coating. (2P)

An antireflective or anti-reflection (AR) coating is a type of optical coating applied to the surface of lenses and other optical elements to reduce back reflection and avoid the creation of ghost images. Many coatings consist of transparent thin film structures with alternating layers of contrasting refractive index. Layer thicknesses are chosen to produce destructive interference in the beams reflected from the interfaces, and constructive interference in the corresponding transmitted beams. The simplest interference anti- reflective coating consists of a single thin layer of transparent material with refractive index

⃗ k ⃗ E ( ⃗ r, t) = ⃗ E

sin(ωt − ⃗ k ⋅ ⃗ r ) ⃗ E

⋅ ⃗ k = 0

⃗ E ( ⃗ r, t) = ⃗ E

[ sin(ωt − ⃗ k ⋅ ⃗ r ) + sin(ωt + ⃗ k ⋅ ⃗ r ]

⃗

E(

r, t) =

E

[ sin(ωt −

k ⋅

r) + sin(ωt +

k ⋅

r) ]

⃗ E ( ⃗ r, t) = 2 ⃗ E

sin(ωt)cos( ⃗ k ⋅ ⃗ r ) cos( ⃗ k ⋅ r) = 0 cos( k ⃗ ⋅ ⃗ r ) = ± 1

⃗ E (z, t) = ( ̂ x + ̂ y)E

exp[i(kz − ωt)]

J = [cos θ, sin θ ] θ

(nf) equal to the square root of the product of the refractive indices of the incident medium (n0) and the substrate (ns) (i.e., nf = . They are designed in such a way that the relative phase shift between the reflected beam at the upper and lower boundaries of the the film is 180°. To do so the optical thickness of the optical coating must be an odd integer multiple of λ/4, where λ is the optimized peak performance wavelength to achieve the desired path difference of λ/2 between the reflected beams.

2. When will a periodic dielectric stack made of two alternating media of thickness d1 and d2, respectively having refractive indexes n1 and n2, will block a normal-incident plane wave of wavelength λ? (2P)

A normal-incident plane wave of wavelength λ will not be transmitted, if the condition is met (photonic band gap at the edge of the Brillouin zone). Here a is the period of the dielectric stack ( ) and is the wave-vector (Bloch vector). One can approximate the Bloch vector using the average refractive index, , i.e.

, to obtain or and . Thus,

the thickness of each layer must be approximately equal to one-quarter of the wavelength in the layer itself.

3 - Optical waveguides

1. Discuss the dispersion relation of a symmetric dielectric slab waveguide (2P)

The dispersion relation of a symmetric dielectric slab (refractive index n1 in a surrounding medium with refractive index n2) is obtained by solving the wave equation or by the ray propagation method (self-consistency condition). The resulting formula is a transcendental equation that can be solved graphically in a qualitative manner. The waveguide supports transverse electric (TE) and transverse magnetic (TM) modes. Furthermore, each of them can be symmetric or antisymmetric with respect to the waveguide symmetry plane. The fundamental mode does not exhibit a cutoff frequency. All modes start with a linear dispersion relation, , and when the propagation constant , the dispersion relation is again approximately linear . This reflects the fact that near cutoff the EM field is mainly in the surrounding medium, whereas for large wavevectors the field is mainly confined in the slab. For even TE modes the cutoff frequency is , with m the mode number, d the slab thickness and the numerical aperture of the waveguide.

2. Discuss the dispersion relation of a surface plasmon-polariton wave at the interface between air and a Drude metal. (2P)

Surface plasmon-polaritons are surface electromagnetic waves that exist at the interface between a dielectric and a metal with dielectric constants of and , respectively. The propagation is along the interface between the two media and the wave vector obeys the

following dispersion relation . We set , the real

dielectric function of a Drude metal, where is the so-called plasma frequency.

The dispersion curve is almost linear for , with , and it flattens for , with . This condition corresponds to the frequency of a surface plasmon, which represents an upper limit for the dispersion relation. When the dielectric constant of the surrounding dielectric medium changes, for example from water to air,

n

n

 )

k = π /a

a = d

+ d

k

n

= n

d

+ n

d

k ≃ 2π n

/λ n

d

+ n

d

= λ /2 d

≃ λ /(4n

) d

≃ λ /(4n a

)

β ≃ ωn

/c β → ∞

β ≃ ωn

/c

ν

= c

2dNA m NA = n

− n

ϵ

ϵ

k = ω

c

ϵ

ϵ

ϵ

+ ϵ

ϵ

= 1 − ω

/ω

ω

k → 0 k ≃ ω ϵ

/c k → ∞ ω ≃ ω

/ 1 + ϵ

ϵ

the dispersion relation will change accordingly. This finds application in sensing, for instance in surface plasmon-polariton resonance biosensors.

4 - Optical nanoparticles

1. Describe the optical properties of a metal nanoparticle in terms of its quasi-static polarizability. (2P)

For their small dimension, metal nanoparticles can be investigated in the quasi-static approximation. For example, the optical properties of a spherical nanoparticle of radius a can be understood by the following polarizability formula , where and respectively are the dielectric functions of the nanoparticle and of the surrounding medium. When the nanoparticle is illuminated by a plane wave, the scattered power Psca

and the absorbed power Pabs are simply related to the polarizability as follows, and , where is the intensity of the incident wave.

When the polarizability largely increases and gives rise to a resonance in scattering and absorption that is typically in the optical spectral range for noble-metal particles.

2. How would you interpret a metal nanoparticle as an optical resonator? (2P)

One can associate a quality factor to the nanoparticle and a mode volume to the near field, based on the fact that it corresponds to confined electromagnetic energy. While an optical resonator is typically characterized by a large quality factor and a diffraction-limited mode volume, a metal nanoparticle exhibits a small quality factor combined with a nanoscale mode volume. Because the Purcell factor is proportional to the ratio between the quality factor and the mode volume, one finds that a nanoparticle can give rise to very large Purcell factors.

Calculation

1 - Fabry-Perot cavity

Given a reflectivity R = 0.99, a Fabry-Perot cavity is used as an interference filter, transmitting light of wavelength 1000 nm at normal incidence. The full width half maximum (FWHM) is about Δλ = 1 nm.

1. Find the spacing of the plate. (2P)

In the absence of absorption in the spacing medium, the transmission of a Fabry-Perot

cavity is given by , where , with k the wave vector, d

the mirror spacing and the angle of incidence. For normal incidence that defines an optical resonator, . Using small angle approximation of Sin x ≈ x and the condition for the FWHM, i.e. , one obtains . Therefore, for

α = 4π a

ϵ

− ϵ

ϵ

+ 2ϵ

ϵ

ϵ

P

= Ik

| α |

/(4π) P

= IkIm(α) I ϵ

+ 2ϵ

≃ 0

I = I

1 +

sin

δ δ = 2kdcosθ cosθ θ = 1

I = I

/2 δ

= 1 − R

R ≃ 0.01

normal incidence, the overall phase change in the FWHM is equal to

. Solving for d, we find nm nm.

2. Find the change in mirror spacing to shift the central wavelength to 1010 nm. (2P)

The condition for the transmission maximum is , m = 0, 1, 2,… for the given resonance mode . The change in the mirror spacing for the desired wavelength shift is given by nm.

3. The spacing is now fixed at a value that gives maximum intensity at 1000 nm at normal incidence. Find the free spectral range and the cavity finesse. (2P)

The free spectral range reads MHz, where m/s is the speed

of light. The cavity finesse reads .

4. If the reflectivity drops to R = 0.9, what is the change in the cavity finesse? (2P)

The finesse would now read , i.e. a reduction of an order of magnitude.

Δδ = 4π d Δλ

λ

= 2δ

d = 5000 π ≃ 1591,5

δ = 2π m m

= 2 d

λ ≃ 3

Δd = m

Δλ /2 ≃ 15

ν

= c

2d ≃ 94 c ≃ 3 × 10

ℱ = π R

1 − R ≃ 313 ℱ = π R

1 − R ≃ 30