1. WAVELENGTH/FREQUENCY

The mathematical relation that relates frequency and wavelength of light is given by:

c=fλ

From which follows that the wavelength λ given frequency f can be expressed as:

λ= cf

We need to find a wavelength λ1 that corresponds to some frequency f 1 that is 100GHz lower than a

given frequency f 0.

λ1=cf 1

f 1=f 0−100GHz

The difference between new and original wavelength is given by:

∆ λ=λ1−λ0=cf 1

− cf 0

=c ( f 0−f 1 )f 1 f 0

≈cf 02 ( f 0−f 1 )=λ0

2 f 0−f 1c

Substituting the wavelengths we obtain:

For λ0=1300nm

∆ λ=(1300∗10−9 )2∗100∗109

299792458m=5.637∗10−10m=0.5637nm

For λ0=850nm

∆ λ=(850∗10−9 )2∗100∗109

299792458m=2.41∗10−10m=0.241nm

From which we obtain that the rule of thumb for those windows is:

In the telecom window near 1300nm rule of thumb is that 100GHz corresponds to 0.5637nm.

In the telecom window near 800nm rule of thumb is that 100GHz corresponds to 0.241nm.

2. QUARTER AND HALF WAVE PLATES

a) Describe briefly how half-wave and quarter-wave plates work

A waveplate works by shifting the phase between two perpendicular polarization components of the light wave. A typical waveplate is simply a birefringent crystal with a carefully chosen orientation and thickness. The crystal is cut into a plate, with the orientation of the cut chosen so that the optic

axis of the crystal is parallel to the surfaces of the plate. This results in two axes in the plane of the cut: the ordinary axis, with index of refraction no, and the extraordinary axis, with index of refraction ne. The ordinary axis is perpendicular to the optic axis. The extraordinary axis is parallel to the optic axis. For a light wave normally incident upon the plate, polarization component along the ordinary axis travels through the crystal with a speed vo = c/no, while the polarization component along the extraordinary axis travels with a speedve = c/ne. This leads to a phase difference between the two components as they exit the crystal. When ne < no, as in calcite, the extraordinary axis is called the fast axisand the ordinary axis is called the slow axis. For ne > no the situation is reversed.

Depending on the thickness of the crystal, light with polarization components along both axes will emerge in a different polarization state. The waveplate is characterized by the amount of relative phase, Γ, that it imparts on the two components, which is related to the birefringence Δn and the thickness L of the crystal by the formula

Γ=2 π ∆nLλ0

b) If ∆ n (difference between „fast“ and „slow“ axes) is 10^-3 for a material with n=1.5, calculate the minimum thickness of a ½ and ¼ wave plate for light with a wavelength λ=500nm. These are called „zero-order“ waveplates.

By substituting the Γ=π we obtain a condition for zero order-half-wave plates

π=2π ∆nL0 HW

λ0

Where L0HW is the zero order half-waveplate length.

L0HW=λ02∆n

=2.5∗10−4m=0.25mm

By substituting the Γ=π2

we obtain a condition for zero order-quarter-wave plates

π2=2π ∆n L0QW

λ0

Where L0QW is the zero order half-waveplate length.

L0QW=λ04 ∆n

=1.25∗10−4m=0.125mm

c) Zero-order waveplates are often difficult to make because they are very thin. To make it thicker and easier to manufacture, we can make the optical path difference = (N+1/2)λ for a half-wave plate and (N+1/4)λ for a quarter wave plate where N is any integer. IF N=5 what are the new thickness for the waveplates?

Multiple order half-waveplates for N=5:

(N+ 12)2π=

2 π ∆n L5HW

λ0

11 π=2π ∆ nL5 HW

λ0

L5HW=11λ02∆ n

=11L0HW=2.75mm

Multiple order quarter-waveplates for N=5:

(N+ 14 )2π=

2 π ∆n L5QW

λ0

2142π=

2 π ∆n L5QW

λ0

L5QW=212

λ02∆n

=10.5 L0HW=2.625mm

d) Waveplates clearly do not work properly for all wavelengths. For b) (zero order waveplates) calculate the wavelength that i) the half waveplate would turn into a quarter waveplate.

Condition for zero order-quarter-wave plates:

π2=2π ∆n L0QW

λ0

Substituting the half waveplate length L0HW and introducing a new wavelength λ1

π2=2π ∆n L0HW

λ1

It follows:

λ1=4 ∆n L0HW=4∗0.001∗0.25mm=10−6m=1000nm

e) Do the same for multiple order half waveplate and make a conclusion about the usable wavelength ranges for the zero-order halfwave plate compared to the multiple order waveplates.

Condition for multiple order-quarter-wave plates:

(N+ 14 )2π=

2 π ∆n L5QW

λ0

Substituting the half waveplate length L5HW and introducing a new wavelength λ1

(N+ 14 )2π=

2 π ∆n L5HW

λ1

It follows:

λ1=∆n L5HW

(N+ 14 )

=0.001∗2.75mm

(5+ 14 )=5.2381∗10−7m=523.8095nm

From which is seen that the wavelength sensitivity is increased many times in comparison to the zero order waveplates. Although the waveplates are easier to manufactore, they are much more sensitive to the input parameters in comparison to the zero order waveplates.

3. IMAGING AND LENSES

a) What is the Airy disc size for this lens, for visible light (λ=500nm¿

Angular size of the Airy disc is given by Rayleigh criterion for barely resolving objects that are point sources of light. The image seen from satellite, is that the center of the Airy disk for the first object occurs at the first minimum of the Airy disk of the second. This means that the angular resolution of a diffraction limited system is given by the formula:

sin (θ )=1.22 λD

Which for small angles θ can be written as:

θ=1.22 λD

Substituting the wavelength and the lens diameter we obtain:

θ=1.22 500∗10−9

4∗10−2 =1.525∗10−5 rad=3.14 arc seconds

Airy disk radius r expressed as distance between two objects on the earth:

sin (θ )=θ= rdearth

r=θf=1.525∗10−5∗100∗103=1.525m

b) How big would this spot size be if imaged onto the earth 100km away?

The most convenient way to measure the spot size is to ignore relatively smaller outer rings of the Airy pattern and to approximate the central lobe with a Gaussian profile, such that

I (q )=I 0' exp (−q2

2σ2). Where I 0

' is the irradiance at the center of the pattern. q represents the

radial distance from the center of the pattern, σ is the Gaussian RMS width (in one dimension).

If we equate the peak amplitude of the Airy pattern and the Gaussian profile, that is I 0=I 0' , and

find the value of the σ giving the approximation to the pattern we get:

σ=0.42 λN

Where N is the f-number = f/D. On the other hand if we wish to enforce that the Gaussian profile has the same volume as does the Airy pattern, then this becomes

σ=0.45 λN=0.45∗500∗10−9∗4∗10−2

5∗10−2=1.8∗10−7m

c) What is the resolution of the camera?

By taking the Rayleigh criterion for resolution, we obtain the the angular resolution is

θ=1.22 500∗10−9m

4∗10−2m=1.525∗10−5rad

Which corresponds to 1.525m distance between two objects on earth.

d) Is your friend exaggerating?

Since the distance between letters in news paper is on the order of magnitude of a centimeter, it follows that it is far from possible to read the news paper with this satellite camera. My friend is exaggerating.

4. FIBER BRAGG GRATINGSa) What is the bandwidth of the grating in nm?

Strength κ can be expressed as:

κ=π δ neff

λ=2c m−1

From which δ neff can be calculated as:δ neff=κλ=2∗10−2∗1550∗10−9=3.1∗10−8

Grating period Λ is:

Λ= λ2neff

=1550∗10−9

2∗1.5=516.67 nm

We need to determine in which grating limit regime are we in.

λL=1550∗10

−9

2∗10−2=7.75∗10−7

Since δ neff ≪λL

We're in weak grating limit regime. Bandwidth can be calculated as:

∆ λ=λ2

neff L=

(1550∗10−9 )2

1.5∗2∗10−2 =8.008∗10−11=0.08nm

b) What is the Reflectivity (in %) and Transmission (in dB)?

r=tanh2κL=tanh (2∗2 )=0.9987

t=1−r=0.0013

t [dB ]=10 log( t1 )=−66.14dB

c) If this grating were written in birefringent fiber where the difference in refractive index for the two polarizations ∆ n=10−3, what is the separation in grating wavelengths for the two polarizations?

λSlow=2neff ,slow∗Λ

λFast=2nef f , fast∗Λ

∆ λ=λslow− λfast=2∆n Λ=2∗10−3∗516.67 nm=1.03nm

d) Will this grating be any good for filtering unpolarised light?

No. Grating wavelength separation is an order of magnitude larger than bandwidth earlier calculated.

5. OPTICAL DATA STORAGE

a) What is the spot size for CD players?

N ACD=sinθCD=1.22λCDDCD

DCD=1.22λ

N ACD

=780nm0.45

∗1.22=2.114∗10−6m

ACD=DCD2

4π=3.506∗10−12m2

b) What is the spot size for BluRay players?

N ABR=sin θBR=1.22λBR

DBR

DBR=1.22λBR

N ABR

= 405nm0.85

∗1.22=5.812∗10−7m

ABR=DBR2

4π=2.653∗10−13m2

c) Calculate the ratio of the focal spot area for the Blu-Ray disc player compared to CDs (ie Ratio=Abr/Acd)

Ratio=ABR

ACD

=0.0758

d) Asssuming that the total information on the disc scales inversely with spot area, what would this predict for the number of Gigabytes for a BluRay disc (CD ROMS are 700MB(1 MB = 1million bytes)?

CBR=CCD

Ratio=700MB0.0758

=9231.8MB=9.01GB

e) Is the spot size reduction itself enough to explain the increased data storage of BluRay discs?

No.

f) What else could account for the difference?

Spot separation can also be reduced as the uncertainity is now smaller. Which means that there is a lot of unused spot space in CD compared to BluRay which adds an additional factor of improvement.