ECE 5/6358 HW 1 (150 pts)
First name, Last name

Student ID:

Han Q Le(c)

Below are excerpted works from HW 1 of two individuals from the class. Many have have very good answers, and over the semester, we will take turn to cite the works of many others. This time, two people’s works are selected as A and B below, highlighted in light blue. Instructor’s comments are in light green. However, there are some mistakes and errors that virtually the whole class did (including selected works). These are high-lighted in light red.

1.(50 pts) Interest in optics or optoelectronics

Write ~ 0.5-1 page (figures & references included) on a topic in contemporary optics/photonics science and technology that interests you, or you are curious about (a reason you take this course).

2.(50 pts) Periodic phenomena

Consider three physical phenomena:
- phenomenon a: periodic in time:  u= Cos[2 π f t]    ( u is just a variable representing amplitude)
- phenomenon b: periodic in space:  v= Cos[k x]
- phenomenon c: periodic in both space and time:  q= Cos[k x-2 π f t]

Below is the code that calculates and plots the three phenomena. You don’t have to do anything but executing the code, vary various parameters, and observe. Write all you think you learn about the phenomena, especially “c”.
Remember: to see a time phenomenon, run variable time; to see a space-periodic phenomenon, sample (measure) over a range of space; and do both time and space for “c”

Code for calculation (no need to see, just Shift+ENTER to execute - delete the output - NOT THE CODE - after done)

Manipulate[
xarr=Range[0,xmax,0.1];
Grid[{{"a",Style["Cos[2π f t]",24,FontFamily->"Times New Roman"]
,
Graphics[{Hue[0.14,0.3,1]
,Rectangle[{-0.25,-1.25},{10.25,1.25}]
,Red, PointSize[0.015]
,Table[Point[{u,Cos[2 π f t]}],{u,xarr}]}
,PlotRange->{{-0.25,10.25},{-1.25,1.25}}
,ImageSize->525]}
,{"b",Style["Cos[k x]",24,FontFamily->"Times New Roman"]
,Graphics[{Hue[0.47,0.3,1]
,Rectangle[{-0.25,-1.25},{10.25,1.25}]
,
Red, PointSize[0.015]
,Table[Point[{u,Cos[k u]}],{u,xarr}]}
,PlotRange->{{-0.25,10.25},{-1.25,1.25}}
,ImageSize->525]
}
,{"c",Style["Cos[k x-2π f t]",24,FontFamily->"Times New Roman"]
,Graphics[{Hue[0.28,0.3,1]
,Rectangle[{-0.25,-1.25},{10.25,1.25}]
,Red, PointSize[0.015]
,Table[Point[{u,Cos[k u-2 π f t]}],{u,xarr}]}
,PlotRange->{{-0.25,10.25},{-1.25,1.25}}
,ImageSize->525]
}},Frame->All]

,{f,0,1},{k,0,2 π},{t,0,10}
,{xmax,0,10,0.1}]

Example A

Answer : A - when varying the time with not changing frequency nothing happened because the first one is a function depending in frequency and time.The frequency needed to be changed to some value and then varying time we could see that both a and c were moving because they
are dependant on time.

However, b showed no change because it is a function of space and does not
depend on time.B - when we varied the value of x (space) with k set to some value we
noticed a movement in b & c because they are both dependent on space
and change value if x change.A showed no movement (because it is only dependent of time).

In case of c we noticed that no matter which variable we are varying it is always moving whether it was space or time it is in a constant movement and that is why it is differnet from the two functions it is always propagating (through time and space)

Example B

Instructor’s comment:

The essence of science is observation, looking for clues, and interpretation. Seeing with a critical eye and not seeing with what your preconceived notion in the brain tells you. Example B is excellent on the last sentence except for a nomenclature error: A and B are NOT waveform. Only C is qualified as “waveform”. This appears to be a semantic issue, as author B might use “waveform” as just a word for “entity, quantity, or phenomenon, or even mathematical function”. In physics, do not confuse a function with “waveform”. Wave has a unique and specific meaning, and this course is about EM waves in the frequency range of (10^14) Hz (100’s of THz).

App that maybe helpful to HW - use to gain concepts and understanding to do Problem 3 (this is NOT a problem of the HW)

3.(50 pts+10 pts bonus) Frequency and wavelength

- We have a group of students who pick 3 wavelengths: 0.65 μm (red DVD laser), 0.532 μm (green laser pointer), 0.405 μm (BluRay laser) (a wavelength, if not specified, is always meant to be in vacuum).
- We have another group of students who pick 2 frequencies: 527 THz  and 384 THz. (1 THz=(10^12) Hz),

Each question below is worth 10 pts - Answer thoroughly and thoughtfully.

3.1 Hypothetical waves

If we could (hypothetically) make waves that could be from all arbitrary combinations of wavelengths and frequencies, how many waves can we make and what do they look like when traveling. Just run the code and discuss the single most significant observation. (if someone tells you that you can say ONLY ONE thing that you notice about the waves below, if it is correct, you will get $1M, and incorrect, $0. Only one. If you make more than one point, you will get zero. What would you say?).

Code (no need to see, just Shift+ENTER to execute - delete the output - NOT THE CODE - after done)

Manipulate[
Show[Plot[Release[Flatten[Table[Cos[k[[i]] x - ω[[j]] t]
+3*((i-1)*mω+j-1),{i,nk} ,{j,mω}],1]]
,{x,0,8}
,PlotStyle->Table[{Thickness[0.01],Hue[((m-1)/(nwaves+1)),0.7,1]},{m,nwaves}]]
,Graphics[{PointSize[0.05],Thickness[0.01],Dashed
,If[TrueQ[SHOWPT],Table[xpos=t*ω[[j]]/k[[i]];{Hue[(((i-1)*mω+j-1)/(nwaves+1)),0.5,1]
,Point[{xpos,1+3*((i-1)*mω+j-1)}]
,Line[{{xpos,0},{xpos,1+3*(nk*mω-1)}}]},{i,nk} ,{j,mω}]
,dummypt]
}]
,Frame->True, FrameStyle->White
,AspectRatio->1, ImageSize-> 500
,Background->Black]
,Grid[{{Control[{{t,0,"time"},0,0.04, ImageSize-> 120}]
,Button[Dynamic[If[TrueQ[SHOWPT],hide,show]],SHOWPT=!SHOWPT]}}]
,Initialization:>{
λ={0.65,0.532,0.405};k=2 π/λ;
f={384.,527.};ω=2 π f;
nk=Length[k]; mω=Length[ω];
nwaves=nk * mω;SHOWPT=False;
show=Show[,ImageSize-> 120];
hide=Show[,ImageSize-> 100];
dummypt={Hue[0,0,1,0],Point[{0,0}]};
}
]

Example A

Answer:
Hypothetically speaking if we could create 6 waves depending in the given input frequencies and wavelengths they will be combined as follows:
{(0.65, 527), (0.65,384), (0.532,527), (0.532,384), (0.405,527), (0.405,384)}
Then their speed would be calculated by multiplying the frequency by the wavelength for each one. 
We notice when we run the code that we get 6 waves traveling at different speeds.
The one thing I could say about the waves that they have different propagation velocity.

Example B

Instructor’s comment:

Watching the movement, with the location of the point on each wave vs. time, can’t one tell what is the most obvious? Science is about observation. The most obvious thing here is that the fixed points on the waves moves at different speeds.

3.2 Identify the waves

All the 6 waves you see above have the form: Cos[k x - ω t] where k is ((2π)/λ) and ω is 2 π f. Denote them from 1 (top) to 6 (bottom), Can you match each wave with its  λ and f? For example:
                               wave 1 corresponds to {0.532 μm, 384 THz} (incorrect, but just an example)

Example A (same as above because it explains both)

Example B

Instructor’s comment:

Example B is perfect. Example A is correct, but it didnot specify the order (bottom to top or top to bottm). Here, for the time being, it is given the benefit of the doubt just like many other works in the class that also give a correct list without a clarity of the order.

Science is about observation. What is wavelength? what is the operational meaning of it? If one sees it, does one recognize what it is? Look at the chart below, what does one automatically recognize about the wavelength? They come in pairs, top two have the same wavelength, then the next two, and the last two. Which wavelength is the shortest? (smallest), which one is next? etc... As we know the wavelengths are: “ 0.65 μm (red DVD laser), 0.532 μm (green laser pointer), 0.405 μm (BluRay laser)” It’s obvious that the top ones are 0.405, then 0.532, and the bottoms: 0.65/

OK. We identify them into 3 groups by wavelength, but what about within each group? If we animate and even just look at one location such as x=0, it is obvious that within each pair, the top one oscillates faster than the bottom one, which means it has higher frequency. This tests the operational understanding of “frequency”, which is the rate of oscillation: the number of cycles per unit time.
Thus, it is straight forward to identify the wave: by operationally looking at its wavelength and frequency.

What does “operational knowledge” mean? Here is an example: John and Bob walk along the seaside of a town. Suddenly, there is a siren of “tsunami” warning. John has the academic knowledge of “tsunami”. He knows that tsunami comes from Japanese word for “harbor wave”, which is a big wave of water that moves in from the sea. John thinks he wants to stay to watch it. Bob however, immediately runs for his life. Bob is said to have an “operational knowledge” of tsunami. John is said to have a “wiki or Internet knowledge” of tsunami.

In the above, if a person recognizes that the spatial periodicity of the top plot is shorter than the lower ones, hence it must correspond to shorter wavelength, then the person is said to have an “operational knowledge” of wavelength. Likewise, if the person knows faster oscillation means higher frequency, that’s an operational knowledge of frequency.

3.3 Speed

We learn that the wavespeed is (λ/T)= λ f (See the app above again about the wave speed not sure).

Below is the calculation of the 6 wave speeds. The unit is μm/ps.

Grid[Join[{{"λ","f","speed"}},ReverseSortBy[Flatten[Table[{v,u,u*v},{u,f},{v,λ}],1],Last]] ,BaseStyle->{24,FontFamily->"Calibri"}, Frame-> All,Background->{{,,},{Hue[0.3,0.4,1],,,,,,}}]

Watch the waves in 3.1 again, Can you tell which wave with what wavelength and frequency just by watching? What is the principle behind the approach that you use to identify each wave?

Example A

Answer:
Based on the observation from question 3.1 and the calculated velocity in 3.2 the waves can be easily defined if we show point and notice which ones are the fastest and which are the slowest. 
we can see based on the point shown on each wave that the yellow one in the fastest with speed (342.55) and the light blue one

3.4 Given that propagation is such a fundamental property, the speed must be a constant of nature (otherwise nature would be a mess)

The speed of light is 299.792458 μm/ps in vacuum, independent of frequency (in 1 ps, light travels about 1/3 of a millimeter, or ~ 6-10 times the width of our hair). Can you calculate the correct frequencies for the 3 wavelengths above and watch how each wave propagates? You only have to calculate f, The code for the wave propagation is given, what is the single most significant feature in the way the waves propagate?

Code (no need to see, just Shift+ENTER to execute - delete the output - NOT THE CODE - after done)

Manipulate[
Plot[Release[Table[Cos[k[[i]]( x - cL t) ]+3 (i -1),{i,Length[k]}]]
,{x,0,5}
,PlotStyle->Table[{Thickness[0.01],Hue[((i-1)/(Length[k]+1)),0.7,1]},{i,Length[k]}]
,PlotRange->{-1.5,9.5}
,Frame->True, FrameStyle->White
,FrameLabel->{Style["space (μm)",20]}
,Epilog->Inset[Grid[{{"time (fs)= ",1000*t}},BaseStyle->{18,Yellow}]
,ImageScaled[{0.5,0.92}]]
,AspectRatio->0.6
,Background->Black]
,{{t,0,"time (ps)"},0,0.02}
,Initialization:>{
λ={0.65,0.532,0.405};k=2 π/λ;
cL=299.792458;
hue={0.04,0.36,0.7};
f=cL/λ;ω=2 π f;
}
]

Example A

Answer :
  The wave speed = the wave length * the wave frequency
The wave speed =  c (the speed of light)
The fwavelengths = { .65, .405, .532}
The wave frequency = the wave speed / the wavelegth
f1 = c/λ
= 299.792458*10^6 / 0.65*10^-6
f1 = 461.219 THz.
     the same equation applies to f2 & f3 
the outputs :
   f2 = 740.228 THz
f3 = 563.519 THz
 the frequencies are { 461.219, 740.228, 
   563.519} for wavelengths = {0.65, 0.405, 0.532} respectively. 
    cl = 299.79

3.5 Speed of light measurement

If you look up any physical quantity, such as those listed on this NIST website:
https://physics.nist.gov/cgi-bin/cuu/Category?view=html&Frequently+used+constants.x=69&Frequently+used+constants.y=29

almost every quantity has some experimental uncertainty. For the speed of light however, it is exactly 299.792458 μm/ps. No error. Why is that?

Example A

Answer:
It is one of the fundamental concepts of the universe we live in. 
The second is the Standard International ( SI ) unit of time. One second is the time that elapses during 9,192,631,770 (9.192631770 x 10 9 ) cycles of the radiation produced by the transition between two levels of the cesium 133 atoms.  Therefore, the second is an absolute reference with no uncertainty, and we measure other stuff using the physical definition of a second as a reference point. The speed of light measures the distance the light travels per second.  The speed of light units of measurement is (m/s). In physics, any number that has units attached to it can have any value that it wants.
To express the speed of light (m/s), one must first define what a meter is and what a second is. Hence, the speed of light definition is related to the definition of length and time. 
Most physics matters are constants that do not have units of measurements (constants that appear as simple numbers). One of these numbers is the fine structure constant, which is a combination of the speed of light, the Planck’s constant, and something known as the permittivity of free space. It has a value of about 0.007. Technically, the speed of light can be whatever we want because it contains units, and we need to define those units. On the other hand, the speed of light can only be precisely what it is, because if we were to change the speed of light, we would change the fine structure constant, but the universe in which we live has chosen the fine structure constant to be about 0.007, not another number. Therefore, the speed of light is constant and has no value of uncertainty. 
It is a fundamental concept of the universe we live in.

Example B

3.6 (bonus) Gravitational wave vs light wave speed

In the original paper (and on LIGO website), people actually published the experimental uncertainty about the difference between (Subscript[c, L]), the speed of light, and (Subscript[c, G]), the speed of gravitational wave. Theoretically, they have to be equal for General Relativity to make sense. What is the experimental relative uncertainty of their difference and how was it determined? (read the materials and report).

A good news is that you did look thinner for a moment when the space-compress part of the gravitational wave passed through you (but it stretched you out also for its space-dilation part).

Example A

Answer:
The Generalized  Uncertainty Principle.  According to that principle the uncertainties of the particle’s location and momentum could not be established to arbitrary precision. In the gravitational wave event the GUP dimension  parapemeters were constrained. Researchers were able to derive a standard energy-momentum  despersion relation and calculate the difference between the propagatiing speed of gravitions and the speed of light.

The proposed version of GUP by MAngano & Man was:
                         δxδp≥\:0304h[1+β(p)2]/2
were δx & δp are the uncertanties for position and momentum. 
β = β0 l^2/n^2 
 β0  is called GUP parameter. 
The speed of gravition in gravitational wave incident was given  by researchers to be:

vg= ∂ω/∂p
                  = c^2 * p / (c^4 * m^2 +c^2*p^2)
=c √1−(m^2c^4/ωg^2)                                                  ≈c(1−mg2^c^4ωg^2)

where ωgand mgare the energy and rest mass of gravitons,respectively.Considering that h=4.136×10−15eV·s and c=3×108m/s,then the difference between the speed of gravitons υgand the speed of light ccan be expressed as:

δv= c- vg = mg^2 c^5/ 2ωg^2
δv= 5.6 * 10^(-12) m/s.

Lack of citation

Example B

Can be better citation but good enough.

Instructor’s comment

Two things:
1- search for correct sources of information
2- presentation (simplification, description in your own words) and citation.

Here is an example how to condense complex information into layman’s words: When the gravitational wave (GW) was detected, almost simultaneously, a sudden burst of extremely high energy photons (known as γ-ray burst or GRB) was detected coming from the same angular direction of the sky as the GW. Both phenomena are best theoretically accounted for as the effects of two black holes merging. If the GW and GRB were purely coincident that had not had anything to do with each other, the chance would be only 1 out of 20,000,000. Hence, it is a better hypothesis to surmise that both must have come from the same black hole merging event. If so, the speed can be compared, and the result is that the speed of the GW and that of the gamma-ray, which is the speed of light, differs no more than between -3×(10^(-15)) and 7×(10^(-16))  of the speed of light.