ECE 5/6358 HW 2A

Part A - medium and phase

Background info that can be used to do HW (you don’t have to do anything, just read )

In previous HW, we study just the basic about frequency and wavelength. Now we learn about light in a dielectric medium. The Maxwell’s equations for EM field in a linear dielectric media show that it behaves just like in vacuum except for one thing: the speed. The wave equations in vacuum is:
            (∇^2)E -(1/(c^2))((∂^2)/(∂ (t^2)))E=0 or for plane wave:                (((∂^2)E)/(∂(z^2)))-(1/(c^2))(((∂^2)E)/(∂ (t^2)))==0 
and in a linear dielectric medium:
            (∇^2)E -((μ ϵ)/(c^2))((∂^2)/(∂ (t^2)))E=0 or for plane wave:                (((∂^2)E)/(∂(z^2)))-(((((Sqrt[μ ϵ])/c)))^2)(((∂^2)E)/(∂ (t^2)))==0 
The speed in vacuum is c, the speed in the dielectric is (c/(Sqrt[μ ϵ])) if (Sqrt[μ ϵ])≡n is real. We’ll see later that μ ϵ can be complex and will learn about that.

1. Wave in medium

1.1 Frequency

Consider a lightwave of wavelength λ=0.5 μm (always assume to be in vacuum if it is not specified). What is its frequency in THz?

1.2 Lightwave in glass

Consider a plane wave of the light in Q 1.1 traveling from vacuum or air (we can treat air the same as vacuum) into a medium like glass, which has index of refraction  (Sqrt[μ ϵ])≡n=1.5 as show below.

What is the speed of light in glass? Recalling that speed of light in vacuum or air is: 299.792458 μm/ps

1.3 Your thought (very important - it doesn’t matter you get it correctly or not)

Given an incident lightwave in air as discussed above, how many possible solutions of lightwave in glass (that corresponds to the incident lightwave) do you think exists? 1, 2, 3, ... infinite?
Remember this: A single plane wave has a unique amplitude, frequency, velocity, and phase. Here, you ONLY KNOW the velocity, nothing else. Hence, shouldn’t you have infinite plane waves that correspond to the incident wave?

1.4 Plot plane waves

Plot waves in two media: the wave in air and waves in the glass medium.

Assume the E field in air is described as: E=Cos[((2π)/λ)z-2 π f t] (amplitude =1). The wave is at normal incident. Usually, there is reflection at the interface, but for simplicity, assume that there is antireflection coating at the interface so that 100% of the light from air enters the glass, in other words, the wave amplitude in glass is also =1.

Hint: Remember: the wave in air travels with speed of light c, the wave in the glass travels with the speed you find in Q. 1.2. The wavelength in air is 0.5 μm, the frequency is what you found in Q. 1.1 above. The wavelength and the frequency in the glass are not known and you have to find out. You have to find all possible solutions of the lightwave in glass, as discussed in 1.3.

demo

1.5 Discussion (very important)

After finishing Q. 1.4, reconsider Q. 1.3 again: Given an incident lightwave in air, how many possible solutions you can find for waves in glass? 1, 2, 3, ... infinite?

2. Devil’s glass door

Have you ever been a victim of walking straight into a clear glass door or wall?

Consider a glass so clear with 100% transmission that it is hard to know if it is there or not. Click on the demo below to see.

Demo

A planewave propagating as shown above. One part of the wave enters a blackbox that may or may NOT contain a glass slab like the one you work on in prob. 1 above, which has index of refraction 1.5 and with 100% transmission (only the devil inside knows to put a slab there or not). Since the slab is completely transparent, regardless whether it is there or not, the output light is exactly as bright as the input light, and since the slab is perfectly parallel, the output beam is exactly in the same direction as the input, regardless.

2.1 Is there anyway for you to find out if there is a glass slab in the box?

Write down your thinking here. You must write and demonstrate sufficient thinking to get credit. It doesn’t have to be correct answer. The credit is for you to do the thinking, and not necessarily to get the correct answer right away.

2.2 Plot the waves

Use your results in prob. 1, Q. 4 above. Plot the wave from air, through the glass slab in the box, and exiting the slab to be compared with the wave that does NOT go through the black box. Choose whatever slab thickness that best demonstrates your understanding.

See hint in this App and the answer below.

Full answer (should open ONLY if you try your best and still cannot find the answer) you must delete this and rewrite the answer in the manner that you understand. Don’t bother to copy and paste what written here. There will be a lot of overlap as you try to re-write the proof, but that’s OK as long as you understand what you write.

Part 1: wave in glass slab

Let the front glass slab be at plane (Subscript[z, 0]) (in the app above, (Subscript[z, 0])= 1 μm). Of course we can choose (Subscript[z, 0])=0, but I want you learn to derive the phase term in the most general form possible.

Let the incident wave be:   (Subscript[E, inc])[z,t]=(e^(i (k z-ω t)))    (2.2.1)
where                            k=(ω/c)= ((2 π )/λ)  (here λ=0.5 μm)   (2.2.2)
We will rewrite in a more convenient form to use later:
                                        (Subscript[E, inc])[z,t]=(e^(i ω ((z/c)- t)))    (2.2.3)

The wave in the glass slab is:       (Subscript[E, slab])=(Subscript[A, g]) (e^(i ((Subscript[k, g]) z-(Subscript[ω, g]) t +ϕ))) ≡(Subscript[A, g]) (e^(i  ϕ )) (e^(i ((Subscript[k, g]) z-(Subscript[ω, g]) t )))    (2.2.4)
Note that we assume that we know nothing about the frequency in glass, (Subscript[ω, g]), amplitude and phase (Subscript[A, g]) (e^(i  ϕ )). However we do have a relationship between (Subscript[k, g]) and (Subscript[ω, g])
                                   (Subscript[k, g])=((Subscript[ω, g])/(Subscript[v, g])) =n((Subscript[ω, g])/c)   . Remenber that (Subscript[v, g])=(c/n)              (2.2.5)
Hence:                                     (Subscript[E, slab])=(Subscript[A, g])  (e^(i  ϕ )) (e^(i  (Subscript[ω, g ])(n (z/c)- t )))    (2.2.6)

To derive unknown quantities, we again apply Faraday law of induction, or the 1st Maxwell’s equation as discussed in the class,  
                           

Let’s apply the tangential E-field continuity condition at the boundary using (2.2.3) and (2.2.6)
             At (Subscript[z, 0])    in air:      (Subscript[E, inc])[(Subscript[z, 0]),t]=(e^(i ω (((Subscript[z, 0])/c)- t)))   ; in glass:     (Subscript[E, slab])=(Subscript[A, g]) (e^(i  ϕ ))(e^(i  (Subscript[ω, g ])(n ((Subscript[z, 0])/c)- t )))   
             
They must be equal to each other at all time:
                (Subscript[E, inc])[(Subscript[z, 0]),t]=(e^(i ω (((Subscript[z, 0])/c)- t))) =(Subscript[A, g])(e^(i  ϕ ))(e^(i  (Subscript[ω, g ])(n ((Subscript[z, 0])/c)- t )))  =(Subscript[E, slab]) [(Subscript[z, 0]),t]      (2.2.7)
or:                         (e^(i  ((ω (Subscript[z, 0]))/c)))(e^(-i ω  t)) =(Subscript[A, g])(e^(i  ϕ )) (e^(i  (n (((Subscript[ω, g ]) (Subscript[z, 0]))/c))))  (e^(-i  (Subscript[ω, g])t))     
because (e^(-i ω  t)) ≢0 , we can divide it from both side:
                              (e^(i  ((ω (Subscript[z, 0]))/c))) =(Subscript[A, g])(e^(i  ϕ )) (e^(i  n (((Subscript[ω, g ]) (Subscript[z, 0]))/c)))  (e^(i  (ω-(Subscript[ω, g]))t))              (2.2.8)

Since  (e^(i  ((ω (Subscript[z, 0]))/c)))  is constant of time, Eq. (2.2.8) would be false if 
                                                 ω-(Subscript[ω, g])≠ 0                              (2.2.9)
because then the term (Subscript[A, g])(e^(i  ϕ )) (e^(i  n (((Subscript[ω, g ]) (Subscript[z, 0]))/c)))  (e^(i  (ω-(Subscript[ω, g]))t)) will vary vs. time unless:
                                         (Subscript[A, g])(e^(i  ϕ )) (e^(i  n (((Subscript[ω, g ]) (Subscript[z, 0]))/c)))  =0                         (2.2.10)
But if    (Subscript[A, g])(e^(i  ϕ )) (e^(i  n (((Subscript[ω, g ]) (Subscript[z, 0]))/c)))=0  , then 
                                               (e^(i  ((ω (Subscript[z, 0]))/c))) =0                     (2.2.11)
which is false, because we assume that there is an incident wave and it is not zero.
Hence:                                 (Subscript[A, g])(e^(i  ϕ )) (e^(i  n (((Subscript[ω, g ]) (Subscript[z, 0]))/c)))  ≠0         (2.2.12)
and (2.2.9) must be false, and
                                            ω-(Subscript[ω, g]) =0                       (2.2.13)

this shows that the frequency in glass MUST be the same as the frequency in air.

Apply Eq. (2.2.13) to (2.2.8)
                           (e^(i  ((ω (Subscript[z, 0]))/c))) =(Subscript[A, g])(e^(i  ϕ )) (e^(i  n ((ω (Subscript[z, 0]))/c)))               (2.2.14)         
or                        (e^(i  (((ω (Subscript[z, 0]))/c)-n ((ω (Subscript[z, 0]))/c)))) =(Subscript[A, g]) (e^(i  ϕ ))             (2.2.15)        
Substitute                    ((ω )/c)≡k   in (2.2.15)
                            (e^(i  k (Subscript[z, 0]) (1-n))) =(Subscript[A, g]) (e^(i  ϕ ))             (2.2.16)

We have solve the unique solution of wave in the glass slab: from (2.2.6)
                         (Subscript[E, slab])=(Subscript[A, g]) (e^(i  ϕ ))  (e^(i  (Subscript[ω, g ])(n (z/c)- t ))) 
           = (e^(i  k (Subscript[z, 0]) (1-n)))  (e^(i  (Subscript[ω,  ])(n (z/c)- t )))              (2.2.17)

It can also be written:
                         (Subscript[E, slab]) = (e^(i  k (Subscript[z, 0]) (1-n)))  (e^(i (n k z -ω t )))=(e^(i  k (Subscript[z, 0]) )) (e^(i (n k (z-(Subscript[z, 0]))-ω t )))      (2.2.18)
Do you see what the significance of (2.2.18)? discuss.

Part 2: wave exiting the glass slab into air

To continue, we MUST calculate the wave solution that exits the glass slab back side into air.
We do exactly the same thing as above, but this time we won’t bother to repeat gory details.
The wave in air back side must have the form:
                                   (Subscript[E, exit])= (Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i (k z-ω t)))                      (2.2.19)
Given the slab thickness be d, the E field at exit plane (Subscript[z, 0])+d is:
                    (Subscript[E, exit])[(Subscript[z, 0])+d,t]= (Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i (k ((Subscript[z, 0])+d)-ω t)))    (2.2.20)
The slab E-field at the exit plane is, from (2.2.18)
                    (Subscript[E, slab])[(Subscript[z, 0])+d,t] =(e^(i  k (Subscript[z, 0]) )) (e^(i (n k (((Subscript[z, 0])+d)-(Subscript[z, 0]))-ω t ))) =(e^(i  k (Subscript[z, 0]) )) (e^(i (n k d-ω t )))     (2.2.21)
Again, Faraday’s law dictates the continuity:
                                   (e^(i  k (Subscript[z, 0]) )) (e^(i (n k d-ω t )))=(Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i (k ((Subscript[z, 0])+d)-ω t)))         (2.2.22)
or:                              (e^(i  k (Subscript[z, 0]) )) (e^(i n k d))=(Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i k ((Subscript[z, 0])+d)))         
or:                                        (e^(i n k d))=(Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i k d))                 
or:                                       (Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) = (e^(i k d (n-1)))                         (2.2.23)
Hence, the exit wave is (subsitute 2.2.23 into 2.2.19)
                           (Subscript[E, exit])= (Subscript[A, exit]) (e^(i (Subscript[ϕ, exit]))) (e^(i (k z-ω t)))=(e^(i k d (n-1))) (e^(i (k z-ω t)))      (2.2.24)
Compare this wave vs. the one in (2.2.1)
                                                         (Subscript[E, inc])[z,t]=(e^(i (k z-ω t)))    (2.2.1)
we see that the difference is the well-known result: the phase shift of a wave by a dielectric slab of thickness d and index n is
                                                            k d (n-1)

The problem is solved: the exit wave has a phase shift k d(n-1) that can be measured by using interferometry, which we will learn later. Notice that the phase shift is independent of (Subscript[z, 0]): the phase shift depends only on the intrinsic property of the slab and independent where you put it along the beam.

2.3 Answer 2.1 again.

Even if you answer the same, it is OK. But you get extra credit for saying if doing Q. 2.2 gives you new ideas how to answer it.