ECE 5/6358 HW 2B (100 pts)
First name, Last name

Student ID:

Han Q Le(c)

Part B

3. (50 pts) Two-plane-wave interference

Download this app, use only the part of two plane waves with interference.

The light intensity pattern on wall x (the x plane),

is proportional to the time averaged of ((|E|)^2) we discussed in lecture:
                    ((I=|⟨(Overscript[S, →])⟩|=|(c/(8 π))Re[ (Overscript[E, ->]) ×(Overscript[(SuperStar[H]), ->])]|
=(c/(8 π))(Sqrt[(ϵ/μ)])| (Overscript[E, ->]) |)^2)     
 and is called the interference patterm.

3.1  (20 pts) Interference

Does the period of the interference pattern depend mainly on the light wavelength (in vacuum as well as in a medium, of which you can change the dielectric constant ϵ), or frequency? Since wavelength and frequency are correlated:   λ f=v=(c/n), you might say “both”, but think carefully, pick ONLY one that is the main factor, what would that be?

hint

It’s the wavelength. The proof is the follow. The E field of each beam (polarized in the z direction):
          (Subscript[E, plus]) = (Subscript[a, P]) (e^(i ((Subscript[k, x])x+(Subscript[k, y])y-ω  t)))   ; (Subscript[E, minus]) = (Subscript[a, M]) (e^(i (-(Subscript[k, x])x+(Subscript[k, y])y-ω  t)))
Both beams:         (Subscript[E, total])=(Subscript[E, plus])+(Subscript[E, minus]) =((Subscript[a, P]) (e^(i (Subscript[k, x])x))  +(Subscript[a, M]) (e^(-i (Subscript[k, x])x)) ) (e^(i ((Subscript[k, y])y-ω  t)))  
The wall perpendicular with x is for any arbitary y=(Subscript[y, 0]). 
                  (Subscript[E, total])[y=(Subscript[y, 0])]=((Subscript[a, P]) (e^(i (Subscript[k, x])x))  +(Subscript[a, M]) (e^(-i (Subscript[k, x])x)) ) (e^(i ((Subscript[k, y])(Subscript[y, 0])-ω  t)))  
The intensity is proportional to ((| (Overscript[E, ->]) |)^2)
Does the frequency matter:?  ((|(e^(i ((Subscript[k, y])(Subscript[y, 0])-ω  t)))|)^2) = 1 (in Re[ (Overscript[E, ->]) ×(Overscript[(SuperStar[H]), ->])], the conjugate pair (e^(-i ω t)) and (e^(i ω t)) becomes = 1), i. e. no time-dependence: the intensity pattern is static, it doesn’t change with time. This is the reason the answer “both wavelength and frequency” is wrong. Only wavelength matters.

Do you know how the app plots the interference pattern? It plots this:
                      ((|((Subscript[a, P]) (e^(i (Subscript[k, x])x))  +(Subscript[a, M]) (e^(-i (Subscript[k, x])x)) )|)^2)

First, let’s change (Subscript[k, x]) to q to avoid the subscript:

Expand[((Subscript[a, P]) (E^(I q x))  +(Subscript[a, M]) (E^(-I q x)))((Subscript[a, P]) (E^(-I q x))  +(Subscript[a, M]) (E^(I q x))) ]

((Subscript[a, M]^2)+(Subscript[a, P]^2)) (1+(((Subscript[a, M]) (Subscript[a, P]))/((Subscript[a, M]^2)+(Subscript[a, P]^2))) ExpToTrig[(E^(2 I q x))+(E^(-2 I q x))])

Define:

1+ Cos[2 q x]FullSimplify[((2 (Subscript[a, M]) (Subscript[a, P]))/((Subscript[a, M]^2)+(Subscript[a, P]^2)))  /. (Subscript[a, M])-> (1/2)(1-ρ) /.  (Subscript[a, P])->(1/2) (1+ρ)]

Or:             1+((1-(ρ^2))/(1+(ρ^2))) Cos[2 q x]

Note that the interference contrast is largest when ρ=0: 2 beams have equal amplitude.

If we let (Subscript[k, x])=q=((2π n)/λ) Cos[θ] , then:
            interference pattern = 1+((1-(ρ^2))/(1+(ρ^2))) Cos[2 π ((2 n)/λ) Cos[θ] x]
and the period of the pattern is: Λ= (λ/(2 n Cos[θ])): it’s wavelength, nothing to do with frequency

Where do we find applications?

What’s the point of all this: it is good to have an intuitive understanding, or gut feeling, but real experimental work requires rigorous calculation.

3.2  (15 pts) Opposite traveling wave pattern

Observe the E field oscillating pattern on the x-wall. discuss its behavior when the two beams have equal amplitude, and what happens when one beam is stronger than the other?

3.3  (15 pts) Phase

Does the relative phase of the two beams matter?

4. (50 pts) 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.

4.1 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.

4.2 What do the waves (in air and in glass) look like?

Calculate and 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)))    (4.2.1)
where                            k=(ω/c)= ((2 π )/λ)  (here λ=0.5 μm)   (4.2.2)
We will rewrite in a more convenient form to use later:
                                        (Subscript[E, inc])[z,t]=(e^(i ω ((z/c)- t)))    (4.2.3)

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 )))    (4.2.4)
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)              (4.2.5)
Hence:                                     (Subscript[E, slab])=(Subscript[A, g])  (e^(i  ϕ )) (e^(i  (Subscript[ω, g ])(n (z/c)- t )))    (4.2.6)

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 (4.2.3) and (4.2.6)
             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]      (4.2.7)
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))              (4.2.8)

Since  (e^(i  ((ω (Subscript[z, 0]))/c)))  is constant of time, Eq. (4.2.8) would be false if 
                                                 ω-(Subscript[ω, g])≠ 0                              (4.2.9)
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                         (4.2.10)
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                     (4.2.11)
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         (4.2.12)
and (4.2.9) must be false, and
                                            ω-(Subscript[ω, g]) =0                       (4.2.13)

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

Apply Eq. (4.2.13) to (4.2.8)
                           (e^(i  ((ω (Subscript[z, 0]))/c))) =(Subscript[A, g])(e^(i  ϕ )) (e^(i  n ((ω (Subscript[z, 0]))/c)))               (4.2.14)         
or                        (e^(i  (((ω (Subscript[z, 0]))/c)-n ((ω (Subscript[z, 0]))/c)))) =(Subscript[A, g]) (e^(i  ϕ ))             (4.2.15)        
Substitute                    ((ω )/c)≡k   in (4.2.15)
                            (e^(i  k (Subscript[z, 0]) (1-n))) =(Subscript[A, g]) (e^(i  ϕ ))             (4.2.16)

We have solve the unique solution of wave in the glass slab: from (4.2.6)
                         (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 )))              (4.2.17)

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 )))      (4.2.18)
Do you see what the significance of (4.2.18)? 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)))                      (4.2.19)
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)))    (4.2.20)
The slab E-field at the exit plane is, from (4.2.18)
                    (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 )))     (4.2.21)
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)))         (4.2.22)
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)))                         (4.2.23)
Hence, the exit wave is (subsitute 4.2.23 into 4.2.19)
                           (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)))      (4.2.24)
Compare this wave vs. the one in (4.2.1)
                                                         (Subscript[E, inc])[z,t]=(e^(i (k z-ω t)))    (4.2.1)
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 was discussed in Prob. 3. 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.