Phase Change and Reflection of Electromagnetic Waves

[Jahnavi]

Homework Statement

Note : There are no minus signs in first two options .

Homework Equations

The Attempt at a Solution

The wave is propagating in +z direction whereas the electric field is varying in x direction . On reflection , there is a phase change of π . Also wave starts traveling in -z direction .As a consequence (kz - ωt) becomes (kz + ωt) .

So reflected wave is E = - E₀icos (kz + ωt)

But this is wrong .

 

[Charles Link]

Working the problem in its entirety, $\\$ $E_{incident}(a,t)=E_o \hat{i} \cos(ka-\omega t)$ .$\\$ $E_{reflected}(a,t)=-E_{incident}(a,t)$, and $\\$ $E_{reflected}(z,t)=E_{reflected} (a,t-\frac{|z-a|}{c} )$, (in the region $z<a$ ) . $\\$ I think this gives $E_{reflected}(z,t)=-E_o \hat{i} \cos(2ka-\omega t-kz)$ once you work through the algebra. $\\$ Note: It's normally against the PF rules to supply a complete solution to a homework question, but this one is really rather subtle, and it would be difficult to lead the OP to the solution, without actually showing the solution. $\\$ Anyway, that's the solution that I computed.

 

[rude man]

I believe the OP's solution is correct.

 

[Charles Link]

I believe the OP's solution is correct.

@rude man Please study my solution in detail, and I think you will see where the OP misses what is basically a $-2ka$ term that needs to be included in the OP's solution. $\\$ Editing: I specialized in Optics (Diffraction and Interference theory, etc.) and Spectroscopy as a graduate student in college, so that it would be somewhat embarrassing if I didn't have it correct. :)

 

[Jahnavi]

The correct answer given is E = E₀icos (kz + ωt) i.e option B) .

There is no minus sign .

Charles , this question is from chapter "EM wave" of 12th grade High school book . Do you think the question setter is using some simplified assumptions so that the factor of -2ka is missing ?

Even then , I am getting a minus sign whereas answer given has no minus sign .

 

[Charles Link]

The correct answer given is E = E₀icos (kz + ωt) i.e option B) .

There is no minus sign .

Charles , this question is from chapter "EM wave" of 12th grade High school book . Do you think the question setter is using some simplified assumptions so that the factor of -2ka is missing ?

Even then , I am getting a minus sign whereas answer given has no minus sign .

IMO, the author of the book probably has limited expertise in Optics and Interference and Diffraction theory. It isn't real surprising that he might have an incorrect answer for such a problem. Unless a physics person actually specializes in this kind of thing, it would be quite easy for them to get an incorrect result for the solution. If they thought the problem was identical to a water wave bouncing off a wall, then $E_{reflected}(a,t)=E_{incident}(a,t)$. In addition, it would be easy for someone who doesn't specialize in this type of thing to miss the $-2 k a$ phase term that results when the barrier is not at $z=0$. I give them credit for what I think was an honest mistake or two in their solution. $\\$ As for the minus sign out in front, there are two ways to explain it: There is an electromagnetic boundary condition that $E_{air \, total}(a,t)=E_{incident}(a,t)+E_{reflected}(a,t)=E_{transmitted}(a,t) =0$, so that $E_{reflected}(a,t)=-E_{incident}(a,t)$. For an alternate derivation, the Fresnel reflection coefficient $\rho=(n1-n2)/(n1+n2)$. For a metal, $n2$ has a very large complex term to it,(and $n1=1$) , so that $\rho=-1$ (or very nearly so), where $\rho=\frac{E_{reflected}(a,t)}{E_{incident}(a,t)}$.

 

[Delta2]

I also think option B is correct (without the minus sign). You saying that on reflection there is a phase change of pi. I just disagree with that, there is no phase change with reflection.

 

[Abhishek kumar]

The correct answer given is E = E₀icos (kz + ωt) i.e option B) .

There is no minus sign .

Charles , this question is from chapter "EM wave" of 12th grade High school book . Do you think the question setter is using some simplified assumptions so that the factor of -2ka is missing ?

Even then , I am getting a minus sign whereas answer given has no minus sign .

You have not included phase shift pi in your calculation

 

[Charles Link]

The minus sign is equivalent to a phase change of $\pi$: $\cos(\omega t+\pi)=-\cos(\omega t)$.

 

[Abhishek kumar]

The minus sign is equivalent to a phase change of $\pi$: $\cos(\omega t+\pi)=-\cos(\omega t)$.

Minus is here for change in direction of electric field after reflection

 

[Charles Link]

I could add one additional input: If they were an r-f (radio frequency) engineer, they might call a perfectly reflecting wall one that had infinite impedance. In that case, the reflected wave does not experience this $\pi$ phase change (i.e. there is no minus sign), but they still missed the $-2ka$ phase term. $\\$ Note: Basically the index of refraction $n$ for an optics person gets replaced by $1/Z$ in these formulas for an r-f engineer, where $Z$ is the characteristic impedance of the medium. A material (the wall) with very high $Z$ has a reflection without any phase change or inversion of the amplitude. $\\$ In any case, I took the statement of the problem to be that the wall consisted of a highly reflective mirror. The terminology "optically inactive" is somewhat unclear. $\\$ Editing: Additional note on the above: I googled this just to verify, and the google confirmed that, in general, you won't encounter an impedance $Z$ that is greater than the $Z$ of free space, so that except for exceptional cases, you will find any time you have a high reflectivity of a transverse electromagnetic wave off of a reflector in free space, you will get the minus sign or $\pi$ phase change. Alternatively, for a reflection of a transverse r-f wave in a coaxial cable that occurs off of a termination that is of a high impedance, you will get a reflection without the minus sign or phase change.

 

[Delta2]

Ok Charles and Kumar (and Rudeman) you might be right regarding the phase change of $\pi$ but I think in high school physics, reflection (of any wave, whether its mechanical or EM) is being taught as just changing the direction of propagation of wave, not the phase of the wave.

 

[Charles Link]

Ok Charles and Kumar (and Rudeman) you might be right regarding the phase change of $\pi$ but I think in high school physics, reflection (of any wave, whether its mechanical or EM) is being taught as just changing the direction of propagation of wave, not the phase of the wave.

The direction of the wave is determined by the relative sign between the $kz$ term and the $\omega t$ term. If they are different, the wave travels in the positive z direction (to the right), e.g. ($\cos(kz-\omega t)$). If they are the same, the wave travels to the left (e.g. $\cos( kz+\omega t)$).

 

[Jahnavi]

Ok Charles and Kumar (and Rudeman) you might be right regarding the phase change of $\pi$ but I think in high school physics, reflection (of any wave, whether its mechanical or EM) is being taught as just changing the direction of propagation of wave, not the phase of the wave.

Change of phase is not an issue in this problem

There is definitely a change of phase of π since the reflection is from an optically inactive surface . I am assuming this to be optically denser medium . Right @Charles Link ?

And since the wave is traveling in opposite direction , (kz-wt) changes to (kz+wt) . Taking both of these in consideration , I am getting an extra minus sign which the given answer doesn't have .

 

[rude man]

@rude man Please study my solution in detail, and I think you will see where the OP misses what is basically a $-2ka$ term that needs to be included in the OP's solution. $\\$ Editing: I specialized in Optics (Diffraction and Interference theory, etc.) and Spectroscopy as a graduate student in college, so that it would be somewhat embarrassing if I didn't have it correct. :)

Agreed! I missed the part about reflection at z=a .

The only possibility was E_r = -E₀cos(ωt + kz) but this doesn't work either because cos(ωt - ka) - cos(wt + ka) ≠ 0 for every ka (wavelength) as required.

The given choices seem to have been meant for a=0.

 

[Charles Link]

Change of phase is not an issue in this problem

There is definitely a change of phase of π since the reflection is from an optically inactive surface . I am assuming this to be optically denser medium . Right @Charles Link .

And since the wave is traveling in opposite direction , (kz-wt) changes to (kz+wt) . Taking both of these in consideration , I am getting an extra minus sign which answer doesn't have .

Posts 6 and 11 are the best explanation I can offer for what I think is definitely an incorrect answer that the author provides. :)

 

[Jahnavi]

What is the difference between my answer E = - E₀icos (kz + ωt) and the book answer E = E₀icos (kz + ωt) ?

What difference does a minus sign make ?

 

[Charles Link]

What is the difference between my answer E = - E₀icos (kz + ωt) and book answer E = E₀icos (kz + ωt) ?

What difference does a minus sign make ?

The minus sign will essentially change the phase of the reflected wave by $\pi$ ($=180^o$), because $-\cos(kz+\omega t)=\cos(kz+\omega t + \pi)$ . (Use $\cos(\theta+\phi)=\cos(\theta) \cos(\phi)-\sin(\theta) \sin(\phi)$ identity to show this). The minus sign is correct, if the reflective material is assumed to be of a higher index $n$.

 

[Jahnavi]

Ok.

I am attaching the solution given in the book . In case you feel there is a printing error in the options then you will be in for a surprise .

Please see if you can understand it .

The confusing part is changing of i to -i in the reflected wave .

 

[Charles Link]

Ok.

I am attaching the solution given in the book . In case you feel there is a printing error in the options then you will be in for a surprise .

Please see if you can understand it .

The confusing part is changing of i to -i in the reflected wave .

It is apparent that the author of the book is not an expert on this topic. It's always very helpful when the book teaches it correctly=this kind of inaccuracies can have a lot of students spinning their wheels unnecessarily. At least the Physics Forums is available that can most often supply the correct result when such a thing is encountered. :)

 

[Jahnavi]

Charles ,

Since there is already confusion , I would like to clarify what does i denote ? I think it just specifies the direction in which electric field is oscillating ( x-axis in this case ) . Right ?

Does -i make any sense ?

 

[Abhishek kumar]

Ok.

I am attaching the solution given in the book . In case you feel there is a printing error in the options then you will be in for a surprise .

Please see if you can understand it .

The confusing part is changing of i to -i in the reflected wave .

Same thing i have mentioned in post 8 and 10.direction of electric and magnetic field will also reverse.

 

[rude man]

What is the difference between my answer E = - E₀icos (kz + ωt) and the book answer E = E₀icos (kz + ωt) ?

What difference does a minus sign make ?

The reflected wave has a - sign because when you ADD Ei and Er the - sign is needed. Superposition is addition so to get cancellation at z=a you need the - sign, otherwise you'd get double instead of zero at z=a.

 

[Jahnavi]

The reflected wave has a - sign because when you ADD Ei and Er the - sign is needed. Superposition is addition so to get cancellation at z=a you need the - sign, otherwise you'd get double instead of zero at z=a.

But presence of minus sign in reflected wave doesn't mean complete cancellation.

E₀cos (kz - ωt) - E₀cos (kz + ωt) ≠ 0

 

[Abhishek kumar]

But presence of minus sign in reflected wave doesn't mean complete cancellation.

E₀cos (kz - ωt) - E₀cos (kz + ωt) ≠ 0

You have to consider at the instant of reflection.it will cancel out

 

[Jahnavi]

You have to consider at the instant of reflection.it will cancel out

At z = a , E₀cos (ka- ωt) - E₀cos (ka + ωt) = 2E₀sin(ka)sin(ωt) ≠ 0

Can you show how they cancel ?

 

[Abhishek kumar]

At z = a , E₀cos (ka- ωt) - E₀cos (ka + ωt) = 2E₀sin(ka)sin(ωt) ≠ 0

Can you show how they cancel ?

You are only considering position not time.when incident wave traveled a unit reflected wave just started

 

[Jahnavi]

You are only considering position not time.when incident wave traveled a unit reflected wave just started

Isn't "t" same in the two wave expressions ?

Aren't we considering the same instant ?

 

[ehild]

Isn't "t" same in the two wave expressions ?

Aren't we considering the same instant ?

Yes, the incident wave and reflected wave add up to zero at the interface at any instant. t is the same in both waves.

 

[Jahnavi]

Thanks !

Thank you @Charles Link .