Reflection/transmission of waves at interfaces

[dyn]

Hi.
I am using r and t to denote the amplitude reflection and transmission coefficients and R and T to denote the power reflection and transmission coefficients.
1. If r = -1 this is total reflection which leads to standing waves. But if the reflected wave has an equal but opposite amplitude why don't the incident and reflected wave cancel at all points leading to zero displacement everywhere ?
2. If r = 1 this is again total reflection but what happens in this case to the overall wave ?
3. When r = 1 , t = 2 but zero power is transmitted ie. T = 0. How is this explained ?

Thanks

 

[Charles Link]

For question 1) ## E_{total}(x,t)=E_o \cos(kx-\omega t)-E_o \cos(kx+\omega t +\phi) ##, where ## \phi=0 ## if the reflection occurs at ## x=0 ##. Work with the trigonometric identity ## \cos{A}-\cos{B}=-
2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2}) ##, and you will see you get a standing wave. ## \\ ## For 2) you can use the trigonometric identity ## \cos{A}+\cos{B}=2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2}) ##, and you will see you also get a standing wave for this case. ## \\ ## For 3) For large ## n_1 ## and small ## n_2 ##, yes ## \rho=(n_1-n_2)/(n_1+n_2) \approx 1 ##, and ## \tau=2 n_1/(n_1+n_2) ## is approximately 2. However, intensity ## I=n E^2 ##, and for large ## n_1 ##, ## E_1 ## will be small for a finite and moderate intensity ## I_1 ##. The result is ## E_2 \approx 2 E_1 ##, making ## E_2 ## somewhat small. Meanwhile intensity ## I_2=n_2 E_2^2 ## will be small because of the small ## n_2 ##. In the limit ## n_1>>n_2 ##, ## R=(n_1-n_2)^2/(n_1+n_2)^2 \approx 1 ##, and ## T=1-R ## will be nearly zero.

 

[tech99]

Hi.
I am using r and t to denote the amplitude reflection and transmission coefficients and R and T to denote the power reflection and transmission coefficients.
1. If r = -1 this is total reflection which leads to standing waves. But if the reflected wave has an equal but opposite amplitude why don't the incident and reflected wave cancel at all points leading to zero displacement everywhere ?
2. If r = 1 this is again total reflection but what happens in this case to the overall wave ?
3. When r = 1 , t = 2 but zero power is transmitted ie. T = 0. How is this explained ?

Thanks

Q1. Notice that, for normal incidence, at a distance of a quarter wavelength from the reflector, the reflected wave has traveled half a wavelength, and together with the 180 deg phase shift on reflection, has a total phase shift of 360 degrees, so it is aiding. The waves traveling in opposite directions can exist separately. The magnetic field is not zero at the reflector and stores energy.
Q2. For any phase of reflection, the forward and reflected waves retain their own existence and can travel over each other. We just add them if we are interested in the field strength at a given position.
If we have a standing wave (with zero traveling wave component), then energy is not traveling but is remaining stored in the system. So no power is being transmitted.

 

[Charles Link]

Q1. Notice that, for normal incidence, at a distance of a quarter wavelength from the reflector, the reflected wave has traveled half a wavelength, and together with the 180 deg phase shift on reflection, has a total phase shift of 360 degrees, so it is aiding. The waves traveling in opposite directions can exist separately. The magnetic field is not zero at the reflector and stores energy.
Q2. For any phase of reflection, the forward and reflected waves retain their own existence and can travel over each other. We just add them if we are interested in the field strength at a given position.
If we have a standing wave (with zero traveling wave component), then energy is not traveling but is remaining stored in the system. So no power is being transmitted.

A standing wave can also occur when the same amount of energy that you are feeding in gets reflected back to you. With a standing wave, there will be locations where nodes occur, so that if you probe the electric field there, (e.g. r-f waves traveling on a transmission line that get reflected back over the same path=there are experimental methods that allow you to insert a probe= a slotted line is used), you will find very minimal electric field at those locations. The electric field is sinusoidal in time everywhere in a standing wave, but at the nodes the amplitude is zero.

 

[dyn]

Thanks. I should have mentioned that at the moment I'm just concerned with mechanical waves eg. waves on a string.
If I have the 1st harmonic on a string and I consider the crest at 1/4 wavelength. At that point the reflected wave would have traveled 3/2 wavelengths which equates to a phase change of 3π/2 ? And also a phase change of π at reflection gives a total phase change of 5π/2 or 3π/2 ? Either way this wouldn't be in phase with the incident wave so wouldn't add constructively.
Where am I going wrong ?

 

[Charles Link]

Thanks. I should have mentioned that at the moment I'm just concerned with mechanical waves eg. waves on a string.
If I have the 1st harmonic on a string and I consider the crest at 1/4 wavelength. At that point the reflected wave would have traveled 3/2 wavelengths which equates to a phase change of 3π/2 ? And also a phase change of π at reflection gives a total phase change of 5π/2 or 3π/2 ? Either way this wouldn't be in phase with the incident wave so wouldn't add constructively.
Where am I going wrong ?

Use the trigonometric identities to get your answer. A wave traveling to the right is written as ## y_R(x,t)=A_R \cos(kx-\omega t +\phi_R) ## and one traveling to the left has ## y_L(x,t)=A_L \cos(kx+\omega t +\phi_L ) ##. ## \\ ## ## y(x,t)_{total}=y_R(x,t)+y_L(x,t) ##. ## \\ ## And a ## \pi ## phase change is the same as a minus sign because ## \cos(\theta+\pi)=-\cos(\theta) ##. With ## \rho=(n_1-n_2)/(n_1+n_2) ## , when ## n_2>n_1 ##, ## \rho ## gets a minus sign, or , as it is often stated, there is a ## \pi ## phase change for an optical reflection off of a more dense medium. ## \\ ## To write this out in detail, for the case where ## n_2>n_1 ##, the reflected wave is ## \\ ## ##y(x,t)=\rho E_o \cos(kx+\omega t)= +|\rho| E_o \cos(kx+\omega t+\pi) =-|\rho| E_o \cos(kx+\omega t) ##. You can work it either way.

 

[dyn]

Thanks. I'm just trying to picture how it works and it still seems like 2 equal but opposite amplitudes adding together to give zero amplitude

 

[Charles Link]

Thanks. I'm just trying to picture how it works and it still seems like 2 equal but opposite amplitudes adding together to give zero amplitude

Both waves=the right going and the left going, are sinusoids=each sinusoid goes both positive and negative. You can think of each as a fixed snapshot of wiggles that travels. One moves to the right and the other to the left. At any location at any instant you add the amount of displacement from each. ## \\ ## Editing this last part: Given ## y(x,0)=y_o(x)=A \cos(\frac{2 \pi}{\lambda} x) ##, which is a snapshot (graph) of the wave for time ## t=0 ##, ## \\ ## we have ## y(x,t)=y_o(x-vt)=A \cos(\frac{2 \pi}{\lambda} (x-vt) ) ##. This is the mathematical way of saying the graph ## y_o(x) ## vs. ## x ## is translated to the right at time ## t ## by the amount ## vt ##. ## \\ ## Alternatively, the sinusoidal signal in time can be given at location ## x=0 ## as ## y(0,t)=g(t)=A \cos(\omega t) ##, where ## \omega=2 \pi f ##. Then, if it travels to the right, ## y(x,t)=g(t-\frac{x}{v} )=A \cos(\omega (t-\frac{x}{v})) ##. ## \\ ## In both cases the result is ## y(x,t)=A \cos(kx-\omega t ) ##, where ## k=\frac{2 \pi}{\lambda} ## and ## \omega=2 \pi f ##. Notice velocity ## v=\lambda f=\frac{\omega}{k} ##.

 

[sophiecentaur]

1. If r = -1 this is total reflection which leads to standing waves.

Any reflection due to a change in impedance can generate a standing wave. The SWR (Standing Wave Ratio) of the pattern is only unity if there is no reflection (a perfect match). In practice, there reflection coefficient will always be within 1 to -1.

Thanks. I'm just trying to picture how it works and it still seems like 2 equal but opposite amplitudes adding together to give zero amplitude

The phases of the two waves will only be equal and opposite at the nodes. Elsewhere the resultant will be non-zero and it will follow the familiar standing wave pattern.

 

[dyn]

So r = -1 generates a standing wave because of the phase change of π at the fixed ends.
r = 1 has no phase change at the ends so shouldn't this lead to destructive interference at all points ?

 

[Charles Link]

So r = -1 generates a standing wave because of the phase change of π at the fixed ends.
r = 1 has no phase change at the ends so shouldn't this lead to destructive interference at all points ?

Because it is two sinusoidal waves of equal amplitudes traveling in opposite directions, you will get a standing wave for ## r=1 ##. Use the trigonometric identity ## cosA+cosB=... ## to analyze it. You won't get a node at the wall when ## r=1 ## like you do for ## r=-1 ##, but you still will get positions where there are nodes with the same spacing between nodes for ## r=1 ## as for ## r=-1 ##.

 

[dyn]

Thanks. And for the case r = 1 ; t = 2 but T = 0 so is there a transmitted wave or not ?

 

[Charles Link]

Thanks. And for the case r = 1 ; t = 2 but T = 0 so is there a transmitted wave or not ?

I explained that in post 2, part 3. There will in general be a transmitted wave=there necessarily must with ## t \approx 2 ##, but (at least for the optical case where ## n_2<<n_1 ##), there is very little energy transmitted. ## T ## will be quite small.

 

[dyn]

thanks for your help

 

[sophiecentaur]

r = 1 has no phase change at the ends so shouldn't this lead to destructive interference at all points ?

The relative phases are different all the way along each wave. Nodes will form where there were antinodes - and so on.