Suppose you send an incident wave of specified shape, g1(z-v1t), down string number 1. It gives rise to a reflected wave, hR(z+v1t), and a transmitted wave, gT(z+v2t). By imposing the boundary conditions 9.26 and 9.27, find hRand gT.

Short Answer

Expert verified

The shape of the reflected wave (hR)is hRu=v2-v1v2+v1g1(-v1t)+k', and the shape of the transmitted wave (gT)is gTu=2v2v1+v2gIv1v2u+k'.

Step by step solution

01

Expression for the transmitted wave and impose boundary conditions:

Write the expression for the transmitted wave.

gt(z-v2t)=g1(z-v1t)+hR(z-v1t) …. (1)

Here, g1(z-v1t)is the shape of the incident wave and is the shape of the reflected wave.

Consider first equation for the boundary condition.

f(0-1,t)=f(0+,t) ….. (1)

Consider the first equation for the boundary condition.

fz0-=fz0+ ….. (2)

02

Determine the shape of the transmitted wave:

Apply the boundary condition from equation (2).

gT(0-v2t)=g1(0-v1t)+hR(0+v1t)gT(-v2t)=g1(-v1t)+hR(v1t)......(4)gT(0-v2t)=g1(0-v1t)+hR(0+v1t)gT(-v2t)=g1(-v1t)+hR(v1t)......(4)

Apply the boundary condition from equation (4).

-1v1gT(-v1t)t+1v1hR(v1t)t=-1v2gT(-v2t)t-g1(-v1t)t+hR(v1t)t=v1v2gT(-v2t)tg1(-v1t)t-hR(v1t)t=v1v2gT(-v2t)t

Integrate on both sides,

g1(-v1t)-hR(v,t)=v1v2gT(-v2t)+k …… (5)

Add equations (4) and (5).

gT-v2t+v1v2gT(-v2t)+k=g1(-v1t)+hR(v1t)+gT(-v1t)-hR(-v1t)gT-v2t1+v1v2+k=gl(-v1t)+gl(-v1t)2gl(-v1t)=gT(-v2t)[v2+v1v2]+kgT(-v2t)=(2v2v2+v1)gl(-v1t)+kl

Here, k1=-k(v2v1+v2).

Let(z-v1t)=(z-v2t)=(z-v1t)=u

Substitute the values in equation (1).

gTu=2v2v1+v2glv1v2u+k1

03

Determine the shape of the reflected wave:

Multiply equation (4) with

v1v2.

role="math" localid="1657700056261" v1v2gT-v2t=v1v2gl-v1t=v1v2hTv1t

Subtract equation (5) from equation (6).

v1v2gT-v2t+k-v1v2gT-v2t=gT-v1t-hRv1t-v1v2gl-v1t-v1v2hRv1tgl(-v1T)-v1v2gl-v1t-hR(-v1T)-v1v2hR-v1t=k1-v1v2gl-v1t-1+v1v2hR-v1t=khRu=v2-v1v1+v2gl-v1t+kl

Therefore, the shape of the reflected wave hRis hRu=v2-v1v1+v2gl-v1t+kl, and the shape of the transmitted wave gTis gTu=2v2v1+v2glv1v2u+kl.

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Most popular questions from this chapter

Consider a rectangular wave guide with dimensions 2.28cm×1.01cm. What TE modes will propagate in this waveguide if the driving frequency is 1.70×1010Hz? Suppose you wanted to excite only one TE mode; what range of frequencies could you use? What are the corresponding wavelengths (in open space)?

Show that the mode TE00 cannot occur in a rectangular wave guide. [Hint: In this case role="math" localid="1657512848808" ωc=k, so Eqs. 9.180 are indeterminate, and you must go back to Eq. 9.179. Show thatrole="math" localid="1657512928835" Bz is a constant, and hence—applying Faraday’s law in integral form to a cross section—thatrole="math" localid="1657513040288" Bz=0 , so this would be a TEM mode.]

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

If you take the model in Ex. 4.1 at face value, what natural frequency do you get? Put in the actual numbers. Where, in the electromagnetic spectrum, does this lie, assuming the radius of the atom is 0.5 Å? Find the coefficients of refraction and dispersion, and compare them with the measured values for hydrogen at 0°Cand atmospheric pressure:A=1.36×10-4,B=7.7×10-15m2 .

(a) Formulate an appropriate boundary condition, to replace Eq. 9.27, for the case of two strings under tension T joined by a knot of mass m.

(b) Find the amplitude and phase of the reflected and transmitted waves for the case where the knot has a mass m and the second string is massless.

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