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[The naive explanation for the pressure of light offered in section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here's another account, due originally to Planck.] A plane wave travelling through vaccum in the z direction encounters a perfect conductor occupying the region z0, and reflects back:

E(z,t)=E0[coskz-ωt-coskz+ωt]x^,(z>0)

  1. Find the accompanying magnetic field (in the region (z>0))
  2. Assuming B=0inside the conductor find the current K on the surface z=0, by invoking the appropriate boundary condition.
  3. Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq.9.64).

Short Answer

Expert verified
  1. The magnetic field isB=E0ccoskz-ωt+kz+ωty^.
  2. The current K on the surface isK=2E0μ0ccosωtx^
  3. The magnetic force per unit area isf=ε0E0 and it is twice the pressure in Eq. 9.64.

Step by step solution

01

Expression for the electric field for (z>0):

Write the expression for the electric field for z>0.

E(z,t)=E0[coskz-ωt+coskz+ωt]x^

Here k is the wave number, ωis the angular frequency and t is the time.

02

Determine the accompanying magnetic field

(a)

Since, E×Bpoints in the direction of propagation, write the equation for the magnetic field.

B=E0ccoskz-ωt+coskz+ωty^

Therefore, the magnetic field isB=E0ccoskz-ωt+coskz+ωty^.

03

Determine the current K on the surface z=0 :

(b)

It is known that K×-z^=1μ0B.

Substitute B=E0ccoskz-ωt+coskz+ωty^in the above equation.

K×-z^=1μ0E0ccoskz-ωt+coskz+ωty^=E0μ0c2cosωty^K=2E0μ0ccosωty^

Therefore, the current K on the surface isK=2E0μ0ccosωty^.

04

Determine the magnetic force per unit area on the surface:

(c)

Write the expression for the force per unit area

f=K×Bavg

Substitute K=2E0μ0ccosωtx^expression and Bavg=cosωty^in the above expression.

f=2E02μ02ccosωtx^×cosωty^f=2ε0E02cos2ωtz^

It is known that the time average of cos2ωtis 12. Hence, the above eqution becomes,

f=2ε0E0212z^f=ε0E02

This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector.

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Most popular questions from this chapter

(a) Show directly that Eqs. 9.197 satisfy Maxwell’s equations (Eq. 9.177) and the boundary conditions (Eq. 9.175).

(b) Find the charge density, λ(z,t), and the current, I(z,t), on the inner conductor.

Question:Equation 9.36 describes the most general linearly polarized wave on a string. Linear (or "plane") polarization (so called because the displacement is parallel to a fixed vector n) results from the combination of horizontally and vertically polarized waves of the same phase (Eq. 9.39). If the two components are of equal amplitude, but out of phase by (say,δν=0,δh=90°,), the result is a circularly polarized wave. In that case:

(a) At a fixed point, show that the string moves in a circle about the axis. Does it go clockwise or counter clockwise, as you look down the axis toward the origin? How would you construct a wave circling the other way? (In optics, the clockwise case is called right circular polarization, and the counter clockwise, left circular polarization.)

(b) Sketch the string at time t =0.

(c) How would you shake the string in order to produce a circularly polarized wave?

Light from an aquarium (Fig. 9.27) goes from water (n=43)through a plane of glass (n=32)into the air n=1. Assuming it’s a monochromatic plane wave and that it strikes the glass at normal incidence, find the minimum and maximum transmission coefficients (Eq. 9.199). You can see the fish clearly; how well can it see you?

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Consider the resonant cavity produced by closing off the two ends of a rectangular wave guide, at z=0and at z=d, making a perfectly conducting empty box. Show that the resonant frequencies for both TE and TM modes are given by

role="math" localid="1657446745988" ωlmn=cπ(ld)2+(ma)2+(nb)2(9.204)

For integers l, m, and n. Find the associated electric and magnetic fields

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