PHYS 110B - HW #5Fall 2005, Solutions by David PaceEquations referenced equations are from GriffithsProblem statements are paraphrased

[1.] Imagine a prism made of lucite (n = 1.5) whose cross-section is a quarter circle of radius a. Asshown below, one flat side of the prism rests on a table, while light is incident normal to the otherflat side. There is a region, between point P and point Q, on the table that is not illuminated by lightfrom the prism. Find the distance betwen O and Q (note: O is the lower left corner of the prism, youare not being asked for the distance PQ).

O P Q

Solution

We are concerned with the light rays that exit the prism. Be-gin by noting that since the original rays are incident per-pendicular to the left surface (i.e. normal incidence whereθI = 0), the transmission angle of the light into the prismwill also be θT = 0. This means that the light rays inside theprism will be parallel to the tabletop, as shown in the figureto the right. This says nothing about the amplitude of theserays inside the prism; for the moment we have only deter-mined their orientation within it.

To solve this problem we will need to know the transmission angle of the light rays as they exit theprism. This angle, θT , may be determined from Snell’s law,

sin θT

sin θI=

n1

n2Eq. 9.100 (1)

where we are given the indices of refraction (since it is not given explicitly, assume that n1 = 1because the prism is either surrounded by air or vacuum) and need to solve for θI before calculatingθT .

In order to determine θI we will need to figure out the geometry of the reflection/transmissionphenomena occurring as the rays exit the prism. See figure 1 for this explanation.

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Figure 1: An arbitrary light ray exiting the prism. Since this is a semi-circular prism, the normal to therounded surface is the radial vector originating at O. The length of this vector is the prism radius, a. Theangle of incidence is shown where the incident ray ( ~EI ) strikes the surface. Other angles also equal toθI are known from basic geometry. The angle of transmission, θT , is shown for the transmitted ray ~ET .Note that Snell’s law tells us θT > θI , which is represented in this figure.

Once we know the transmission angle that pro-duces the shortest OQ length, we can use thetriangle to the right to determine that length.This is possible because once we know thetransmission angle we can solve for the inci-dent angle (or the other way around, if we canever find the transmission angle it is probablybecause we have already determined the inci-dent angle). Once the incident angle is known,we can apply the law of cosines to the triangleat right and determine the length in question.

The smallest possible value of θT is zero. This occurs for the bottom most ray in the prism. Thebottom ray strikes the interface at normal incidence, meaning that the transmitted ray emerges atnormal incidence (θT = 0) and never touches the tabletop. From the figures and Snell’s law we knowthat the other rays emerge from the prism directed down toward the table. There is a maximumvalue of θT , however, as required by Snell’s law,

θT = sin−1

[n1

n2sin(θI)

](2)

2

but the argument of the sin−1 function must be less than or equal to one,

n1

n2sin(θI) ≤ 1 (3)

1.51

sin(θI) ≤ 1 (4)

sin(θI) ≤ 23

(5)

Using sin(θI) = 2/3 allows us to solve for the largest possible θT ,

sin(θT,max) =32

sin(θI,max) (6)

=32· 23

(7)

θT,max =π

2= 90o (8)

The maximum transmission angle is 90o. Such a limit makes physical sense because a 90o transmis-sion angle from a circular surface means the transmitted ray is tangent to the surface. A transmissionangle greater than this value would mean that the transmitted ray is still in the prism, which is notpossible. We now know that 0 ≤ θT ≤ π/2.

The first ray to strike the table will be the one with θT = θT,max. One conceptual argument for thisfact is: the bottom ray is known never to reach the table. The top most ray (at θT = θT,max) definitelyreaches the table. All of the rays with a transmission angle between these two extremes must strikethe table between them. As such, the top most ray is the one hitting the table at the point Q.

The ray featuring a transmission angle of 90o is tangent to the prism surface. That means we havea right triangle in which the length OQ is the hypotenuse. This is why the problem asks for thislength instead of PQ. One side of this triangle has length equal to a, and another side whose lengthL may be written as (using figure 1),

L = OQ sin(θI) (9)

=23OQ (10)

The length of the hypotenuse in this right triangle is calculated by,

OQ2 = a2 +

(23OQ

)2

(11)

(1− 4

9

)OQ

2 = a2 (12)

OQ =

√9a2

5(13)

=3a√

5(14)

≈ 1.34a (15)

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[2.] The polarization of an electromagnetic wave is determined according to the direction of its electricfield. Show that when a wave strikes an interface between two media, the reflected and transmittedwaves have the same polarization. Let the polarization vectors of these waves be nT = cos(θT ) x +sin(θT ) y and nR = cos(θR) x + sin(θR) y for the transmitted and reflected waves respectively.Consider the boundary conditions in this system to show that θT = θR.

Solution

Regardless of the possible changes in polarization, this is still a problem of reflection and transmis-sion. Consider the case of normal incidence because that will provide the easiest solution (if youwere not sure which type of incidence to use and asked about this problem, then Professor Carteror myself would have told you to use normal incidence).

The components of the electric and magnetic fields are everywhere parallel to the interface and weuse the following boundary conditions (the following are (iii) and (iv) from Eq. 9.74),

~E‖1 = ~E

‖2 (16)

1µ1

~B‖1 =

1µ2

~B‖2 (17)

As usual, set the direction of the incident wave propagation to be in the +z direction.

The electric fields for all of the waves may be written in the form,∼~E (z, t) =

∼Eo ei(kz−ωt)n (18)

where n is the polarization vector. Note that all of our other work in reflection/transmission utilizedn = x (e.g. Eqs. 9.75 - 9.77). The same method is employed here, now using x→ n.

The incident wave is,∼~EI (z, t) =

∼EoI ei(k1z−ωt)x (19)

∼~BI (z, t) =

1v1

∼EoI ei(k1z−ωt)y (20)

where I can set the polarization of this wave as I wish. The point of this exercise is to show that thereflected and transmitted rays have the same polarization, the incident wave’s polarization does notmatter.

The reflected wave is written,∼~ER (z, t) =

∼EoR ei(−k1z−ωt)nR (21)

=∼EoR ei(−k1z−ωt)(cos θR x + sin θR y) (22)

∼~BR (z, t) =

1v1

∼EoR ei(−k1z−ωt)(−z × nR) (23)

=1v1

∼EoR ei(−k1z−ωt)(sin θR x− cos θR y) (24)

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where the cross product is used because the magnetic field must be perpendicular to both the electricfield and the direction of propagation.

The transmitted wave is,∼~ET (z, t) =

∼EoT ei(k2z−ωt)nT (25)

=∼EoT ei(k2z−ωt)(cos θT x + sin θT y) (26)

∼~BT (z, t) =

1v2

∼EoT ei(k2z−ωt)(z × nT ) (27)

=1v2

∼EoT ei(k2z−ωt)(− sin θT x + cos θT y) (28)

Condition (15) provides the following (the interface is placed at z = 0 so the exponential dependencefactors out of all the terms),

∼EoI x+

∼EoR (cos θR x + sin θR y) =

∼EoT (cos θT x + sin θT y) (29)

∼EoR sin θR =

∼EoT sin θT (30)

where (29) is the y component of (28), the usefulness of which will be shown soon.

Condition (16) gives,

1µ1

(1v1

∼EoI y +

1v1

∼EoR (sin θR x− cos θR y)

)=

1µ2

(1v2

∼EoT (− sin θT x + cos θT y)

)This time keep the x component terms,

1µ1v1

∼EoR sin θR = − 1

µ2v2

∼EoT sin θT (31)

∼EoR sin θR = −β

∼EoT sin θT (32)

where β is given in (51).

One way to proceed from this point is to subtract (31) from (29). The result is,

0 = (1 + β) sin θT (33)

This is satisfied when either β = −1 or the sine term is zero. Using the definition of β, where thepropagation speeds are positive definite, this condition could be met if one of the permeabilities isnegative. Such novel materials are studied and have unique properties, but they are not a part ofthis course and you are not expected to apply the physics of such materials. We generally assumeall permeabilities are close to that of free space and certainly not negative.

Our solution is sin θT = 0. For transmission we limit the angle to 0 ≤ θT ≥ π/2 (if the anglewas greater than π/2 then the wave would be propagating back through the glass instead of beingtransmitted, just as in problem [1.]). This means that we have shown θT = 0. Now, going back toeither (29) or (31) and considering that the amplitude terms and β are non-zero we see that sin θR =sin θT = 0 and therefore θR = θT = 0. Replacing this in the equations for all of the original waveequations shows that they all have the same polarization.

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[3.] Consider the problem of using a submerged light source to illuminate the surface of a body of water.If the light is placed at a depth d = 1 m below the surface, determine the surface area of the lightseen on the surface. The relative permittivity of water at optical frequencies is εr = 1.77.

Solution

Treat the light as a point source. A point source will create a circular pattern on the surface of thewater because its light is emitted equally well in all directions. The emitted light is then going fromwater to air, meaning we have a reflection/transmission problem. Solve for the index of refractionof water,

n ∼=√

εr Eq. 9.70 (34)

=√

1.77 (35)

Figure 2 sets up the geometry of the problem and determines the method for solving.

Figure 2: For a point light source, all of the emitted light share the same origin. A few selectrays are shown here to illustrate the process that will determine the area of light seen on thewater’s surface. At the surface (solid horizontal line) we have a situation where n1 > n2,meaning that θT > θI . As we consider light rays that strike the surface further away fromthe source, we notice that θI increases and θT must increase as well. Once θT = 90o, there isno transmitted light. The position where this happens will determine the illuminated areaon the surface.

6

The radius, r, of the illuminated water surface area isgiven as part of the triangle shown to the left when θI = θc

is the critical angle (zero transmission). This critical angleis solved for using Snell’s law, (1), when θT = 90o.

sin θI =n2

n1sin θT (36)

=1√1.77

(37)

where I leave this in terms of sin θI because,

sin θI =r

a(38)

and the length of r is determined by,

a2 = d2 + r2 (39)

r2 =(

r2

sin2 θI

)− d2 (40)

r2

[1

sin2 θI− 1]

= d2 (41)

r2 =1

0.77(42)

(43)

employing d = 1.

The area of the illuminated surface, A, is,

A = πr2 (44)≈ 4.08 m2 (45)

[4.] Solve for the reflection (R) and transmission (T ) coefficients for normal incidence of light waves onthe interface between two materials with arbitrary permeabilities µ1 and µ2 (repeat what was donein lecture, but don’t assume µ = µo). Also show that R + T = 1.

Solution

Begin with the definition of the coefficients,

R =IR

IIEq. 9.86 (46)

T =IT

IIEq. 9.87 (47)

The intensities mentioned above are,

I =12εvE2

o (48)

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Define region 1 to contain the incident and reflected waves (the left side of the interface with theincident wave moving to the right) and region 2 to contain the transmitted wave. The reflectioncoefficient becomes,

R =12ε1v1E

2oR

12ε1v1E2

oI

(49)

=E2

oR

E2oI

(50)

Now consider the following general relation between the reflected and incident amplitudes,

∼EoR =

(1− β

1 + β

)∼EoI Eq. 9.82 (51)

β ≡ µ1v1

µ2v2(52)

If we divide the incident amplitude from both sides in (50) and then square the result, the complexfactors will become unity. Therefore,( ∼

EoR∼EoI

)2

=(

1− β

1 + β

)2

(53)

E2oR

E2oI

=(

1− β

1 + β

)2

(54)

Returning to (49) we have solved for the reflection coefficient,

R =(

1− β

1 + β

)2

(55)

Following the same method to solve for T ,

T =12ε2v2E

2oT

12ε1v1E2

oI

(56)

=ε2v2

ε1v1

(EoT

EoI

)2

(57)

The amplitude ratio is found from,

∼EoT =

(2

1 + β

)∼EoI (58)

( ∼EoT∼EoI

)2

=(

21 + β

)2

(59)

(EoT

EoI

)2

=(

21 + β

)2

(60)

8

Insert this result into (56).

T =ε2v2

ε1v1

(2

1 + β

)2

(61)

The velocities are related to ε and µ according to,

v =1√

εµEq. 9.68 (62)

Use this to simplify the expression for T .

T = β

(2

1 + β

)2

(63)

Making use of,µ1µ2

µ1µ2· ε2v2

ε1v1=

v21

v22

· µ1v2

µ2v1=

µ1v1

µ2v2= β (64)

Expressing everything in terms of β allows us to easily show R + T = 1.

R + T =(

1− β

1 + β

)2

+ β

(2

1 + β

)2

(65)

=(1− β)2 + 4β

(1 + β)2(66)

=1− 2β + β2 + 4β

1 + 2β + β2= 1 (67)

[5.] Problem 9.18 from Griffiths

(a) Let there be an amount of free charge in a piece of glass. Roughly how long will it take for thischarge to entirely migrate to the surface?

(b) How thick should you make the silver coatings to protect yourself from 1010 Hz microwaves?Bear in mind that silver is expensive so you want to balance the protection it affords with a minimalcost.

(c) What are the wavelength and propagation speed of 1 MHz radio waves in copper? Comparethis to the wavelength and propagation speed of these waves in air (which you may approximateas vacuum).

Solution

(a) The characteristic time, τ , for free charge to get to the surface of its conductor is,

ρf (t) = e−(σ/ε)tρf (0) Eq. 9.120 (68)

where τ ≡ ε/σ.

Table 7.1 in Griffiths gives the resistivity of glass, and the conductivity σ is the inverse of the resis-tivity. Therefore, σ ≈ 10−12 Ω/m.

9

The permittivity of glass is ε = εoεr, where εr is the relative permeability. Glass is non-magnetic sowe can write its index of refraction as,

n ∼=√

εr Eq. 9.70 (69)

The index of refraction of glass is given in Griffiths to be n = 1.5 (p. 391). The permittivity of glassis,

ε = εo(1.5)2 ≈ 2× 10−11 (70)

The characteristic time we seek,

τ ≈ 2× 10−11

10−12≈ 20 s (71)

(b) Since silver is expensive we want to make these coatings as thin as possible while still blockingthe microwaves. The skin depth for microwaves in silver is a measure of how far these wavescan penetrate into the metal. If the metal is thinner than the skin depth, then the waves will passcompletely through it. Find this distance and then make the coating slightly thicker.

Skin depth is given by d ≡ 1/κ where,

κ ≡ ω

√εµ

2

[√1 +

( σ

εω

)2− 1

]1/2

Eq. 9.126 (72)

ω is the angular frequency of the wave and σ is the conductivity of the material into which the waveis propagating.

Take the resistivity of silver from Table 7.1 in Griffiths to find that σ = 6.29 × 107. The angularfrequency is ω = 2πf = 2π × 1010. We can use ε ≈ εo = 8.85× 10−12.

ε ω = 0.56 (73)

σ

εω= 1.12× 108 ≫ 1 (74)

Due to (73) we can simplify κ as,

κ = ω

√εµ

2

[√( σ

εω

)2− 1

]1/2

(75)

= ω

√εµ

2

[ σ

εω

]1/2(76)

=√

ωµσ

2(77)

The permeability of conductors is approximately that of vacuum. Check Table 6.1 in Griffiths toverify this.

The skin depth is,

d =√

2ωµσ

= 6.4× 10−7 m (78)

so the coatings should be made a little thicker than this.

10

(c) Wavelength, λ, can be written in terms of wave vector as,

λ =2π

k(79)

In a conductor the wave vector is given by,

k ≡ ω

√εµ

2

[√1 +

( σ

εω

)2+ 1

]1/2

Eq. 9.126 (80)

As in part (b) we get the conductivity of copper using Table 7.1. Also, the permittivity may beapproximated as εo.

σ = 6× 107 (81)

εω = εo(2π × 106) = 5.56× 10−5 (82)

σ

εω= 1.08× 1012 ≫ 1 (83)

We can skip to the same simplifications made in part (a) due to (82).

k =√

ωµσ

2(84)

= 15390.60 (85)

λ = 4.08× 10−4 m (86)

The propagation speed is,

v =ω

kEq. 9.129 (87)

= 408.25ms

(88)

In vacuum the speed of an electromagnetic wave is always c = 3.0 × 108 m/s. The correspondingwavelength is,

λ =2π

k=

2πc

ω(89)

= 300 m (90)

[6.] Problem 9.21 from Griffiths

Solve for the reflection coefficient of light striking an air-silver interface. Use µ1 = µ2 = µo, ε1 = εo,σ = 6× 107, and ω = 4× 1015.

Solution

11

The reflection coefficient is defined according to (49). A relationship between the complex ampli-tudes is given in,

∼EoR =

1−∼β

1+∼β

∼EoI Eq. 9.147 (91)

where∼β is now a complex quantity.

∼β =

µ1v1

µ2ω

∼k2 Eq. 9.146 (92)

∼k2 = k + iκ Eq. 9.125 (93)

Arrange this into a useful form,

( ∼EoR∼EoI

)2

=

1−∼β

1+∼β

2

(94)

(EoR

EoI

)2

=

1−∼β

1+∼β

1−∼β∗

1+∼β∗

= R (95)

where the starred variable means complex conjugate.

From previous work in problem 9.18 we know that silver is a good conductor and can show that ithas the property k ≈ κ. Using (76) and (83) we can write,

k2 = κ2 =√

ωµσ

2(96)

∼β =

µ1v1

µ2ω

√ωµσ

2(1 + i) (97)

= µ1v1

√σ

2ωµ2(1 + i) (98)

Now make the numeric substitutions given in the problem to find the prefactor in (97), noting thatv1 = vair ≈ c.

µ1v1

√σ

2ωµ2= µoc

√6× 107

2(4× 1015)µo(99)

= 29 (100)

∼β = 29(1 + i) (101)

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The reflection coefficient is,

R =

1−∼β

1+∼β

1−∼β∗

1+∼β∗

(102)

=(

1− 29(1 + i)1 + 29(1 + i)

)(1− 29(1− i)1 + 29(1− i)

)(103)

=−57 + 292(2)59 + 292(2)

(104)

= 0.93 (105)

This shows that 93% of the light incident on the air-silver interface is reflected.

[7.] Consider light traveling from x = −∞ (in the x direction), incident normally on a plate of glass withthickness a. The plate if parallel to the yz plane, with one face at x = 0 and the other at x = a. Theglass has index of refraction n, and is surrounded by vacuum (n = 1). The incident wave has itselectric field in the y direction.

(a) Write expressions for the electric and magnetic fields in the three regions: x < 0, 0 < x < a, andx > a. Note: in the first region there will be an incident and reflected wave, in the second regionthere will be a transmitted wave and a reflected wave (reflected off of the back surface of the glass),and in the third region there will only be a transmitted wave.

(b) Using the appropriate boundary conditions, solve for the transmission coefficient, T (here youare interested in the energy transmitted into the vacuum region beyond the glass). Also solve forthe reflection coefficient R (this will be the energy reflected in the first region).

(c) Plot T and R as a function of ka, where k is the incident wave vector. Interpret your result phys-ically. In particular, comment on how R can equal zero in this case - isn’t there always a reflectedwave from the first interface?

Solution

(a) Write out all of the waves in a general manner. Beginning with the incident and reflected wavesin region 1, extending from x < 0,

∼~EI (x, t) =

∼EoI ei(k1x−ωt) y

∼~BI (x, t) =

1v1

∼EoI ei(k1x−ωt) z (106)

∼~ER (x, t) =

∼EoR ei(−k1x−ωt) y

∼~BR (x, t) = − 1

v1

∼EoR ei(−k1x−ωt) z (107)

In region 2, from 0 < x < a, I define a right-going (subscript r) and left-going wave (subscript l),

∼~Er (x, t) =

∼Eor ei(k2x−ωt) y

∼~Br (x, t) =

1v2

∼Eor ei(k2x−ωt) z (108)

∼~El (x, t) =

∼Eol ei(−k2x−ωt) y

∼~Bl (x, t) = − 1

v2

∼Eol ei(−k2x−ωt) z (109)

where the right-going wave is the transmitted wave from the x = 0 interface and the left-goingwave is the wave reflected from the x = a interface.

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In region 3, extending to x > a, there is only the final transmitted wave.

∼~ET (x, t) =

∼EoT ei(k1x−ωt) y

∼~BT (x, t) =

1v1

∼EoT ei(k1x−ωt) z (110)

where the k and v values in this region are the same as in region 1 because they are both vacuum.

This is a normal incidence problem so we apply the boundary conditions of (15) and (16). Thepermeability of glass is approximately equal to that of vacuum, so (16) becomes,

~B‖1 = ~B

‖2 at z = 0 (111)

~B‖2 = ~B

‖3 at z = a (112)

Applying the boundary conditions at z = 0 provides (the exponential dependence factors out as inother similar problems),

∼EoI +

∼EoR =

∼Eor +

∼Eol (113)

1v1

(∼EoI −

∼EoR

)=

1v2

(∼Eor −

∼Eol

)(114)

∼EoI −

∼EoR =

v1

v2

(∼Eor −

∼Eol

)(115)

where the reason for writing (114) (and (117) below) will become apparent soon.

At z = a only the angular frequency parts of the exponentials factor out when the boundary condi-tions are applied,

∼Eor eik2a+

∼Eol e−ik2a =

∼EoT eik1a (116)

1v2

(∼Eor eik2a−

∼Eol e−ik2a

)=

1v1

∼EoT eik1a (117)

∼Eor eik2a−

∼Eol e−ik2a =

v2

v1

∼EoT eik1a (118)

The expressions in (112), (114), (115), and (117) relate the five amplitudes in this problem. We treatthe incident wave amplitude as a known parameter (assuming that we have purposely shot thisincident wave at the glass). This set then consists of four equations with four unknowns.

Begin by rewriting (117),

v1

v2

[∼Eor eik2a−

∼Eol e−ik2a

]=

∼EoT eik1a (119)

Now let β = v1/v2 and then divide (118) by (115),

β(∼Eor eik2a−

∼Eol e−ik2a

)∼Eor eik2a+

∼Eol e−ik2a

= 1 (120)

which removes the transmitted wave dependence.

14

Performing further algebra allows∼Eol to be written directly in terms of

∼Eor,

∼Eol =

β − 1β + 1

ei2k2a∼Eor (121)

and to make the algebra easier to write out I will further define,

γ =β − 1β + 1

ei2k2a (122)

Place (120) into (112) and (114) to obtain,

∼EoI +

∼EoR =

∼Eor + γ

∼Eor = [1 + γ]

∼Eor (123)

∼EoI −

∼EoR = β

[∼Eor − γ

∼Eor

]= [1− γ] β

∼Eor (124)

Divide (122) by (123) and then solve for∼EoR,

∼EoI +

∼EoR

∼EoI −

∼EoR

=1 + γ

β(1− γ)(125)

∼EoR =

1 + γ + β(γ − 1)1 + γ − β(γ − 1)

∼EoI (126)

Place (125) into (122),

∼EoI +

1 + γ + β(γ − 1)1 + γ − β(γ − 1)

∼EoI = (1 + γ)

∼Eor (127)

∼Eor =

21 + γ − β(γ − 1)

∼EoI (128)

Now put (127) into (120),

∼Eol = γ

∼Eor (129)

= γ

[2

1 + γ − β(γ − 1)∼EoI

](130)

Solve for the remaining amplitude by placing (129) and (127) into (115),

∼EoT eik1a =

2∼Eor

1 + γ − β(γ − 1)eik2a + γ

[2∼EoI

1 + γ − β(γ − 1)

]e−ik2a (131)

∼EoT =

eik2a + γe−ik2a

eik1a

[2∼EoI

1 + γ − β(γ − 1)

](132)

With this amplitude we have solved for the fields in all of the regions.

(b) The reflection and transmission coefficients are defined in (45) and (46). Since the reflected andtransmitted rays were are considering here exist in vacuum, their intensities share the same ve-locities and permittivities. The solutions here are then given entirely by the ratios of the squareamplitudes.

15

For R,

R =[1 + γ + β(γ − 1)1 + γ − β(γ − 1)

]2

(133)

=

1 + β−1β+1ei2k2a + β

(β−1β+1ei2k2a − 1

)1 + β−1

β+1ei2k2a − β(

β−1β+1ei2k2a − 1

) ·1 + β−1

β+1e−i2k2a + β(

β−1β+1e−i2k2a − 1

)1 + β−1

β+1e−i2k2a − β(

β−1β+1e−i2k2a − 1

) (134)

where the complex conjugates have been taken for the squared complex terms.

Multiplying all of this out gives,

R =2(1− cos(2k2a))

(1 + β)4 + (1− β)4 − 2(1 + β)2(1− β)2 cos(2k2a)(135)

Rewrite k2 in terms of k1 using,

kI =n1

n2kT Eq. 9.92 (136)

k2 = nk1 (137)

and then R is,

R =2(1− cos(2ank1))

(1 + β)4 + (1− β)4 − 2(1 + β)2(1− β)2 cos(2ank1)(138)

The transmission coefficient is found in a similar way,

T =(

eik2a + γe−ik2a

eik1a

[2

1 + γ − β(γ − 1)

])2

(139)

=

(eik2a + β−1

β+1ei2k2ae−ik2a

eik1a

[2

1 + β−1β+1ei2k2a − β(β−1

β+1ei2k2a − 1)

])(140)

·

(e−ik2a + β−1

β+1e−i2k2aeik2a

e−ik1a

[2

1 + β−1β+1e−i2k2a − β(β−1

β+1e−i2k2a − 1)

])(141)

=

(1 + β−1

β+1

)eik2a

eik1a·

(1 + β−1

β+1

)e−ik2a

e−ik1a(142)

·

[2

1 + β−1β+1ei2k2a − β(β−1

β+1ei2k2a − 1)

][2

1 + β−1β+1e−i2k2a − β(β−1

β+1e−i2k2a − 1)

](143)

=16β2

(β + 1)2 [(1 + β)4 + (β − 1)4 − 2 cos(2k2a)(1 + β)2(β − 1)2](144)

=16β2

(β + 1)2 [(1 + β)4 + (β − 1)4 − 2 cos(2ank1)(1 + β)2(β − 1)2](145)

(c) We can rewrite β so that it is in terms of only the index of refraction of the glass,

β =v1

v2=

c

n1· n2

c= n2 = n (146)

16

where I have applied n1 = 1 for the vacuum region and n2 is the index of refraction of the glass.

The cases in which R = 0 are found when the numerator in (137) is equal to zero. This leads to,

2ank1 = 2mπ (147)

ak1 =mπ

n(148)

The plot below is for a case of n = 1.5, which is a commonly used value for glass. The coefficientsmay not always add to unity because of rounding errors in my approximation. This indicates thatpurposely making the glass a certain thickness will result in zero reflection. Furthermore, this thick-ness, a, is related to the wavelength of the light inside the glass. This relation will be discussedfurther when we consider dipole radiation.

0 0.25π 0.5π 0.75π π 1.25π 1.5π

0.05

0.1

0.15

0.2

T

R

17