PHYx485. Solutions to Assignment 1

October 13, 2017, 07:51

Preliminaries

Constants used in this document:

h = 6.62× 10−34J · s = 4.14× 10−15eV · s (1)

c = 3.0× 108 m/s (2)

1 eV = 1.602× 10−19J (3)

Z0 = 377 Ω (4)

kB = 1.380× 10−23 JK−1 (5)

a = 5.879× 1010Hz

K( in Wein’s Law) (6)

b = 2 900 µm ·K ( in Wein’s Law) (7)

ε0 = 8.885× 10−12 C

Vm(8)

e = 1.6× 10−19C (9)

me = 9.11× 10−31 kg (10)

Problem 1

a) P = 100 W , ` = 1 m

I =100 W

4π(1 m)2= 7.96

W

m2(11)

Erms =|E0|√

2=√Z0I = 55

V

m(12)

b) P = 100 MW , λ = 694.3 nm, ϕ = 10µm

I =100× 106W

π (10µm/2)2= 1.27× 1018 W

m2 (13)

|E0| =√

2Z0I = 3.1× 1010 Vm (14)

c1) (For the light bulb in part a) ) Assume λmax =

693 × 10−9m as the wavelength at which Iλ(λ) has its

maximum. First, use Wein’s Law to find the temperature,

then find the irradiance Iλ(λ):

T =b

λmax= 4200 K (15)

Iλ(λ) =2πhc2

λ51

ehc/(λkBT ) − 1= 1.6× 1013 W

m3 (16)

To find the irradiance on a 1nm bandwidth, we do

I1nm =

∫1nm

Iλ(λ)dλ ≈ Iλ(λ) ∆λ

= (1.6× 1013 Wm3 )(1nm) = 1.6× 104 W

m2 (17)

Finally, we get the photon flux, the photon density, and

the number of photons per cubic wavelength, i.e.,

Φ =I1nmEphoton

= 5.7× 1023 m−2s−1, (18)

n = Φ/c = 1.9× 1014 m−3, (19)

photons

λ3= nλ3 = 6.3× 10−5, (20)

respectively.

CLARIFICATION: Do not confuse Iν(ν) with Iλ(λ);

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PHYx485F. Assignment Solutions

these are related throughout

Iλ(λ) = Iν(ν)ν

λ(21)[

Wm3

]=[

WHz m2 · Hz

m

](22)

If you mistakenly assumed that λmax = 693 × 10−9m as

the wavelength at which Iν(ν) has its maximum, then

you might have obtained the following. Taking, νmax =

c/λmax, Wein’s Law you would give T = 7388 K. Then

the irradiance

Iν(ν) =2πhν3

c21

ehc/(λkBT ) − 1(23)

= 2.4× 10−7 WHz m2 (24)

(compare Eq.16 and Eq.23) Then the irradiance per 1nm

bandwidth would be

Iνmax(ν) ≈ Iνmax

(ν) ∆ν (25)

∆ν =c

λmax− c

∆λ+ λmax= 6.23× 1011Hz (26)

Iνmax(ν) ≈ 1.5× 105 Wm2 (27)

then

Φ =I1nmEphoton

= 5.2× 1023 m−2s−1, (28)

n = Φ/c = 1.7× 1015 m−3, (29)

photons

λ3= nλ3 = 5.7× 10−4, (30)

Problem 2

part a) [Problem 3.1] To find the λ and ν at which

ρλ(λ) and ρν(ν) have their respective maximum for a given

temperature, we just use Wein’s Law. The idea is that

you would derive Wein’s Law both in wavelength and in

frequency versions. You can find the step-by-step details

in the Wikipedia.

λmax =b

T→ 10µm (31)

νmax = a T → 18 THz (32)

where a and b are constants given in the first page of this

document; T is measured in Kelvins.

part b) [Problem 3.2] Take the Lorentz force expres-

sion, and split it into its electric and magnetic parts

F = qE + q v ×B (33)

Fe = qE, Fm = q v ×B (34)

Then consider a planewave electric field propagating along

the z-direction

E = E0ekz−ωtx, B = B0e

kz−ωty;B0 = Eo/c (35)

Then find the ratio of the two forces,

FeFm∼ c

|v|(36)

Problem 3

Given λ = 589nm, and 2.7× 1018atoms/m3 at T= 200 C,

find absorption.

Follow your textbook, Milonni–Eberly, pages 107 and 121.

Specifically, see Equations (3.9.11) and (3.13.8)–(3.13.10).

You will get δD ∼ 1.5 GHz, σ ∼ 6 × 10−16 m2 and

a ∼ 1.6× 103m−1

Problem 4

The dielectric function in the presence of free charges is

ε(ω) = 1−ω2p

ω2 + iωγ, ω2

p =Ne2

ε0me(37)

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PHYx485F. Assignment Solutions

where ωp is the plasma frequency, N the volumetric con-

centration of free charges of charge e and mass me; γ is the

collision rate (billard like) of the free charges. Assuming

that this collision rate of free charges in the ionosphere is

much less than 100 MHz (due to low concentration of the

free charges and other particles in general), we have

n2 = ε ≈ 1−ω2p

ω2→ ω2

p = ω2(1− n2) (38)

N = 4π2 ε0m

e2(1− n2)f2 (39)

where ω = 2πf . Taking n = 0.9 and f = 108 Hz, we get

N ∼ 3.36× 106 electrons/cm3.

Positive charges contribute a term ∼ Ω2p/ω

2 to the

dielectric function, where Ωp = Ne2/(ε0M). Because

M >> me, this contribution is much smaller than that

coming from electrons.

Problem 5

The absorption

a(ν) =Ne2

4πε0mc

δν0(ν − ν0)2 + (δν0)2

; δν0 =ν

4π(40)

at ν = ν0, consider the oscillator strength that is the frac-

tion of oscillators responce to frequency ν0

a(ν) =Ne2

4πε0mc

f

δν0= 1400 m−1 (41)

from the line shape, and 2δν0 = 1014s−1 → δν0 = 5 ×

1013s−1, with N ≈ 1023 /m3 from Eq. 40, f = 0.1331 Hz.

The lifetime is then

τ =1

2πδν0= 3.18× 10−15s (42)

The real part

n = nb +Ne2

4πε0ω0

ω0 − ω(ω − ω0)2 + δ2/4

; nb = 1.55 (43)

with

ω∓ = ω0 ∓ δ/2 (44)

where ω− = ωmax and ω+ = ωmin, hence dn/dω = 0

Then

Vφ =c

n(ω); Vg =

dk

dω; (45)

Vφ − VgVφ

=ω

n(ω)

dn(ω)

dω(46)

so

dn(ω)

dω=

Ne2

4πε0ω0

[2(ω − ω0)2

[(ω − ω0)2 + δ/4]2− 1

(ω − ω0)2 + δ2/4

](47)

with ω0 = 2πν0 and γ = 4πδ0,

Vφ − VgVφ

=Ne2

4πε0ω0

[2(ω − ω0)2

[(ω − ω0)2 + δ/4]2− 1

(ω − ω0)2 + δ2/4

](48)

For ν = 7× 1014 Hz, ν0 = 6.8× 1014 Hz,∣∣∣∣Vφ − VgVφ

∣∣∣∣ = 0.035% (49)

first calculate n(ω)|ν = 1 × 1014 Hz, then substitute into

Eq. 48.

For ν = 9× 1014 Hz,∣∣∣∣Vφ − VgVφ

∣∣∣∣ = 0.003% (50)

Problem 6

part a) Start from

1

c

∂Iν∂t

+∂Iν∂z

= hν∂N2

∂t+ hνhν

∂N1

∂t(51)

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PHYx485F. Assignment Solutions

Consider constant brightness. Then get

hν∂N2

∂t=hν

cIνS(ν)B12(N1 − g1/g2 ·N2) (52)

hν∂N1

∂t= hν/(4π)A21N2S(ν) (53)

combining the previous two equations you can get the

equation of radiative transfer.

part b), c), d) You need to solve the equation of ra-

diative transfer for Iν(z) for different cases.

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