# 978-0123869449 Chapter 2

Document Type

Homework Help

Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

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CHAPTER 2

2.1 Show that for an electromagnetic wave traveling through a dielectric (m1=n1), impinging on the interface

with another, optically less dense dielectric (n2<n1), light of any polarization is totally reﬂected for incidence

angles larger than θc=sin−1(n2/n1).

Hint: Use equations (2.105) with k2=0.

Solution

w′

tw′′

tcos θ2=0.

The last of these relations dictates that either w′′

t=0 or cos θ2=0 (w′

t=0 is not possible since—from the first

relation—this would imply η0n1sin θ1=0 which is known not to be true).

w′′

t=0:Substituting this into the second relation leads to w′

t=η0n2, and the first leads to n1sin θ1=n2sin θ2.

Since n1>n2this is a legitimate solution only for sin θ1≤(n2/n1), or θ1≤θc=sin−1(n2/n1).

e

rk=in2

1w′′

t+n2

2w′

icos θ1

,e

r⊥=w′

icos θ1+iw′′

t

ρk=e

rke

r∗

k=ρ⊥=e

r⊥e

r∗

⊥=1.

19

20 RADIATIVE HEAT TRANSFER

2.2 Derive equations (2.109) using the same approach as in the development of equations (2.89) through (2.92).

Hint: Remember that within the absorbing medium, w=w′−iw′′ =w′ˆ

s−iw′′ ˆ

n; this implies that E0is not a

vector normal to ˆ

s. It is best to assume E0=Ekˆ

ek+E⊥ˆ

e⊥+Esˆ

s.

i=0) and reﬂected wave

(w′′

r=0,w′

r=−w′

i). The contribution from Esvanishes for incident and reﬂected wave. Using the same

vector relations as given for the nonabsorbing media interface, one obtains

w′

i(Ei⊥−Er⊥) cos θ1=(w′

tcos θ2−iw′′

t)Et⊥(2.2-B)

Similarly, from equation (2.77),

E0×ˆn =(Ekˆek+E⊥ˆe⊥+Esˆs)×ˆn =−Ekˆe⊥cos θ+E⊥ˆ

t−Esˆe⊥sin θ

and

(Eik+Erk) cos θ1=Etkcos θ2+Ets sin θ2(2.2-C)

or

(Eik+Erk) cos θ1=Etkcos θ2−iw′′

tsin2θ2

w′

t−iw′′

tcos θ2=Etk

w′

tcos θ2−iw′′

t

w′

t−iw′′

tcos θ2

.

=η2

0m2

2cos θ1(1 +e

rk),

e

rk=w′

i(w′

tcos θ2−iw′′

t)−η2

0m2

2cos θ1

w′

,

22 RADIATIVE HEAT TRANSFER

2.3 Find the normal spectral reﬂectivity at the interface between two absorbing media. [Hint: Use an approach

similar to the one that led to equations (2.89) and (2.90), keeping in mind that all wave vectors will be complex,

but that the wave will be homogeneous in both media, i.e., all components of the wave vectors are colinear

with the surface normal].

2.4 A circularly polarized wave in air is incident upon a smooth dielectric surface (n=1.5) with a direction of

45◦oﬀnormal. What are the normalized Stokes’ parameters before and after the reﬂection, and what are the

degrees of polarization?

Solution

From the definition of Stokes’ parameters the incident wave has degrees of polarization

the sign giving the handedness of the circular polarization. With Erk=Eikrkand Er⊥=Ei⊥r⊥, from equa-

tions (2.50) through (2.53):

Since EikE∗

ik=Ei⊥E∗

i⊥[from equation (2.51)] and ρ=r2.

Similarly,

Qr

Ir

=ρk−ρ⊥

ρk+ρ⊥

,Vr

Ir

=2rkr⊥

ρk+ρ⊥

Vi

Ii

.

From Snell’s law

sin θ2=sin θ1

; cos θ2=s1−sin2θ1

=r1−0.5

n1cos θ2+n2cos θ1

Vr

24 RADIATIVE HEAT TRANSFER

2.5 A circularly polarized wave in air traveling along the z-axis is incident upon a dielectric surface (n=1.5).

How must the dielectric-air interface be oriented so that the reﬂected wave is a linearly polarized wave in the

y-z-plane?

2.6 A polished platinum surface is coated with a 1 µm thick layer of MgO.

(a) Determine the material’s reﬂectivity in the vicinity of λ=2µm (for platinum at 2 µmmPt =5.29 −6.71 i,

for MgO mMgO =1.65 −0.0001 i).

(b) Estimate the thickness of MgO required to reduce the average reﬂectivity in the vicinity of 2 µm to 0.4.

What happens to the interference eﬀects for this case?

R=0.6149.

(b) The cos in the numerator ﬂuctuates between −1<cos <+1. The average value for Ris obtained by

dropping the cos-term. Then

r2

23(1 −r2

12)=0.4−0.24532

0.79082(1 −0.24532)=0.5782,

d=−1

ln τ=−1

ln τ2=−ln 0.5782

0.79082[1 −1.6×0.24532]=0.6013

26 RADIATIVE HEAT TRANSFER

and

d=−ln 0.6013

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