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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 n2 this is a legitimate solution only for sin ฮธ1 โค (n2 /n1 ), or ฮธ1 โค ฮธc = sinโ1 (n2 /n1 ).
cos ฮธ2 = 0 (ฮธ2 = ฯ/2): Substituting the first relation into the second gives
w00
t = ฮท0
q
n21 sin2 ฮธ1 โ n22 ,
i.e., a legitimate nonzero solution for n21 sin2 ฮธ1 โn22 โฅ 0 or ฮธ1 โฅ ฮธc . Inspection of the reflection coefficients,
equations (2.109), shows that
e
rk =
2 0
in21 w00
t + n2 wi cos ฮธ1
2 0
in21 w00
t โ n2 wi cos ฮธ1
,
e
rโฅ =
w0i cos ฮธ1 + iw00
t
w0i cos ฮธ1 โ iw00
t
Since, in both reflection coefficients, there are no sign changes within the real and imaginary parts, it
follows readily that
ฯk = e
rke
rโk = ฯโฅ = e
rโฅe
rโโฅ = 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 = w0 โ iw00 = w0 sฬ โ iw00 nฬ; this implies that E0 is not a
vector normal to sฬ. It is best to assume E0 = Ek eฬk + Eโฅ eฬโฅ + Es sฬ.
Solution
0
00
Inside the absorbing medium wt = w0t โ iw00
t = wt sฬt โ iwt nฬ, and the electric field vector does not lie in a plane
normal to sฬ. Thus, we assume a general three-dimensional representation, or
E0 = Ek eฬk + Eโฅ eฬโฅ + Es sฬ.
Following the development for nonabsorbing media, equations (2.77) through (2.88), then leads to
ฮฝยตH0 = w ร E0 = (w0 sฬ โ iw00 nฬ) ร (Ek eฬk + Eโฅ eฬโฅ + Es sฬ).
= 0) and reflected wave
This formulation is valid for the transmitted wave, but also for the incident wave (w00
i
0
0
(w00
r = 0, wr = โwi ). The contribution from Es vanishes for incident and reflected wave. Using the same
vector relations as given for the nonabsorbing media interface, one obtains
ฮฝยตH0 = w0 (Ek eฬโฅ โ Eโฅ eฬk ) โ iw00 (Ek eฬโฅ cos ฮธ โ Eโฅ tฬ + Es eฬโฅ sin ฮธ).
For the interface condition (with ฮฝยต the same everywhere)
ฮฝยตH0 ร nฬ = w0 (Ek tฬ + Eโฅ eฬโฅ cos ฮธ) โ iw00 (Ek cos ฮธtฬ + Eโฅ eฬโฅ + Es sin ฮธtฬ).
Thus, from equation (2.78)
w0i (Eik tฬ + Eiโฅ eฬโฅ cos ฮธ1 ) โ w0i (Erk tฬ + Erโฅ eฬโฅ cos ฮธ1 )
= w0t (Etk tฬ + Etโฅ eฬโฅ cos ฮธ2 ) โ iw00
t (Etk cos ฮธ2 tฬ + Etโฅ eฬโฅ + Ets sin ฮธ2 tฬ)
or
00
w0i (Eik โ Erk ) = (w0t โ iw00
t cos ฮธ2 ) Etk โ iwt sin ฮธ2 Ets
w0i (Eiโฅ โ Erโฅ ) cos ฮธ1
=
(w0t cos ฮธ2 โ iw00
t ) Etโฅ
(2.2-A)
(2.2-B)
Similarly, from equation (2.77),
E0 ร nฬ = (Ek eฬk + Eโฅ eฬโฅ + Es sฬ) ร nฬ = โEk eฬโฅ cos ฮธ + Eโฅ tฬ โ Es eฬโฅ sin ฮธ
and
(Eik + Erk ) cos ฮธ1 = Etk cos ฮธ2 + Ets sin ฮธ2
Eiโฅ + Erโฅ = Etโฅ
(2.2-C)
(2.2-D)
These four equations have 5 unknowns (Erk , Ers , Etk , Etโฅ , and Ets ), and an additional condition is needed,
e.g., equation (2.23) or equation (2.64). Choosing equation (2.23) we obtain, inside the absorbing medium,
w ยท E0 = 0 = (w0t sฬ โ iw00
t nฬ) ยท (Etk eฬtk + Etโฅ eฬโฅ + Ets sฬ)
= w0t Ets + iw00
t (Etk sin ฮธ2 โ Ets cos ฮธ2 ).
Eliminating Etโฅ from equations (2.2-B) and (2.2-D), with e
rโฅ = Erโฅ /Eiโฅ , gives
rโฅ ),
w0i (1 โ e
rโฅ ) cos ฮธ1 = (w0t cos ฮธ2 โ iw00
t )(1 + e
or
e
rโฅ =
which is identical to equation (2.109).
w0i cos ฮธ1 โ (w0t cos ฮธ2 โ iw00
t )
w0i cos ฮธ1 + (w0t cos ฮธ2 โ iw00
t )
,
(2.2-E)
CHAPTER 2
21
Now, eliminating Ets from equations (2.2-A) and (2.2-C) [multiplying equation (2.2-C) by iw00
t and adding]:
0
w0i (Eik โ Erk ) + iw00
t cos ฮธ1 (Eik + Erk ) = wt Etk .
(2.2-F)
Eliminating Ets from equations (2.2-C) and (2.2-E) leads to
Ets =
โiw00
t Etk sin ฮธ2
0
wt โ iw00
t cos ฮธ2
๏ฃฎ
๏ฃน
2
iw00
w0 cos ฮธ2 โ iw00
๏ฃฏ๏ฃฏ
๏ฃบ๏ฃบ
t sin ฮธ2
t
๏ฃฏ
๏ฃบ๏ฃบ = Etk t
(Eik + Erk ) cos ฮธ1 = Etk ๏ฃฏ๏ฃฐcos ฮธ2 โ 0
.
w โ iw00 cos ฮธ ๏ฃป
w0 โ iw00 cos ฮธ
t
t
2
t
t
2
Using this to eliminate Etk from equation (2.2-F), with e
rk = Erk /Eik , gives
w0i (1 โ e
rk ) + iw00
rk ) = w0t cos ฮธ1 (1 + e
rk )
t cos ฮธ1 (1 + e
w0t โ iw00
t cos ฮธ2
w0t cos ฮธ2 โ iw00
t
w0i (w0t cos ฮธ2 โ iw00
rk )
t )(1 โ e
0 0
00
0
00
= wt (wt โ iwt cos ฮธ2 ) โ iw00
rk )
t (wt cos ฮธ2 โ iwt ) cos ฮธ1 (1 + e
= ฮท20 m22 cos ฮธ1 (1 + e
rk ),
e
rk =
2 2
w0i (w0t cos ฮธ2 โ iw00
t ) โ ฮท0 m2 cos ฮธ1
2 2
w0i (w0t cos ฮธ2 โ iw00
t ) + ฮท0 m2 cos ฮธ1
,
which is the same as equation (2.109).
It is a simple matter to show that other conditions give the same result. For example, from equation (2.64)
n21 (Eik eฬik ยท nฬ + Erk eฬrk ยท nฬ) = m22 (Etk eฬtk ยท nฬ + Ets sฬ ยท nฬ)
or
n21 (Eik โ Erk ) sin ฮธ1 = m22 (Etk sin ฮธ2 โ Ets cos ฮธ2 ), etc.
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].
Solution
Equations (2.19) and (2.20) remain valid for incident, reflected and transmitted waves, with w = w0 โ iw00 =
(w0 โiw00 ) nฬ for all three cases. From equation (2.31) wยทw = (w0 โiw00 )2 nฬยทnฬ = ฮท20 m2 it follows that w0 โiw00 = ยฑฮท0 m.
Thus
w0i โ iw00
i = ฮท0 m1 ,
w0r โ iw00
r = โฮท0 m1
(reflected wave is moving in a direction of โ nฬ),
w0t โ iw00
t = ฮท0 m2 .
From equations (2.23) and (2.24), it follows that the electric and magnetic field vectors are normal to nฬ, i.e.,
tangential to the surface, say E0 = E0 tฬ. Then, from equation (2.77)
(Ei + Er ) tฬ ร nฬ = Et tฬ ร nฬ,
or
Ei + Er = Et
From equation (2.25) ฮฝยตH0 = w ร E0 = (w0 โ iw00 ) E nฬ ร tฬ, and from equation (2.78)
n1 (Ei โ Er ) = m2 Et .
Substituting for Et and dividing by Ei , with e
r = Er /Ei :
m1 (1 โ e
r) = m2 (1 + e
r)
or
e
r =
m1 โ m2
m1 + m2
and
ฯn = e
re
rโ =
(m1 โ m2 )(m1 โ m2 )โ
(n1 โ n2 ) + i(k1 โ k2 ) 2
=
(m1 + m2 )(m1 + m2 )โ
(n1 + n2 ) + i(k1 + k2 )
ฯn =
(n1 โ n2 )2 + (k1 โ k2 )2
(n1 + n2 )2 + (k1 + k2 )2
CHAPTER 2
23
2.4 A circularly polarized wave in air is incident upon a smooth dielectric surface (n = 1.5) with a direction of
45โฆ off normal. What are the normalized Stokesโ parameters before and after the reflection, and what are the
degrees of polarization?
Solution
From the definition of Stokesโ parameters the incident wave has degrees of polarization
Qi
Ui
=
= 0,
Ii
Ii
Vi
= ยฑ1,
Ii
the sign giving the handedness of the circular polarization. With Erk = Eik rk and Erโฅ = Eiโฅ rโฅ , from equations (2.50) through (2.53):
Ir = Eik Eโik r2k + Eiโฅ Eโiโฅ r2โฅ = Eik Eโik (ฯk + ฯโฅ ) =
1
(ฯk + ฯโฅ ) Ii
2
Since Eik Eโik = Eiโฅ Eโiโฅ [from equation (2.51)] and ฯ = r2 .
Similarly,
Qr = Eik Eโik r2k โ Eiโฅ Eโiโฅ r2โฅ = Eik Eโik (ฯk โ ฯโฅ )
Ur = Eik Eโiโฅ rk rโฅ + Eiโฅ Eโik rโฅ rk = Ui rk rโฅ = 0
Vr = i(Eik Eโiโฅ โ Eiโฅ Eโik ) rk rโฅ = Vi rk rโฅ
ฯk โ ฯโฅ
Qr
=
,
Ir
ฯk + ฯโฅ
2rk rโฅ Vi
Vr
=
.
Ir
ฯk + ฯโฅ Ii
From Snellโs law
sin ฮธ1
; cos ฮธ2 =
sin ฮธ2 =
n2
s
sin2 ฮธ1
1โ
=
n22
r
0.5
=
1โ
1.52
r
7
,
9
and from equations (2.89) and (2.90)
โ
โ
n1 cos ฮธ2 โ n2 cos ฮธ1
7/9 โ 1.5 1/2
rk =
= โ0.0920, ฯk = 0.0085
= โ
โ
n1 cos ฮธ2 + n2 cos ฮธ1
7/9 + 1.5 1/2
โ
โ
n1 cos ฮธ1 โ n2 cos ฮธ2
1/2 โ 1.5 7/9
rโฅ =
= โ
= โ0.3033, ฯโฅ = 0.0920
โ
n1 cos ฮธ1 + n2 cos ฮธ2
1/2 + 1.5 7/9
Qr
0.0085 โ 0.0920
Ur
=
= โ0.8315,
=0
Ir
0.0085 + 0.0920
Ir
Vr
2ร0.0920ร0.0085
=ยฑ
= ยฑ0.5556
Ir
0.0085 + 0.0920
Since the perpendicular polarization is much more strongly reflected, the resulting wave is no longer circularly
polarized, but to a large degree linearly polarized (in the perpendicular direction).
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?
Solution
From equations (2.50) through (2.53) it follows that
Qr /Ir = 1, Ur = Vr = 0 (i.e., linear polarization), if
either Erk or Erโฅ vanish. From Fig. 2-9 it follows that
rโฅ , 0 and, therefore Erโฅ , 0 for all incidence directions, while rk = 0 for ฮธ = ฮธp (Brewsterโs angle),
or
n2
ฮธp = tanโ1
= tanโ1 1.5 = 56.31โฆ .
n1
The resulting wave is purely perpendicularpolarized, i.e., eฬโฅ must lie in the y โ z plane, or
eฬk must be in the xโz plane. Therefore, the surface
may be expressed in terms of its surface normal as
.โ
nฬ = ฤฑฬ sin ฮธp โ kฬ cos ฮธp = (ฤฑฬ โ 1.5kฬ) 3.25.
x
Ei
z
n
p
p
Er
CHAPTER 2
25
2.6 A polished platinum surface is coated with a 1 ยตm thick layer of MgO.
(a) Determine the materialโs reflectivity in the vicinity of ฮป = 2 ยตm (for platinum at 2 ยตm mPt = 5.29 โ 6.71 i,
for MgO mMgO = 1.65 โ 0.0001 i).
(b) Estimate the thickness of MgO required to reduce the average reflectivity in the vicinity of 2 ยตm to 0.4.
What happens to the interference effects for this case?
Solution
(a) The desired overall reflectivity must be calculated from equation (2.124) after determining the relevant
reflection coefficient. From equation (2.122)
e
r12 =
1 โ m2
1 โ n2
1 โ 1.65
= โ0.2453
‘
=
1 + m2
1 + n2
1 + 1.65
since k2 1, and r12 = 0.2453. e
r23 may also be calculated from equation (2.122) or, more conveniently, from
equation (2.126):
r223 =
(1.65 โ 5.29)2 + 6.712
= 0.6253 or r23 = 0.7908.
(1.65 + 5.29)2 + 6.712
Since the real part of e
r12 0 (numerator) and <(r23 ) < 0 (denominator) ฮด23 lies in the second quadrant, ฯ/2 < ฮด23 < ฯ,
or ฮด23 = 2.8364. Also ฮถ12 = 4ฯ ร 1.65 ร 1 ยตm/2 ยตm = 10.3673, and
cos [ฮด12 ยฑ (ฮด23 โ ฮถ12 )] = cos [ฯ ยฑ (2.8364 โ 10.3673)] = โ0.3175.
Also ฮบ2 d = 4ฯ ร 10โ4 ร 1 ยตm/2 ยตm = 2ฯ ร 10โ4 and ฯ = eโฮบ2 d = 0.9994 ' 1. Thus
R=
0.24532 + 2ร0.2453ร0.7908ร(โ0.3175) + 0.79082
1 + 2ร0.2453ร0.7908ร(โ0.3175) + 0.24532 ร0.79082
R = 0.6149.
(b) The cos in the numerator fluctuates between โ1 < cos < +1. The average value for R is obtained by
dropping the cos-term. Then
Rav =
or
ฯ2 =
r212 + r223 ฯ2
1 + r212 r223 ฯ2
Rav โ r212
r223 (1 โ r212 )
d = โ
,
=
0.4 โ 0.24532
= 0.5782,
0.79082 (1 โ 0.24532 )
โ ln 0.5782
1
1
ln ฯ = โ
ln ฯ2 =
= 43.6 ยตm.
ฮบ2
2ฮบ2
4ฯ ร 10โ4 ยตmโ1
More accurate is the averaged expression, equation (2.129)
Rav = ฯ12 +
ฯ23 (1 โ ฯ12 )2 ฯ2
1 โ ฯ12 ฯ23 ฯ2
or
ฯ2 =
=
Rav โ ฯ12
Rav โ ฯ12
=
2
ฯ23 (Rav โ ฯ12 )ฯ12 + (1 โ ฯ12 )
ฯ23 1 โ (2 โ Rav )ฯ12
0.4 โ 0.24532
= 0.6013
0.79082 [1 โ 1.6 ร 0.24532 ]
26
RADIATIVE HEAT TRANSFER
and
d=
โ ln 0.6013
= 40.47 ยตm
4ฯ ร 10โ4 ยตmโ1
For such a large d, it follows that ฮถ2 ' 40 ร 10.3673 ' 450. A full interference period is traversed if ฮถ2 ' 450 ยฑ ฯ.
Around ฮป = 2 ยตm this implies a full period is traversed between 2 ยตm ยฑ 0.014 ยตm. Such interference effects
will rarely be observed because (i) the detector will not respond to such small wavelength changes, and (ii)
the slightest inaccuracies in layer thickness will eliminate the interference effects.
Note: since incoming radiation at ฮป0 = 2 ยตm has a wavelength of ฮป = ฮป0 /n1 = 2/1.65 = 1.2. ยตm, mPt should
really be evaluated at 1.21 ยตm.

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