1051-733-20082
Solution Set #4
1. The resonant frequency for lead glass is in the ultraviolet region fairly near to the visible
region and the resonant frequency for fused silica is far into the ultraviolet region. Use
the dispersion equation to make a rough sketch of n [λ] in the visible region of the
spectrum.
The resonance for lead glass is in the near ultraviolet, so its refractive index and dispersion are both larger than for fused silica, whose resonance is farther into the ultraviolet
spectrum. Both the refractive index and dispersion of lead glass are larger than those
for fused silica:
1
2. Crystal quartz has refractive indices of n = 1.557 at λ = 410.0 nm and n = 1.547 at
λ = 550.0 nm. Calculate the first two coefficients (only) in Cauchy’s approximation for
the refractive index:
C2 C4
n = C1 + 2 + 4 + · · ·
λ
λ
to calculate C1 and C2 and estimate n at λ = 610.0 nm.
The equations for the indices are:
C2
= 1.557
(410.0 nm)2
C2
= 1.547
n [550.0 nm] = C1 +
(550.0 nm)2
n [410.0 nm] = C1 +
You can solve these two simultaneous equations in any way you want: I prefer putting
the equation in matrix-vector form and then inverting the matrix:
#∙
"
¸
¸ ∙
1 (410.01nm)2
C1
1.557
=
1.547
1 (550.01nm)2
C2
∙
C1
C2
¸
=
"
∙
1
1
1
(410.0 nm)2
1
(550.0 nm)2
#−1 ∙
1.557
1.547
¸
−1.2507
2.2507
∼
=
3.783 5 × 105 nm2 −3.783 5 × 105 nm2
∙
¸
1.534 5
∼
=
3783.5 nm2
¸∙
1.557
1.547
¸
C1 ∼
= 1.5345
C2 ∼
= 3783.5 nm2
n [610.0 nm] ∼
= 1.534 5 +
3783.5 nm2 ∼
= n [610.0 nm] ∼
= 1.5447
(610.0 nm)2
: 1. 544 7n.b.,
n [610.0 nm] < n [550.0 nm] < n [410.0 nm] as expected for normal dispersion
2
3. The critical angle for a certain oil is found to be θcritical = 33◦ 330 of arc. Find the
Brewster angle for both external and internal reflections.
Brewster’s angle is that for which there is no TM reflection. An “external reflection”
is one for which n1 < n2 and an internal reflection has n1 > n2 (dense to rare). For
an internal reflection at the critical angle, there is no transmission from a dense to a
rare medium. From the notes, the expression satisfied by the critical angle is:
∙ ¸
−1 n2
= 33◦ 330
θC = sin
n1
For the internal reflection with n1 < n2 :
=⇒
=⇒
n2
= sin [33◦ 330 ] ∼
= 0.552 66 for internal
n1
θB = tan−1 [sin [33◦ 330 ]] ∼
= 28◦ 560
= 0.504 88 radians ∼
= 28.93◦ ∼
For the external reflection, n1 > n2 =⇒
n1
1
∼
=
= 1.8094
n2
sin [33◦ 330 ]
∙ ¸
∙
¸
1
−1 n1
−1
∼
= tan
θB = tan
= 61◦ 40
= tan−1 [1.8094] ∼
= 1.0659 radians ∼
= 61.07◦ ∼
n2
sin [33◦ 330 ]
3
4. Consider an air-medium interface. Determine the value of n of the medium for which
Brewster’s angle is equal to the critical angle.
If we tried to equate the two angles for the same configuration of reflection — i.e., both
are “dense to rare,” then we would have to solve:
∙ ¸
∙ ¸
−1 n2
−1 n2
θC
= θB = tan
=
sin
n1
n1
−1
−1
=⇒ sin [n] = tan [n]
=⇒ n = 0
So this result is not very interesting. We can find a value if the critical angle is for an
internal reflection and Brewster’s angle for an external
n1
θC
=
=
=⇒
1
¸
∙ ¸
n2
−1 n1
= θB = tan
sin
n1
n2
∙ ¸
1
= tan−1 [n]
sin−1
n
−1
∙
This can be solved directly, graphically, or by iteration.
To solve directly, construct the triangles with these tangents and sines:
The triangles are similar, so the ratios of the sides must be equal:
n
1
=√
sin [θ]
=
n
1 + n2
√
=⇒ n2 = 1 + n2
=⇒ n4 − n2 − 1 = 0p
√
− (−1) ± 1 − 4 (1) (−1)
1± 5
2
=
=⇒ n =
2
2
The negative sign yields a negative value for n2 , so take the plus sign:
r
√
√
5
1+ 5 ∼
1
+
2
n =
=⇒ n =
= 1.272
2
2
Not that it matters for this problem, but n2 is the golden ratio of 1.618.
4
To solve graphically, plot the two curves:
∙ ¸
−1 1
= tan−1 [n]
sin
n
angle
2.0
1.5
1.0
0.5
0.0
0.0
sin−1
£1¤
n
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
n
in red and tan−1 [n] in blue, showing the intersection near n = 1.27
angle 0.94
0.92
0.90
0.88
0.86
1.25
1.26
1.27
1.28
1.29
1.30
n
Two graphs are shown; the right-hand graph is a magnified view for 1.25 < n < 1.30,
which shows the intersection near n = 1.272.
5
To solve by interation, make a trial solution, check the results, and increment or
decrement: start with n = 1.5 :
∙ ¸
1
−1
∼
sin
= 0.7297
1.5
tan−1 [1.5] = 0.9828
so we need to decrease the arctangent, which means the argument has to be smaller,
say n = 1.3:
∙ ¸
1
−1
∼
sin
= 0.877 6
1.3
tan−1 [1.3] = 0.9151
The arctangent is still too large, try n = 1.25:
∙
¸
1
−1
∼
sin
= 0.927 3
1.25
tan−1 [1.25] = 0.896 1
Too far the other way, try 1.27
∙
¸
1
∼
sin
= 0.9066
1.27
tan−1 [1.27] = 0.9038
−1
Just a bit larger, say 1.272:
∙
¸
1
∼
sin
= 0.9046
1.272
tan−1 [1.272] = 0.9046
−1
n∼
= 1.272
so both methods yield the same value for n.
6
5. Consider glass with refractive index n = 1.52. A nonabsorbing film of MgFl with
n = 1.38 and thickness τ = λ40 is deposited on the glass. Monochromatic light with
wavelength λ0 is incident normal to the film+glass. Determine
(a) the reflectance of the surface without the film;
R=
µ
n1 − n2
n1 + n2
¶2
=
µ
1 − 1.52
1 + 1.52
¶2
= R∼
= 0.0426
(b) the reflectance of the air-film interface;
R=
µ
1 − 1.38
1 + 1.38
¶2
= R∼
= 0.0255
somewhat smaller because the ∆n is less.
(c) the reflectance of the film-glass interface;
R=
µ
1.38 − 1.52
1.38 + 1.52
¶2
= R∼
= 0.00233
a lot less.
(d) the reflectance of the two-interface system
Here we need to consider the thickness of the film, which causes interference of
the light reflected from the two surfaces. The light is incident normally so that
θ0 = 0. Since the thickness is a quarter wave in air, then the phase change of
the light that travels through the film, reflects, and travels back includes the phase
change of π at the second external reflection
∆φ =
2πnf ilm
2π · 1.38
λ0
= 1.38π
· 2τ =
·2·
λ0
λ0
4
From notes, the irradiance of the reflected beam is:
¶2 µ
¶2
µ
n1 n3 − n22
1 · 1.52 − 1.382
Ir
=
=
= 0.0126
R=
I0
n1 n3 + n22
1 · 1.52 + 1.382
7
6. The refractive index of water is n = 1.33. Plot the reflectances of water for both TE
and TM polarizations between θ0 = 0◦ and θ0 = 90◦
For external reflection (rare to dense), n1 = 1, n2 = 1.33
RT E
RT M
q
⎞2
2
2
cos [θ0 ] − (1.33) − sin [θ0 ]
⎠
q
= ⎝
2
2
cos [θ0 ] + (1.33) − sin [θ0 ]
q
⎞2
⎛
2
+ (1.33) cos [θ0 ] − (1.33)2 − sin2 [θ0 ]
⎠
q
= ⎝
2
2
2
+ (1.33) cos [θ0 ] + (1.33) − sin [θ0 ]
⎛
θB = tan−1 [1.33] ∼
= 0.926 ∼
= 53.06◦
θC = sin−1 [1.33] (undefined)
R
1.0
0.8
0.6
0.4
0.2
0.0
0
10
20
30
40
50
60
70
80
90
theta
RT E (solid black), RT M (solid red) for external reflection (rare to dense) showing
Brewster’s angle at θ0 = tan−1 [1.33] ∼
= 53. 062◦
For internal reflection (dense to rare), n1 = 1.33, n2 = 1.0
8
RT E
RT M
q
⎞2
(1.33) cos [θ0 ] − (1)2 − (1.33)2 sin2 [θ0 ]
⎠
q
= ⎝
2
2
2
2
(1.33) cos [θ0 ] + − (1) − (1.33) sin [θ0 ]
q
⎞2
⎛
+12 cos [θ0 ] − (1)2 − (1.33)2 sin2 [θ0 ]
⎠
q
= ⎝
2
2
2
2
+ (1) cos [θ0 ] + (1) − (1.33) sin [θ0 ]
⎛
θB
θC
R
2
∙
¸
1
∼
= tan
= 0.645 ∼
= 36.9◦
1.33
∙
¸
1
−1
∼
= sin
= 0.851 ∼
= 48.8◦
1.33
−1
1.0
0.8
0.6
0.4
0.2
0.0
0
10
20
30
40
50
60
70
80
90
theta
RT E (solid black), RT£ M (solid
red) for internal reflection (dense to rare) showing
¤
£ 1Brewster’s
¤
−1
1
−1
◦
∼
∼ 48.8◦
angle at θB = tan
and
TIR
for
both
polarizations
if
θ
>
sin
36.9
=
1.33
1.33 =
9
7. You are probably familiar with the mirage due to warm air above hot pavement. We
can also interpret the bending of the light as total internal reflection. If the refractive
index of air at the observer’s eye is the standard value of n = 1.00029 and if the
apparent mirage is at angle θ = 88.7◦ measured from the normal, find the refractive
index of the air immediately above the pavement.
This is an example of total internal reflection where the index of the cooler air above
the road (n = 1.00029) is larger than the unknown index of the warmer air at the road
surface. The equation to be satisfied is:
nair · sin θ
=
=⇒
1.00029 · sin [88.7◦ ] = nunknown · sin [90]
nunknown = 1.00029 · sin [88.7◦ ] ∼
= 1.00003 < 1.00029
10
8. Unpolarized light is reflected from a plane surface of fused silica glass of index n =
1.458.
(a) Determine the critical and polarizing (Brewster) angles.
n1 = 1
n2 = 1.458
The critical angle is an internal reflection but the Brewster angles are both internal
and external
∙ ¸
∙
¸
1
−1 n1
−1
∼
= sin
θC = sin
= 43.30◦
= 0.7558 radians = θC ∼
n2
1.458
∙ ¸
∙
¸
1
−1 n1
−1
∼
= tan
θB (internal) = tan
= 0.6012 radians = θB (internal) ∼
= 33.44◦
n2
1.458
∙ ¸
∙
¸
−1 n2
−1 1.458 ∼
= tan
θB (external) = tan
= 55.55◦
= 0.9696 radians = θB (external) ∼
n1
1
(b) Determine the reflectance and transmittance for the TE mode at θ0 = 0◦ and at
θ0 = 45◦ .
Ã
!2
p
n1 cos [θ0 ] − n22 − n21 sin2 [θ0 ]
p
RT E =
n1 cos [θ0 ] + n22 − n21 sin2 [θ0 ]
RT E [θ0 = 0◦ ] ∼
= 0.0347
RT E [θ0 = 45◦ ] ∼
= 0.0821
TT E
p
n22 − n21 sin2 [θ0 ]
=³
´2
p
2
2
2
n1 cos [θ0 ] + n2 − n1 sin [θ0 ]
4n1 cos [θ0 ]
TT E [θ0 = 0◦ ] ∼
= 0.9653
TT E [θ0 = 45◦ ] ∼
= 0.9179
(c) Repeat part (b) for TM light.
The reflectances are the squares of the amplitude reflectance coefficients, so the
choice of convention for rT M does not matter:
Ã
!2
p
+n22 cos [θ0 ] − n1 n22 − n21 sin2 [θ0 ]
p
RT M =
+n22 cos [θ0 ] + n1 n22 − n21 sin2 [θ0 ]
RT M [θ0 = 0◦ ] = 0.0347
RT M [θ0 = 45◦ ] = 0.00674
11
TT M
! Ã
Ãp
!2
n22 − n21 sin2 [θ0 ]
2n1 n2 cos [θ0 ]
p
·
=
n1 cos [θ0 ]
+n22 cos [θ0 ] + n1 n22 − n21 sin2 [θ0 ]
TT M [θ0 = 0] = 0.965
TT M [θ0 = 45◦ ] = 0.9933
(d) Calculate and plot the phase difference between the TM and TE modes for internally reflected light at angles of incidence of θ0 = 0◦ , 20◦ , 40◦ , 50◦ , 70◦ , and 90◦ .
(extra credit given for a itan2 itan2Internally reflected light =⇒ n2 = 1, n1 =
1.458
The reflection coefficients are generally complex numbers that may be written in
the form:
r = |r| exp [+iφ]
where φ is the phase shift:
−1
φ = tan
∙
Im {r}
Re {r}
¸
Note that −π ≤ φ < +π, which means that the full inverse tangent must be used
(the usual form of the inverse tangent calculates values in the range − π2 ≤ φ <
+ π2 .
For the TE case, the amplitude reflectance coefficients in both conventions (Hecht
and Pedrotti) are identical:
p
n1 cos [θ0 ] − n22 − n21 sin2 [θ0 ]
p
rT E =
n1 cos [θ0 ] + n22 − n21 sin2 [θ0 ]
The arguments of the square roots are negative in this case if:
n1
>
=⇒
θ0
>
n2 =⇒ n22 − n21 sin2 [θ0 ] < 0 if 12 < (1.458)2 sin2 [θ0 ]
1 ∼
sin [θ0 ] >
= 0.68587
1.458
∙
¸
1
−1
∼
θC = sin
= 0.75580 radians ∼
= 43.3◦
1.458
The amplitude reflectance is real valued if the incident angle is smaller than the
critical angle, and therefore the phase difference of the reflected light is zero. If
the incident angle exceeds the critical angle, the amplitude reflectance for TE light
is complex, and the phase angle of the reflected light is different. For example,
12
consider the two incident angles θ0 = 0 < θC and θ0 = 50◦ > θC :
rT E [θ0 = 0]
=
=⇒
rT E [θ0 = 50◦ ]
=
=⇒
n1 − n2
1.458 − 1 ∼
=
= 0.186 + 0i
n1 + n2
1.458 + 1
∙
¸
0
−1
φT E [θ0 = 0] = tan
=0
0.186
q
◦
1.458 cos [50 ] − 12 − (1.458)2 sin2 [50◦ ]
q
≡ 0.560 − 0.828i
1.458 cos [50◦ ] + 12 − (1.458)2 sin2 [50◦ ]
∙
¸
◦
−1 −0.828 ∼
φT E [θ0 = 50 ] = tan
= −0.976 radians ∼
= −55.928◦
0.560
The amplitude reflectance for TM light in internal reflectance differs with convention:
n1 p 2
n2 − n21 sin2 [θ0 ]
−n2 cos [θ0 ] +
n2
(rT M )Pedrotti =
n1 p 2
+n2 cos [θ0 ] +
n2 − n21 sin2 [θ0 ]
n2
(rT M )Hecht
n1 p 2
n2 − n21 sin2 [θ0 ]
n2
=
n1 p 2
+n2 cos [θ0 ] +
n2 − n21 sin2 [θ0 ]
n2
q
1.458
+1 · cos [θ0 ] −
12 − (1.458)2 sin2 [θ0 ]
1
q
=
1.458
+1 · cos [θ0 ] +
12 − (1.458)2 sin2 [θ0 ]
1
+n2 cos [θ0 ] −
Recall that these differ by the algebraic sign:
(rT M )Pedrotti = − (rT M )Hecht
In the Hecht convention, the numerator is real AND negative if
q
1.458
<
12 − (1.458)2 sin2 [θ0 ]
+1 · cos [θ0 ]
1
µ
¶2
1
=⇒
cos2 [θ0 ] < 12 − (1.458)2 sin2 [θ0 ]
1.458
µ
¶2
1
=⇒
cos2 [θ0 ] + (1.458)2 sin2 [θ0 ] < 1
1.458
=⇒
=⇒
=⇒
θ0
<
cos2 [θ0 ] + (1.458)4 · sin2 [θ0 ] < (1.458)2
¡
¢
cos2 [θ0 ] + sin2 [θ0 ] + (1.458)4 − 1 · sin2 [θ0 ] < (1.458)2
¡
¢
(1.458)4 − 1 · sin2 [θ0 ] < (1.458)2 − 1
"s
#
2
(1.458)
−
1
∼
sin−1
= 0.60118 radians ∼
= 34.44◦ = θB
4
(1.458) − 1
13
q
1.458
+1 · cos [70 ] −
12 − (1.458)2 sin2 [70◦ ]
1
In the Hecht convention, the numerator is real and negative for 0 < θ0 < θB ,
real and positive for θB < θ0 < θC , and complex for θ0 < θC . In the Pedrotti
convention, the numerator is real and positive for 0 < θ0 < θB , real and negative
for θB < θ0 < θC , and complex for θ0 < θC .
We can now answer the question for the listed angles θ0 = 0◦ , 20◦ , 40◦ , 50◦ , 70◦ , 90◦ .
◦
14
In the Hecht convention:
θ0 = 0 =⇒ φT E = 0 and φT M = −π =⇒ ∆φ [0] = φT E − φT M = +π
θ0 = 20◦ < θB =⇒ φT E = 0 and φT M = −π =⇒ ∆φ [20◦ ] = +π
θ0 = 40◦ =⇒ θB < θ0 < θC =⇒ φT E = 0 and φT M = 0 =⇒ ∆φ [40◦ ] = 0
rT E [θ0 = 50◦ ]
=
=⇒
rT M [θ0 = 50◦ ]
=
φT M [θ0 = 50◦ ]
=
q
1.458 cos [50 ] − 12 − (1.458)2 sin2 [50◦ ]
q
≡ 0.560 − 0.828i
1.458 cos [50◦ ] + 12 − (1.458)2 sin2 [50◦ ]
∙
¸
◦
−1 −0.828 ∼
φT E [θ0 = 50 ] = tan
= −0.976 radians ∼
= −55.928◦
+0.560
q
1.458
◦
12 − (1.458)2 sin2 [50◦ ]
+1 · cos [50 ] −
1 q
∼
= −0.120 19 − 0.992 75i
1.458
2
2
◦
2
◦
+1 · cos [50 ] +
1 − (1.458) sin [50 ]
1
t [−0.120 19, −0.992 75] ∼
= −1.6913 radians ∼
= −96.904◦
◦
∆φ [θ0 = 50◦ ] = −55.928◦ − (−96.904◦ ) ∼
= 0.715 15 ∼
= 0.22764 ∼
= +40.975◦
rT E [θ0 = 70◦ ]
=
=⇒
rT M [θ0 = 70◦ ]
=
φT M [θ0 = 70◦ ]
∼
=
q
1.458 cos [70 ] − 12 − (1.458)2 sin2 [70◦ ]
q
≡ −0.558 23 − 0.829 69i
2
2
◦
2
◦
1.458 cos [70 ] + 1 − (1.458) sin [70 ]
∙
¸
◦
−1 −0.829 69 ∼
φT E [θ0 = 70 ] = tan
= −2.163 radians ∼
= −123. 94◦
−0.55823
q
1.458
+1 · cos [70◦ ] −
12 − (1.458)2 sin2 [70◦ ]
1 q
∼
= −0.881 93 − 0.471 38i
1.458
2
2
◦
2
◦
+1 · cos [70 ] +
1 − (1.458) sin [70 ]
1
−2.6507 radians ∼
= −151.88◦
◦
∆φ [θ0 = 70◦ ] = −123. 94◦ − (−151.88◦ ) ∼
= 0.15522π radians = 0.48764 radians ∼
= +27.93◦
rT E [θ0 = 90◦ ]
=
=⇒
rT M [θ0 = 90◦ ]
=
=⇒
q
12 − (1.458)2 sin2 [90◦ ]
q
≡ −1
2
2
◦
2
◦
1.458 cos [90 ] + 1 − (1.458) sin [90 ]
1.458 cos [90◦ ] −
φT E [θ0 = 90◦ ] = −π radians
q
1.458
◦
+1 · cos [90 ] −
12 − (1.458)2 sin2 [90◦ ]
1 q
∼
= −1
1.458
2
2
+1 · cos [90◦ ] +
12 − (1.458) sin [90◦ ]
1
φT M [θ0 = 90◦ ] = −π radians
∆φ [θ0 = 90◦ ] = 0
15
In the Pedrotti convention:
θ0 = 0 =⇒ φT E = 0 and φT M = 0 =⇒ ∆φ [0] = φT E − φT M = 0
θ0 = 20◦ < θB =⇒ φT E = 0 and φT M = 0 =⇒ ∆φ [20◦ ] = 0
θ0 = 40◦ =⇒ θB < θ0 < θC =⇒ φT E = 0 and φT M = −π =⇒ ∆φ [40◦ ] = π
Note that:
∆φHecht − π = ∆φPedrotti
rT E [θ0 = 50◦ ]
q
1.458 cos [50 ] − 12 − (1.458)2 sin2 [50◦ ]
q
≡ 0.560 − 0.828i
1.458 cos [50◦ ] + 12 − (1.458)2 sin2 [50◦ ]
φ [θ0 = 50◦ ] ∼
= −0.976 radians ∼
= −55.928◦
◦
=
=⇒
=
=
◦
rT M [θ0 = 50 ]
φT M [θ0 = 50◦ ]
TE
− (rT M [θ0 = 50◦ ])Hecht = +0.120 19 + 0.992 75i
+1.4503 radians ∼
= +83.096◦
∆φ [θ0 = 50◦ ] = −55.928◦ − (+83.096◦ ) ∼
= −0.772 36π = −2.43 radians ∼
= −139.02◦
rT E [θ0 = 70◦ ]
q
12 − (1.458)2 sin2 [70◦ ]
q
≡ −0.558 23 − 0.829 69i
2
2
◦
2
◦
1.458 cos [70 ] + 1 − (1.458) sin [70 ]
∙
¸
◦
−1 −0.829 69 ∼
φT E [θ0 = 70 ] = tan
= −2.163 radians ∼
= −123. 94◦
−0.55823
1.458 cos [70◦ ] −
=
=⇒
rT M [θ0 = 70◦ ] ∼
= +0.881 93 + 0.471 38i
◦
φT M [θ0 = 70 ] = t [+0.881 93, +0.471 38] ∼
= +.49085 radians ∼
= +28.124◦
∆φ [θ0 = 70◦ ] = −123. 94◦ − (+28.124◦ ) ∼
= −0.844 8π radians ∼
= −2.654 radians ∼
= −152.06◦
rT E [θ0 = 90◦ ]
=
=⇒
rT M [θ0 = 90◦ ]
=
φT M [θ0 = 90◦ ]
=
q
1.458 cos [90 ] − 12 − (1.458)2 sin2 [90◦ ]
q
≡ −1
2
2
◦
2
◦
1.458 cos [90 ] + 1 − (1.458) sin [90 ]
◦
φT E [θ0 = 90◦ ] = −π radians
q
1.458
◦
+1 · cos [90 ] −
12 − (1.458)2 sin2 [90◦ ]
1 q
= +1
−
1.458
2
2
◦
2
◦
+1 · cos [90 ] +
1 − (1.458) sin [90 ]
1
0 radians
∆φ [θ0 = 90◦ ] = −π radians = 180◦
16
9. Light is dispersed by a glass prism to create the spectrum as shown:
(a) One glass exhibits normal dispersion:
On a sketch of the prism, show the sequence of dispersed colors.
The light out of the prism is dispersed into the classical ROYGBIV spectrum,
where red is deviated least because the refractive index is smallest at longer wavelengths.
17
(b) Now consider a substance that exhibits anomalous dispersion centered at λ =
500 nm using the “undamped” model of the refractive index:
On a sketch of the prism, show the sequence of dispersed colors.
Just check the relative sizes of the refractive index at each wavelength — you can see
that the refractive index for blue is smallest, red is somewhat larger, then orange,
yellow, and green. So the dispersion looks like the drawing:
18