1051-455-20073
Solution Set #3
1. The variation of refractive index with wavelength for a transparent substance (such as glass) may be
approximately represented by the empirical equation due to Cauchy:
n [ ] = A0 +
B0
2
0
where A0 and B0 are empirically determined constants and 0 is the wavelength of light in a vacuum.
2
If A0 = 1:40, B0 = 2:5 104 ( nm) , determine the phase and group velocities at 0 = 500 nm.
n [ = 500 nm]
v [ = 500 nm]
0 0
=
n
2:5 104 ( nm)
2
= 1:5
2
(500 nm)
!0
c
2:99792458 108 m s
=
= =
jk0 j
n
1:5
= 1:40 +
1
= v = 2:0
108
m
s
The “group velocity” may also be called the “modulation velocity” v mod , which is the speed of the
low-frequency modulation. We need to …nd an expression for the modulation velocity in terms of the
numbers we know:
vmod
=
v
=
=)
vmod
=
dn
dk
=
vmod
=
n[ ]
=
vmod
=
=
d!
dk =500 nm
c
!
=) ! = k v = k
k
n
d!
d
c
c dk
k dc
d 1
=
k
=
+
+ kc
dk
dk
n
n dk n dk
dk n
d!
c
k
dn
c
ck dn
c
=
1+
0 + kc
n 2
=
=
dk
n
n
dk
n n2 dk
n
dn d
dn
=
d
dk
d
dk
d
1
=
dn
d
1
2
2
=
dn
d
1
2
n
1
k dn
n dk
2
2
2
dn
dn
c
=v 1+
=
d
2
n d
n
B0
dn
B0
A0 + 2 =)
= 2 3
d
B0
B0
c
c
1+
2 3
1 2
=
n
n
n
n 2
!
2
2:5 104 ( nm)
c
v
c
1 2
=
=
2
1:5
1:7308
1:154
1:5 (500 nm)
v
1
1+
dn
n d
2. For the crown and ‡int glasses given in the notes with the following indices measured at two vacuum
wavelengths:
Line
n for Crown n for Flint
0 [ nm]
C
656:28
1:51418
1:69427
F
486:13
1:52225
1:71748
Approximate the empirical constants A0 and B0 in Cauchy’s equation (given in #1) and use them to
evaluate the refractive index at 0 = 589:59 nm (Fraunhofer’s “D” line); compare the results to the
actual values:
Line
n for Crown n for Flint
0 [ nm]
D
589:59
1:51666
1:70100
Two equations in two unknowns:
nC
= A0 +
nF
= A0 +
B0
2
C
B0
2
F
Many ways to solve; I’ll use matrix inversion.
For the crown glass:
2
1
1
C
A0
B0
2
F
1 2:321 8
1 4:231 5
10
10
1
1
6
6
nC
nF
=
1
1
=
1:9097
10
1:9097
1
10
2
1
1
det
=
C
2
F
nm
nm
2
2
2
C
1
=
2
F
at
D
=
A0
B0
=
6
nm
C
nC
nF
2
F
2
1
1
=
2
2:321 8
4:231 5
4:231 5
6
nm
2:215 8
5:236 4 105 nm2
A0
B0
=
10
10
6
6
nm
nm
2
2
2
2
2:215 8
5:236 4 105 nm2
=
A0
B0
(656:28 nm)
(486:13 nm)
1
2
1
1
=)
10
1
6
nm
2
2:321 8
1:215 8
5:236 4 105 nm2
1:215 8
5:236 4 105 nm2
1:51418
1:52225
1:504 4
4225:8 nm2
= 589:59 nm :
Crown glass: nF = A0 +
B0
2
D
= 1:504 4 +
4225:8 nm2
(589:59 nm)
2
= 1:5166, identical to measurement
For the ‡int glass:
A0
B0
at
D
10
1
=
2:215 8
5:236 4 105 nm2
1:215 8
5:236 4 105 nm2
1:69427
1:71748
=
1:666 1
12154: nm2
= 589:59 nm :
Flint Glass: nF = A0 +
B0
2
D
= 1:666 1 +
2
12154: nm2
(589:59 nm)
2
= 1:701 1 vs. 1:70100
6
nm
2
3. (Fowles 2.5) The electric vector of a wave is given by the real expression:
E [z; t] = E0 x
^ cos [k0 z
! 0 t] + y
^ b cos [k0 z
!0 t +
0]
Show that this is equivalent to the complex-valued expression:
^+y
^ b exp [+i
E [z; t] = E0 x
E [z; t]
= E0 x
^ cos [k0 z
=
0]
exp [+i (k0 z
! 0 t] + y
^ b cos [k0 z
!0 t +
! 0 t)]
0]
Re E0 x
^ exp [+i (k0 z
! 0 t)] + y
^ b exp [+i (k0 z
!0 t +
= E0 Re x
^ exp [+i (k0 z
! 0 t)] + y
^ b exp [+i (k0 z
! 0 t)] exp [+i
= E0 Re
x
^+y
^ b exp [+i
0]
exp [+i (k0 z
0 )]
0]
! 0 t)]
4. Sketch diagrams to show the type of polarizations in #3 for the following cases:
(a)
0
= 0; b = 1
E [z; t]
=
E0 x
^+y
^ exp [+i (k0 z
=
E0
=)
at angle
(b)
0
=
LP with amplitude E0
1
tan 1
= = 45
1
4
=
E0 x
^+y
^ 2 exp [+i (k0 z
=
E0
=)
at angle
0
exp [+i (k0 z
! 0 t)]
p
p
12 + 12 = 2 E0
= 0; b = 2
E [z; t]
(c)
1
1
! 0 t)]
= +4;b =
1
2
exp [+i (k0 z
! 0 t)]
! 0 t)]
p
p
12 + 12 = 2 E0
LP with amplitude E0
2
1:1071 radians = 0:3524 radians = 63:44
tan 1
1
=
1
E [z; t]
E3
=
E0 x
^
=
E0
=
=)
i
h
exp [+i (k0 z ! 0 t)]
y
^ exp +i
4
1
exp [+i (k0 z ! 0 t)]
1 exp +i 4
1
1 exp +i 4
=
1
exp +i
=
4
LHEP with major axis at angle
= tan
1
5. Write down the Jones vectors for the three cases in the previous problem.
Did it within the problems
(a)
E1 =
1
1
E2 =
1
2
(b)
(c)
E3 =
1
1 exp +i 4
3
=
1
exp
i 34
1
exp
i 34
1
= = 45
1
4
6. For the following three Jones vectors:
E1
=
p1
3
E2
=
+i
1
E3
=
1 i
1+i
(a) Determine the type of polarization of each wave;
p1
3
cos [ ]
= E0
sin [ ]
"p #
3
=
= tan 1
1
3
r
p 2 p
=
12 +
3 = 4=2
=
E1
E0
Linearly polarized with amplitude of 2 at angle of 60 =
E2
+i
1
=
= +i
1
h
i
= exp +i
2
1
+i
= +i
1
+i
RHCP with unit amplitude
E3
=
1 i
1+i
=
(1
i)
RHCP with amplitude
1
= (1
1+i
1 i
p
i)
"
1
(1+i) (1+i)
(1 i) (1+i)
#
=
3
1
1 exp +i 2
p
2 exp
h
i
2
i
1
i
2
(b) Find Jones vectors that are orthogonal to each of the three cases and describe the state of polarization.
Find a vector such that the scalar product is zero, where the de…nition of the scalar product for
vectors with complex-valued components includes a complex conjugate
E1 E?
1
=
2
X
n=1
(E 1 )n E ?
1
n
= 0 =) (E 1 )x E ?
1
p
x
+ (E 1 )y E ?
1
p
3
3
p1
=) E ?
=
or
1 =
3
1
1
check by evaluating scalar product
p
p
p
p
= 1
3 +
3 1 =1
3+
3 1 =0
=
E1
E1 E?
1
E2
=
E?
2
=
i
h
exp +i
2
exp +i 2
1
E3
=
E?
3
=
i
h
= exp +i
2
1
1 exp +i 2
+i
1
=
p
2 exp
+i
1
4
h
or
i
2
or
i
1
i
1
i
1
+i
p
3
1
1
+i
y
7. (P3 15-1) Initially unpolarized light passes in turn through three linear polarizers with transmission
axes at 0 , 30 , and 60 , respectively, relative to the horizontal axis. What is the irradiance of the
product light expressed as a percentage of the unpolarized light irradiance?
this is Malus’ law implemented twice, plus recognizing that the irradiance is the squared magnitude
of the amplitude. The …rst polarizer reduces the irradiance by half and the light is linearly polarized
horizontally, so the Jones vector after the …rst polarizer is:
r
I0 1
E1 =
0
2
The Jones matrix for the second polarizer is:
M2
(n:b:; det M2
cos2
=
=
cos
cos
6
sin
6
sin
6
sin2
6
6
=
6
"
p
3
4
1
4
3
p4
3
4
#
0)
The angle of the third polarizer is 60 :
M3 =
cos2
cos
3
cos
3
sin
3
sin2
3
sin
3
3
=
"
The output state is the product of the matrices with the input state:
"
p #"
p #r
3
3
3
1
I0 1
4
4
4
4
p
E 3 = M3 M2 E 1 = p3
3
3
1
0
2
4
I3
4
4
4
1
p4
3
4
p
3
=
8
3
4
3
4
r
#
3
I0
2
1
9 3
9
9
= E3 E3 =
I0
+1 =
I0 : I3 =
I0 = 0:28I0
64 2
3
32
32
hp i
= tan 1
3 = (as it should!)
3
5
p1
3
1
8. (P3 15-3): Since a sheet of Polaroid is not an ideal polarizer, not all the energy of the E-vibrations
parallel to the TA are transmitted, nor are all E-vibrations perpendicular to the transmission axis are
absorbed. Suppose an energy fraction us transmitted in the …rst case and a fraction is transmitted
in the second.
(a) Extend Malus’ law by calculating the irradiance transmitted by a pair of such polarizers with
angle between their transmission axes. Assume initially unpolarized light of irradiance I0 . Show
that Malus’law follows in the ideal case.
If two polarizers oriented at angle
Ideal Malus’ Law
:
: I1 = I0 cos2
We know that half of irradiance of unpolarized light is blocked by an ideal polarizer, so that if a
percentage is passed of the light at angle and a percentage is passed at the orthogonal angle,
then some light gets through in all cases.
Assume that the initial polarizer is oriented along x and the second at the angle relative to x:
Light passed by x-polarizer oriented along x
:
Light passed by x-polarizer oriented along y
:
1
I0
2
1
(I1 )y = I0
2
(I1 )x =
x-axis light passed by polarizer #2 oriented along :
1
(I2x ) = (I1 )x cos2 = I0 2 cos2
2
y-axis light passed by polarizer #2 oriented along :
1
(I2y ) = (I1 )y
cos2
+
= I0 sin2
2
2
x-axis light passed by polarizer #2 in orthogonal direction:
1
(I2x ) + = (I1 )x
cos2
+
=
I0 sin2
2
2
2
y-axis light passed by polarizer #2 in orthogonal direction
1
: (I2x ) + = (I1 )y cos2 ( ) =
I0 cos2
2
2
Total light passed:
1
I0 2 cos2 + 2 sin2 + 2 cos2
2
Check limiting behavior of total light passed: set = 1, = 0 :
I2 =
I2 =
1
I0
2
12 cos2 + 2 1 0 sin2 + 02 cos2
=
I0
cos2 [ ]
2
identical to Malus’ law for ideal polarizer:
I2 = I0
1
cos2 [ ] =
2
I0
2
cos2 [ ]
(b) Let = 0:95 and = 0:05 for a given sheet of Polaroid. Compare the irradiance with that of
an ideal polarizer when unpolarized light is passed through two such sheets have relative angle
between transmission axes of 0 , 30 , 45 , and 90 .
Ideal Malus’ Law : I2 = I0
Realistic Malus’ Law : I2 =
=
1
cos2 [ ]
2
1
I0 0:952 cos2 + 2 0:95 0:05 sin2 + 0:052 cos2
2
1
I0 0:905 cos2 + 0:095 sin2
2
6
0 =) I2 =
=
0:905
I0 = 0:4525I0
2
=
=
=
=
1
I0 0:905 cos2 0 + 0:095 sin2 0
2
=
4
I0
2
=
=
1
=) I2 = I0 0:905 cos2 + 0:095 sin2
6
2
6
6
0:351 25I0
1
I0 0:905 cos2 + 0:095 sin2
2
4
4
0:905 0:095
0:905 0:095
+
= I0
+
= 0:25I0
2
2
4
4
=) I2 =
1
=) I2 = I0 0:905 cos2 + 0:095 sin2
3
2
3
3
0:148 75I0
Just for fun, plot the ideal equation and the realisitic equations for these values of
I0 = 1
Transmission
and
with
0.5
0.4
0.3
0.2
0.1
0.0
0
10
20
30
40
50
60
70
80
90
Angle theta (degrees)
Comparison of the realistic expression for Malus’ Law (black solid line) to the ideal expression
(red dashed line) for a = 0:95, = 0:05. Note that the expressions are equal at = 45 .
Try it again for di¤ erent values of
Transmission
= 0:75;
= 0:25 :
0.5
0.4
0.3
0.2
0.1
0.0
0
10
20
30
40
50
60
70
80
90
Angle theta (degrees)
which shows that the sensitivity of the amount of transmitted light the the angle of the polarizers
has become poor.
7