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1051-455-20073 Solution Set
- 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 [] = A 0 +
B 0
^20
where A 0 and B 0 are empirically determined constants and 0 is the wavelength of light in a vacuum. If A 0 = 1: 40 , B 0 = 2: 5 104 ( nm)^2 , determine the phase and group velocities at 0 = 500 nm.
n [ = 500 nm] = 1 :40 +^2 :^5 ^10
(^4) ( nm)^2 (500 nm)^2
v [ = 500 nm] =
n
jk 0 j
c n
=^2 :^99792458 ^10
(^8) m s 1 1 : 5
= v = 2: 0 10 8 ms
The ìgroup velocityî may also be called the ìmodulation velocityî v (^) mod , which is the speed of the low-frequency modulation. We need to Önd an expression for the modulation velocity in terms of the numbers we know:
vmod = d! dk (^) =500 nm v =
k =)^!^ =^ k^ ^ v ^ =^ k^ ^
c n =) d! dk
d dk
k c n
c n
dk dk
k n
dc dk
n
vmod = d! dk
c n
k n 0 + kc
n ^2
(^) dn dk
c n
ck n^2
dn dk
c n
k n
dn dk
dn dk
dn d
d dk
dn d
dk d
dn d
^2
dn d
^2
vmod = v
n
dn d
^2
= v
n
dn d
c n
n
dn d
n [] = A 0 +
B 0
^2
dn d
B 0
^3
vmod = c n
n
B 0
^3
c n
B 0
n^2
c 1 : 5
2 : 5 104 ( nm)^2 1 : 5 (500 nm)^2
c 1 : 7308
v 1 : 154
- For the crown and áint glasses given in the notes with the following indices measured at two vacuum wavelengths: Line 0 [ nm] n for Crown n for Flint C 656 : 28 1 : 51418 1 : 69427 F 486 : 13 1 : 52225 1 : 71748
Approximate the empirical constants A 0 and B 0 in Cauchyís equation (given in #1) and use them to evaluate the refractive index at 0 = 589:59 nm (Fraunhoferís ìDî line); compare the results to the actual values: Line 0 [ nm] n for Crown n for Flint D 589 : 59 1 : 51666 1 : 70100
Two equations in two unknowns:
nC = A 0 +
B 0
^2 C
nF = A 0 +
B 0
^2 F
Many ways to solve; Iíll use matrix inversion. For the crown glass: 1 C^2 1 F^2
A 0
B 0
nC nF
1 C^2
1 F^2
nC nF
A 0
B 0
1 C^2
1 F^2
1 (656:28 nm) ^2 1 (486:13 nm) ^2
1 2 :321 8 10 ^6 nm ^2 1 4 :231 5 10 ^6 nm ^2
det
1 2 :321 8 10 ^6 nm ^2 1 4 :231 5 10 ^6 nm ^2
= 1 : 9097 10 ^6 nm ^2 1 C^2 1 F^2
1 : 9097 10 ^6 nm ^2
4 :231 5 10 ^6 nm ^2 2 :321 8 10 ^6 nm ^2 1 1
5 :236 4 105 nm^2 5 :236 4 105 nm^2
A 0
B 0
5 :236 4 105 nm^2 5 :236 4 105 nm^2
A 0
B 0
4225 :8 nm^2
at D = 589:59 nm :
Crown glass: nF = A 0 + B^0 ^2 D
= 1:504 4 + 4225 :8 nm
2 (589:59 nm)^2
= 1: 5166 , identical to measurement
For the áint glass: A 0 B 0
5 :236 4 105 nm^2 5 :236 4 105 nm^2
12154 : nm^2
at D = 589:59 nm :
Flint Glass: nF = A 0 +
B 0
^2 D
12154 : nm^2 (589:59 nm)^2
= 1:701 1 vs. 1 : 70100
- For the following three Jones vectors:
E 1 =
p^1 3
E 2 =
+i 1
E 3 =
1 i 1 + i
(a) Determine the type of polarization of each wave;
E 1 =
p^1 3
= E 0
cos [] sin []
= tan ^1
" (^) p 3 1
E 0 =
r 12 +
p 3
p 4 = 2 Linearly polarized with amplitude of 2 at angle of 60 ^ = 3
E 2 =
+i 1
= +i
1 +i
= +i
+i
exp
h +i
i ^1 1 exp
+i 2
RHCP with unit amplitude
E 3 =
1 i 1 + i
= (1 i)
1+i 1 i
= (1 i)
(1+i) (1 i)
(1+i) (1+i)
p 2 exp
h i
i
i
RHCP with amplitude
p 2 (b) Find Jones vectors that are orthogonal to each of the three cases and describe the state of polar- ization. Find a vector such that the scalar product is zero, where the deÖnition of the scalar product for vectors with complex-valued components includes a complex conjugate
E 1 E? 1 =
X^2
n=
(E 1 )n
E? 1
n = 0 =) (E 1 )x
E? 1
x
E? 1
y
E 1 =
p^1 3
=) E? 1 =
p 3 1
p 3 1
or
p 3 1
check by evaluating scalar product E 1 E? 1 = 1
p 3
p 3 1
p 3 +
p 3 1
E 2 =
exp
h +i
i ^1 1 exp
+i 2
exp
h +i
i ^1 +i
E? 2 =
exp
+i 2
+i 1
or
i
E 3 =
p 2 exp
h i
i
+i
E? 3 =
+i 1
or
i
- (P^3 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 Malusí law implemented twice, plus recognizing that the irradiance is the squared magnitude of the amplitude. The Örst polarizer reduces the irradiance by half and the light is linearly polarized horizontally, so the Jones vector after the Örst polarizer is:
E 1 =
r I 0 2
The Jones matrix for the second polarizer is:
M 2 =
cos^2
6
cos
6
sin
6
cos
6
sin
6
sin^2
6
3 4
p 3 p^4 3 4 1 4
(n:b:; det M 2 = 0 )
The angle of the third polarizer is 60 ^ :
M 3 =
cos^2
3
cos
3
sin
3
cos
3
sin
3
sin^2
3
1 4
p 3 p^4 3 4 3 4
The output state is the product of the matrices with the input state:
E 3 = M 3 M 2 E 1 =
1 4
p 3 p 3 4 4
3 4
3 4
p 3 p 3 4 4
1 4
r
I 0 2
r 3 2 I^0
(^) p 1 3 1
I 3 = E 3 E 3 = 9
I 0
I 0 : I 3 = 9
I 0 = 0: 28 I 0
= tan ^1
hp 3
i
(as it should!)
= 0 =) I 2 =
I 0
0 :905 cos^2 0 + 0:095 sin^2
I 0 = 0: 4525 I 0
=) I 2 =
I 0
0 :905 cos^2
= 0 :351 25I 0
=) I 2 =
I 0
0 :905 cos^2
= I^0
+^0 :^095
= I 0
+^0 :^095
= 0: 25 I 0
=) I 2 =
I 0
0 :905 cos^2
= 0 :148 75I 0
Just for fun, plot the ideal equation and the realisitic equations for these values of and with I 0 = 1
0 10 20 30 40 50 60 70 80 90
Angle theta (degrees)
Transmission
Comparison of the realistic expression for Malusí 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 = 0: 75 ; = 0:25 :
0 10 20 30 40 50 60 70 80 90
Angle theta (degrees)
Transmission
which shows that the sensitivity of the amount of transmitted light the the angle of the polarizers has become poor.
