Optical Properties of Solids
Second Edition
Mark Fox
Oxford University Press, 2010
SOLUTIONS TO EXERCISES
These notes contain detailed solutions to the Exercises at the end of each chapter of the book, for the benefit of class instructors. Please note that figures within the solutions are numbered consecutively from the start of the document (e.g. Fig. 1) in order to distinguish them from the figures in the book, which have an additional chapter label (e.g. Fig. 1.1). A similar convention applies to the labels of tables.
The author would be very grateful if mistakes that are discovered in the solu-tions would be communicated to him. He is also very appreciative of comments about the text and/or the Exercises. He may be contacted at the following address:
Department of Physics and Astronomy University of Sheffield Hicks Building Sheffield, S3 7RH United Kingdom. email: [email protected] c ° Mark Fox 2010
Chapter 1
Introduction
(1.1) Glass is transparent in the visible spectral region and hence we can assume α = κ = 0. The reflectivity is calculated by inserting n = 1.51 and κ = 0 into eqn 1.29 to obtain R = 0.041. The transmission is calculated from eqn 1.9 with R = 0.041 to obtain T = 92%.
(1.2) From Table 1.4 we read that the refractive indices of fused silica and dense flint glass are 1.46 and 1.746 respectively. The reflectivities are then calculated from eqn 1.29 to be 0.035 and 0.074 respectively, with κ = 0 in both cases because the glass is transparent. We thus find that the reflectivity of dense flint glass is larger than that of fused silica by a factor of 2.1. This is why cut–glass products made from dense flint glass have a sparkling appearance.
(1.3) We first use eqns 1.25 and 1.26 to convert ˜²r to ˜n, giving n = 3.01 and
κ = 0.38. We then proceed as in Example 1.2. This gives: v = c/n = 9.97 × 107m s−1,
α = 4πκ/λ = 9.6 × 106m−1,
R = [(n − 1)2+ κ2]/[(n + 1)2+ κ2] = 25.6%.
(1.4) The anti–reflection coating prevents losses at the air–semiconductor in-terface, and 90% of the light is absorbed when exp(−αl) = 0.1 at the operating wavelength. With α = 1.3 × 105m−1 at 850 nm, we then find
l = 1.8 × 10−5m = 18 µm.
(1.5) We are given n = 3.68 and we can use eqn 1.19 to work out κ = αλ/4π = 0.083. We then use eqn 1.29 to find R = 0.328. Since αl = 2.6, we do not need to consider multiple reflections and we can just use eqn 1.8 to find the transmission. This gives: T = (1 − 0.328)2 exp(−1.3 × 2) = 0.034.
The optical density is calculated from eqn 1.11 as 0.434 × 1.3 × 2 = 1.1. (1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62 m−1.
We use eqn 1.19 to find κ = αλ/4π = 3.5 × 10−8. We thus have ˜n =
1.33 + i 3.5 × 10−8. The real and imaginary parts of ˜²
r are found from
eqns 1.23 and 1.24 respectively, and we thus obtain ˜²r= 1.77+i 9.2×10−8.
(1.7) The filter appears yellow and so it must transmit red and green light, but not blue. The filter must therefore have absorption at blue wavelengths. (1.8) (a) In the incoherent limit, we just add the intensities of the beams. The
incident light
transmitted light
reflected light
I 0 I 0(1-R1) I 0(1-R1)e -a l I0(1-R1)R2e-a l I 0(1-R1) (1-R2) e -a l I 0(1-R1) R2R1e -2a l I 0R1 I 0(1-R1)R2e -2a l I 0(1-R1)2R2e -2a l I0(1-R1) R2R1e -3a l I 0(1-R1) (1-R2) R2R1e -3a l I0(1-R1) R22R1e-3a l I 0(1-R1) R22R1e -4a l I 0(1-R1)2R22R1e -4a l I0(1-R1) R22R12e-4a l I 0(1-R1) R22R12e -5a l I 0(1-R1) (1-R2) R22R12e -5a l I 0(1-R1) R23R12e -5a lR
1R
2e
-alincident light
transmitted light
reflected light
I 0 I 0(1-R1) I 0(1-R1)e -a l I0(1-R1)R2e-a l I 0(1-R1) (1-R2) e -a l I 0(1-R1) R2R1e -2a l I 0R1 I 0(1-R1)R2e -2a l I 0(1-R1)2R2e -2a l I0(1-R1) R2R1e -3a l I 0(1-R1) (1-R2) R2R1e -3a l I0(1-R1) R22R1e-3a l I 0(1-R1) R22R1e -4a l I 0(1-R1)2R22R1e -4a l I0(1-R1) R22R12e-4a l I 0(1-R1) R22R12e -5a l I 0(1-R1) (1-R2) R22R12e -5a l I 0(1-R1) R23R12e -5a lR
1R
2e
-alFigure 1: Multiple reflections in the incoherent limit, as considered in Exercise 1.8.
in Fig. 1. The transmitted intensity is given by:
It = I0(1 − R1)(1 − R2)e−αl+ I0(1 − R1)(1 − R2)R1R2e−3αl + I0(1 − R1)(1 − R2)R21R22e−5αl+ · · · = I0(1 − R1)(1 − R2)e−αl ¡ 1 + R1R2e−2αl+ (R1R2)2e−4αl+ · · · ¢ , = I0(1 − R1)(1 − R2)e−αl ∞ X k=0 (R1R2e−2αl)k, = I0(1 − R1)(1 − R2)e−αl 1 1 − R1R2e−2αl ,
where we used the identity P∞k=0xk = 1/(1 − x) in the last line. The
transmissivity is thus: T = It
I0 =
(1 − R1)(1 − R2)e−αl
1 − R1R2e−2αl .
(b) We have an air-medium-air situation, and so it will be the case that R1= R2≡ R. We need to compare the exact formulae given in eqns 1.6
and 1.9 with the approximate one that neglects multiple reflections given in eqn 1.8.
(i) With α = 0 the extinction coefficient κ will also be zero. We then calculate the reflectivity from eqn 1.29 to be:
R = (3.4 − 1)2 (3.4 + 1)2 =
2.42
Hence the exact transmission from eqn 1.9 is given by: Texact= 1 − 0.30 1 + 0.30 = 0.70 1.30= 54% .
The approximate result with multiple reflections neglected is found from eqn 1.8 with α = 0:
Tapprox= (1 − R)2= (1 − 0.30)2= (0.70)2= 49% .
Thus by neglecting multiple reflections, the transmission is underestimated by a factor of about 10%.
(ii) If we assume n À κ, we can ignore the terms in κ in the formula for the reflectivity (eqn 1.29) and so just obtain R = 0.30 as in part (a). The exact transmission is calculated from eqn 1.6 to be:
Texact= (1 − 0.30) 2e−1
1 − 0.302e−2 = 18.2% .
The formula with multiple reflections neglected (eqn 1.8) gives: Tapprox= (1 − 0.30)2e−1= 18.0% .
The transmission is thus underestimated by a factor of 1%. (iii) Following the same method as in part (b)(i), we find:
R = 0.772/2.772= 0.077 ,
Texact = (1 − 0.077)/(1 + 0.077) = 85.7% ,
Tapprox = (1 − 0.077)2= 85.2% .
The error is thus −0.6%.
(c) It is intuitively obvious that it is valid to neglect multiple reflections when the absorption is strong (i.e. αl & 1), since the reflected beams will be very weak. Part (b)(ii) confirms this.
For the case of transparent materials (i.e. α = 0), it will be valid to neglect multiple reflections when R is small, since then it will be valid to approximate 1/(1 + R) as (1 − R), and eqn 1.8 with α = 0 will be equivalent to eqn 1.9. The reflectivity will be small when (n − 1) is small (see eqn 1.29), and so we conclude that multiple reflections are insignificant for small (n − 1). This point is illustrated by comparing parts (b)(i) and (iii).
(1.9) In contrast to the previous exercise, we must now add up the electric field amplitudes of the beams and consider their relative phases. Let r and t be the amplitude reflection and transmission coefficients for going from air to the medium, and r0 and t0 be the equivalent quantities for going
from the medium to air. If the absorption coefficient is α, the amplitude will decay by a factor of e−αl/2 during a single pass of the medium, since
I ∝ E2. For simplicity we define x = e−αl/2eiΦ/2, where Φ = 4πnl/λ is
incident
amplitude
transmitted
amplitude
reflected
amplitude
E 0 tE 0 txE 0 txr' E 0 rE 0 tx2r' E 0 tt'x2r'E 0x
= e
if/2e
-al/ 2 tx2r'2E 0 tx3r'2E0 tx3r'3E 0 tx4r'3E 0 tt'x4r'3E 0 tx4r'4E0 tx5r'4E0 tx5r'5E 0 tt'x E 0 tt'x3r'2E 0 tt'x5r'4E 0t
r
t
'r'
incident
amplitude
transmitted
amplitude
reflected
amplitude
E 0 tE 0 txE 0 txr' E 0 rE 0 tx2r' E 0 tt'x2r'E 0x
= e
if/2e
-al/ 2 tx2r'2E 0 tx3r'2E0 tx3r'3E 0 tx4r'3E 0 tt'x4r'3E 0 tx4r'4E0 tx5r'4E0 tx5r'5E 0 tt'x E 0 tt'x3r'2E 0 tt'x5r'4E 0t
r
t
r
t
'r'
t
'r'
Figure 2: Multiple reflections in the coherent limit, as considered in Exercise 1.9.
factor x after one pass through the medium.
(a) With the definitions above, the transmitted amplitude Etis given by (see Fig. 2): Et = tt0xE 0+ tt0x3r02E0+ tt0x5r04E0+ · · · , = tt0xE0(1 + x2r02+ x4r04+ · · · ) , = tt0xE 0 ∞ X k=0 (x2r02)k, = E0 tt 0x 1 − x2r02,
where we used P∞k=0ak = 1/(1 − a) in the last line. The transmitted
intensity is then given by: It∝ Et(Et)∗= |E
0|2 |x|
2|tt0|2
(1 − x2r02)(1 − x2r02)∗.
Now x2= e−αleiΦ, r = −r0(due to the π phase shift on going from a more
we have: T = It I0 = |E t|2 |E0|2 , = (1 − R) 2e−αl (1 − Re−αleiΦ)(1 − Re−αle−iΦ), = (1 − R) 2e−αl (1 − Re−αl(eiΦ+ e−iΦ) + R2e−2αl), = (1 − R) 2e−αl (1 − 2Re−αlcos Φ + R2e−2αl),
where we used eiϕ+ e−iϕ= 2 cos ϕ in the last line.
(b) For the reflectivity, we proceed as in part (a). From Fig. 2 it is apparent that the reflected amplitude Er is given by:
Er = rE 0+ tt0x2r0E0+ tt0x4r03E0+ · · · , = E0(r + r0tt0x2[1 + x2r02+ x4r04+ · · · ]) , = E0 µ r + r 0tt0x2 1 − x2r02 ¶ ,
where we again used P∞k=0ak = 1/(1 − a). On inserting r0 = −r and
tt0 = 1 − r2= 1 − R, we find: Er= E0r µ 1 −x 2(1 − R) 1 − x2R ¶ = E0r µ 1 − x2 1 − x2R ¶ .
Therefore, with x2 = e−αleiΦ, and again using eiϕ+ e−iϕ = 2 cos ϕ, we
find: reflectivity = |E r|2 |E0|2, = R (1 − e−αleiΦ)(1 − e−αle−iΦ) (1 − Re−αleiΦ)(1 − Re−αle−iΦ), = R (1 − 2e −αlcos Φ + e−2αl) (1 − 2Re−αlcos Φ + R2e−2αl).
(c) In the limit with α = 0, the reflectivity and transmission formulae just reduce to the standard ones for a Fabry–Perot etalon (see e.g. Hecht chapter 9):
limα→0(T ) = (1 − R) 2
(1 − 2R cos Φ + R2),
limα→0(reflectivity) = 2R(1 − cos Φ)
(1 − 2R cos Φ + R2).
Note that in this α → 0 limit, conservation of energy requires that the transmission and reflection coefficients must sum to unity. This can be
proven as follows. We have It= T × I
0 and Ir= reflectivity × I0. Hence:
It+ Ir = I0 (1 − R)2 (1 − 2R cos Φ + R2)+ I0 2R(1 − cos Φ) (1 − 2R cos Φ + R2) = I0(1 − R) 2+ 2R(1 − cos Φ) (1 − 2R cos Φ + R2) . = I0(1 − 2R + R 2+ 2R − 2R cos Φ) (1 − 2R cos Φ + R2) , = I0(1 + R 2− 2R cos Φ) (1 − 2R cos Φ + R2), = I0.
(d) If αl À 1, then e−αl¿ 1. Since R ≤ 1, we can neglect the second and
third terms in the denominator, and so the transmission is just: T = (1 − R)2e−αl,
as in eqn 1.8.
(e) For negligible absorption, we expect to observe thin-film fringes. The transmission will be as given in part (c):
limα→0(T ) = (1 − R) 2
(1 − 2R cos Φ + R2).
When Φ = 2mπ, where m is an integer, we have constructive interference, with: T = (1 − R) 2 (1 − 2R + R2) = (1 − R)2 (1 − R)2 = 1 .
Since Φ = 4πnl/λ, this constructive interference condition occurs when 2nl = mλ, that is, the round trip path length is equal to an integer number of wavelengths. The other extreme occurs when Φ = (2m + 1)π:
T = (1 − R)
2
(1 + 2R + R2) =
(1 − R)2
(1 + R)2.
At this point, we have 2nl = (m + 1/2)λ0, so that the waves from each
round trip are out of phase and interfere destructively. Therefore, as the wavelength is scanned, the transmission oscillates between unity and (1 − R)2/(1 + R)2, as in a Fabry–Perot etalon.
(1.10) This problem is meant to be a rough attempt to model the optical prop-erties of GaAs near its band edge. The band gap is taken to be 1.42 eV, as appropriate for GaAs at room temperature, and the form of the ab-sorption is meant to be rough fit to the data in Fig. 3.9, with C chosen to give the correct absorption around 2 eV, namely ∼ 4 × 106m−1.
The reflectivity and transmissivity have to be calculated from the two for-mulae given in parts (a) and (b) of the previous exercise. In principle, for wavelengths below the band edge, we have to work out the extinction co-efficient from the absorption via eqn 1.19, and then work out the reflection
600 700 800 900 1000 0.0 0.2 0.4 0.6 0.8 1.0
T
ra
n
sm
is
si
o
n
o
r
R
e
fle
ct
iv
ity
Wavelength (nm)
Transmission ReflectivityFigure 3: Transmission and reflectivity of a 2µm thick platelet, as considered in Exercise 1.10.
coefficient from eqn 1.29 with both the terms in n and κ included. How-ever, the maximum value of κ in this exercise is only 0.19, which occurs at 600 nm, where α = 4 × 106m−1, and (0.19)2 is negligible compared to
(3.5 − 1)2. Therefore, in fact we can just work out R from the real part of
the refractive index and use R = 2.52/4.52= 0.31 at all wavelengths. The
transmissivity and reflectivity calculated in this way are plotted in Fig. 3. For wavelengths below the band edge (i.e. 870–1000 nm), the medium is transparent, and we observe Fabry–Perot fringes. Peaks in the transmis-sion with T = 1 occur at wavelengths that satisfy 2nl = mλ, (m = integer), i.e. at 875 nm (m = 16), 933 nm (m = 15), and 1000 nm (m = 14). The reflectivity at these wavelengths is equal to zero. Peaks in the reflectivity and minima in the transmission are observed in be-tween the transmission maxima. The minimum value of the transmission is (1 − R)2/(1 + R)2 = 0.28, implying that the maximum value of the
reflectivity is 0.72.
For wavelengths below the band edge, the absorption increases rapidly, and the interference fringes are rapidly damped, with multiple reflections becoming negligible. The reflectivity settles to the value determined by the front surface (i.e. 31%), while the transmission decreases exponentially according to eqn 1.8. The transmissivity at 600 nm where α = 4×106m−1
is (0.69)2exp(−8) = 1.6 × 10−4.
(1.11) In the incoherent limit, the transmissivity of a transparent plate is given by eqn 1.9 as (1 − R)/(1 + R), while the reflectivity is obtained from
eqn 1.29, which becomes R = (n − 1)2/(n + 1)2 when κ = 0. Now: 1 − R = 1 −(n − 1)2 (n + 1)2 = (n + 1)2− (n − 1)2 (n + 1)2 = 4n (n + 1)2. Similarly: 1 + R = 1 +(n − 1) 2 (n + 1)2 = (n + 1)2+ (n − 1)2 (n + 1)2 = 2(n2+ 1) (n + 1)2 . We therefore have: T =1 − R 1 + R = 4n (n + 1)2 × (n + 1)2 2(n2+ 1) = 2n (n2+ 1).
(1.12) The reflection coefficient for the interface between two transparent media with refractive indices of n1and n2 is (see eqn A.55):
R = (n1− n2)2 (n1+ n2)2
.
Note that this reduces to R = (n − 1)2/(n + 1)2 for the case of an
air-medium interface with n1= 1 and n2= n. (cf. eqn 1.29 with κ = 0.) For
the three cases considered here we have:
• air → film: n1= 1, n2= 2.5, so R = (1.5/3.5)2= 18%.
• film → substrate: n1 = 2.5, n2 = 1.5, so R = (2.5 − 1.5)2/(2.5 +
1.5)2= (1/4)2= 6%.
• substrate → air: n1= 1.5, n2= 1, so R = (0.5/2.5)2= 4%.
(1.13) For a thick sample it is appropriate to use eqn 1.8 for the transmission. We take the log10of eqn 1.8 to obtain (using log10x = logex/ loge10):
− log10(T ) = −2 log10(1 − R) + αl/ loge10 .
We can then substitute αl/ loge10 from eqn 1.11 to obtain the required
result.
If the medium is transparent at λ0 then we will have that T
λ0 = (1 − R)/(1 + R) for the incoherent limit, where R is the reflectivity at λ0. We
assume that the reflectivity varies only weakly with wavelength. This is a reasonable assumption for most materials if we choose λ0 sensibly, for
example, just above the absorption edge we are trying to measure. With this assumption, we can work out the value of R from a measurement of the transmission at λ0, and then use this value in the formula derived
in the exercise to work out the optical density from a measurement of the transmission at λ. Measurements of T (λ) and T (λ0) thus allow the
optical density to be determined. The absorption coefficient can then be determined from the optical density by using eqn 1.11.
(1.14) With σ = 6.6 × 107Ω−1m, and ω = 1.88 × 1013rad/s, we find ˜² r =
approximation ²2 À ²1 is a good one. In this approximation, we have (with√i = (1 + i)/√2): ˜ n =p˜²r= (3.97 × 105)1/2 √ i = 445(1 + i) .
By inserting n = κ = 445 into eqn 1.29, we then obtain R = 99.6 %. (1.15) We use the same formula for the complex dielectric constant as in the
previous exercise. With ω = 1.88 × 1013 rad/s, we find
˜ n2= ˜²
r= ²1+ i 2.94 × 105.
Since ²2À ²1, this implies:
˜
n =p˜²r= (2.94 × 105)1/2
√
i = 383(1 + i) .
We then find from eqn 1.19 that α = 4πκ/λ = 4.8 × 107m−1. Beer’s law
means that we set exp(−αl) = 0.5 for a drop in intensity by a factor of 2, giving l = 1.4 × 10−8m = 14 nm.
(1.16) It is apparent from eqn 1.29 that R = 1 when n = 1 and κ = 0. For zero reflectivity we thus require ˜²r= (n + iκ)2= 1.
(1.17) (a) We convert wavelengths to photon energies using E = hc/λ to obtain the energy level scheme shown in Fig. 4. It is thus apparent that 0.294 eV of energy is dissipated during each absorption / emission process. (b) When the quantum efficiency is 100%, every absorbed photon pro-duces a luminescent photon. The ratio of the light energy emitted to that absorbed is then simply given by the ratio of the relevant photon energies. The emitted power is thus (1.165/1.459) × 10 = 8 W, and the dissipated power is 2 W.
(c) For a luminescent quantum efficiency of 50% the number of photons emitted drops by a factor of 2 compared to part (b), and so the light power emitted falls to 4 W. The remaining 6 W of the absorbed power is dissipated as heat. absorption 1.459 eV emission 1.165 eV relaxation (0.294 eV)
Figure 4: Energy level scheme for Exercise 1.17.
(1.18) This is an example of Raman scattering, which is discussed in detail in Section 10.5. Conservation of energy in the scattering process is satisfied when
hνout= hνin− hνphonon. With ν = c/λ, we then find λout= 521 nm.
(1.19) The transmission is given by eqn 1.12, with the wavelength dependence of the scattering cross section given by eqn 1.13. At 850 nm we have 10% transmission, so that N σsl = 2.30. Since σs∝ λ−4, the scattering
cross-section is 11.1 times larger at 850 nm than at 1550 nm, and so we have N σsl = 2.30/11.1 = 0.21 at the longer wavelength, implying a
transmis-sion of 81 %.
In general, the scattering losses decrease as the wavelength increases, and hence the propagation losses decrease. Longer wavelengths are therefore preferable for long range communication systems. At the same time, the fibres start to absorb in the infrared due to phonon absorption. 1550 nm is the longest practical wavelength for silica fibres before phonon absorption becomes significant.
(1.20) We again use eqn 1.12 to calculate the transmission, setting exp(−N σsl) =
0.5. This gives N σsl = 0.69, which implies l = 3.5 m for the given values
of N and σs.
If the wavelength is reduced by a factor of two, Rayleigh’s scattering law (eqn 1.13) implies that σsincreases by a factor of 16. The length required
for the same transmission is thus smaller by a factor of 16: i.e. l = 3.5/16 = 0.22 m.
(1.21) Birefringence is an example of optical anisotropy as discussed in Sec-tion 1.5.1, and also in SecSec-tion 2.5. Ice is a uniaxial crystal, and therefore has preferential axes, making optical anisotropy possible. Water, by con-trast, is a liquid and has no preferential axes. The optical properties must therefore be isotropic, making birefringence impossible.
Chapter 2
Classical propagation
(2.1) We envisage two displaced masses as shown in Fig. 5. The spring is extended by a distance (x1− x2) and so the force on the masses are
±Ks(x1− x2). The equations of motion are therefore
m1d 2x 1 dt2 = −Ks(x1− x2) and m2 d2x 2 dt2 = −Ks(x2− x1) .
Divide the equations by m1 and m2 respectively and subtract them to
obtain: d2 dt2(x1− x2) = −Ks µ 1 m1+ 1 m2 ¶ (x1− x2) .
On defining the relative displacement x = x1− x2 and introducing the
reduced mass µ, where 1/µ = 1/m1+ 1/m2, we then have:
µd
2x
dt2 = −Ksx .
This is the equation of motion of an oscillator of angular frequency (Ks/µ)1/2.
m
1m
2x
1x
2rest
displaced
m
1m
2x
1x
2rest
displaced
Figure 5: Displacement of two masses as described in Exercise 2.1 (2.2) The solution is simpler if complex exponentials are used. We therefore
write the force as the real part of F0e−iωt, and look for solutions of the
form x(t) = x0e−iωt. On substituting into the equation of motion we then
obtain: m(−ω2− iωγ + ω2 0)x0e−iωt= F0e−iωt, which implies: x(t) = F0 m 1 (ω2 0− ω2− iωγ) e−iωt.
The phase factor comes from the factor of [ω2
0− ω2− iωγ]−1. On
multi-plying the top and bottom by the complex conjugate, we find: 1 ω2 0− ω2− iωγ = (ω 2 0− ω2) + iωγ (ω2 0− ω2)2+ (ωγ)2 . On writing this in the form:
a + ib = reiθ≡ r(cos θ + i sin θ) , we then see that the phase factor θ is given by:
tan θ = ωγ (ω2
0− ω2)
.
This implies that the displacement of the oscillator is of the form: x(t) ∼ eiθe−iωt= e−i(ωt−θ),
which shows that the oscillator has a relative phase lag of: θ = tan−1[ωγ/(ω02− ω2)] .
(2.3) By applying the Lorentz oscillator model of Section 2.2.1, we realize that the refractive index will have a frequency dependence as shown in Fig. 2.4, with ω0 corresponding to 500 nm. (i.e. ω0 = 3.8 × 1015rad/s.) For
frequencies well above the resonance, we will just have the contribution of the undoped sapphire crystal:
n∞≡ n(ω À ω0) = 1.77 ,
which implies ²∞= (1.77)2. The refractive index well below the resonance
can be worked out from eqn 2.19. On using the value of N given in the exercise, we find ²st− ²∞= 2.23 × 10−3. We thus have:
nst =
√
²st= [(1.77)2+ 2.23 × 10−3]1/2.
We thus find nst− n∞= 6.3 × 10−4.
(2.4) We again use the Lorentz oscillator model of Section 2.2. The Exercise is similar to Example 2.1, because we are dealing with a relatively small number of absorbers and the overall refractive index will be dominated by the host crystal. We can therefore assume n = 1.39 throughout the Exercise. On the other hand, the host crystal is transparent at 405 nm, and so the absorption will be determined by the impurity atoms. The other factor we have to include is the low oscillator strength of the transition. We therefore modify the first equation in Example 2.1 to:
κ(ω0) =²2(ω0) 2n = N e2 2n²0m0 1 γω0 × f ,
where f = 9 × 10−5 is the oscillator strength. For the absorption line
we have ω0 = 2πc/405 nm = 4.65 × 1015rad/s, and γ = ∆ω = 2π∆ν =
5.15 × 1014s−1. With N = 2 × 1026m−3 and n = 1.39, we then find
κ(ω0) = 8.6 × 10−6. We finally obtain the absorption at the line centre
(405 nm) from eqn 1.19 as 270 m−1.
This Exercise is broadly based on the results presented in the paper by Iverson and Sibley in J. Luminescence 20, 311 (1979).
(2.5) The result of this Exercise works in the limit where the contribution of the particular oscillator to the dielectric constant is relatively small, as in Example 2.1 and the previous exercise. In this limit we have ²2¿ ²1, and
therefore κ(ω0) = ²2(ω0)/2n, where n =√²1, and ²1(ω0) = 1+χ. We then
see from eqn 2.16 that ²2(ω0) = N e2/²0m0γω0, so that the absorption is
(cf. eqn 1.19): α(ω0) = 4πκ(ω0) λ = 4π × ²2(ω0) 2n ÷ (2πc/ω0) = N e2 n²0m0γc.
This shows that it is the linewidth that determines the peak absorption strength per oscillator. The oscillator strength is, of course, also impor-tant.
(2.6) The data can be analysed by comparison with Fig. 2.4 with the assumption that the oscillator strength is unity.
(a) The low frequency refractive index corresponds to√²st. With n = 2.43
for ω ¿ ω0 from the data, we find ²st= 5.9.
(b) The resonant frequency is the mid point of the “wiggle”, i.e. 5.0 × 1012Hz.
(c) The natural frequency is given by eqn 2.2, which implies Ks = µω02.
The reduced mass µ is given by:
1/µ = 1/m1+ 1/m2= 1/23 + 1/35.5 amu−1,
which gives µ = 14 amu = 2.33 × 10−26kg. With ω
0 = 2πν0 = 3.1 ×
1013rad/s, we find K
s = 23 kg s−2. The restoring force is given by F =
−Ksx, which implies |F | = 23 N for x equal to unity.
(d) The oscillator density can be found from eqn 2.19. ²st= 5.9 has been
found in part (a), and ²∞ can be read from the graph for ω À ω0 as
²∞= n2= (1.45)2= 2.10. We thus have ²st− ²∞= 3.8. Using the values
of ω0 and µ worked out previously, we then find N = 3.0 × 1028m−3.
(e) γ is equal to the shift between the maximum and minimum in the re-fractive index in angular frequency units. We can only make a rough esti-mate of γ because the data does not follow a simple line shape. The damp-ing rate depends strongly on the frequency, which is why the resonance line is asymmetric. By comparison with Fig. 2.4 we find ∆ν ∼ 1×1012Hz,
and hence γ = 2π∆ν ∼ 6 × 1012s−1.
(f) The result of Exercise 2.5 tells us that α = N e2/n²
0µγc at the line
centre for a weak absorber. This limit does not really apply here, but we can still use it to get a rough answer. On inserting the values of N , µ and γ found above, and taking n ∼ 2, we find α ∼ 1 × 106m−1.
(2.7) We start by re-writing eqn 2.25 as: 1 vg
= dk dω.
We then substitute k = nω/c and v = c/n to obtain: 1 vg = 1 c µ n + ωdn dω ¶ , = n c µ 1 +ω n dn dω ¶ , = 1 v µ 1 + ω n dn dω ¶ .
The first relationship in eqn 2.26 follows immediately by taking the recip-rocal.
The second relationship in eqn 2.26 is obtained by substituting λ = 2πc/ω so that dn dω = dn dλ dλ dω = − λ2 2πc dn dλ, and hence: vg= v[1 + (2πc/λn)(−λ2/2πc)dn/dλ]−1 = v[1 − (λ/n)dn/dλ]−1.
(2.8) We consider three separate frequency regions.
(i) ω < ω0: In this frequency region ²ris real, and increases with frequency.
Since n =√²r, it is apparent that dn/dω is positive, so that from eqn 2.26
we see that vg< v. Since ²r> 1, n > 1, and hence v = c/n < c. Therefore
vg< c.
(ii) ω0< ω < (ω02+N e2/²0m0)1/2: In this frequency region, ²ris negative.
The refractive index is purely imaginary and the wave does not propagate. This is an example of the Reststrahlen effect discussed in Section 10.2.3. (iii) ω > (ω2
0+ N e2/²0m0)1/2: In this region ²r is positive and increases
with frequency, approaching unity asymptotically. dn/dω is therefore pos-itive, but we cannot use the same line of argument as part (i) because n < 1 and therefore v > c. We must therefore work out vg explicitly using
eqn 2.26. With n =√²r, we obtain:
dn dω = d dω µ 1 + N e 2 ²0m0 1 ω2 0− ω2 ¶1/2 = 1 n N e2 ²0m0 ω (ω2 0− ω2)2 . Hence 1 vg = n c + N e2 nc²0m0 ω2 (ω2 0− ω2)2 , = 1 nc µ n2+ N e2 ²0m0 ω2 (ω2 0− ω2)2 ¶ , = 1 nc µ ²r+ N e 2 ²0m0 ω2 (ω2 0− ω2)2 ¶ . On substituting for ²rfrom the exercise, we find:
1 vg = 1 nc µ 1 + N e 2 ²0m0 ω2 0 (ω2 0− ω2)2 ¶ ,
and hence: vg= nc µ 1 + N e 2 ²0m0 ω2 0 (ω2 0− ω2)2 ¶−1 . The denominator is greater than unity, and n < 1, so vg< c.
(2.9) (a) Consider a dipole p placed at the origin. The electric field generated at position vector r is given by:
E(r) = 3(p · r)r − r
2p
4π²0r5 .
The electric field generated at the origin by a dipole p at position vector r is therefore given by:
E(−r) = 3(p · (−r))(−r) − r
2p
4π²0r5 =
3(p · r)r − r2p
4π²0r5 .
Consider the ith dipole within the sphere illustrated in Fig. 2.8. We assume that the dipole is oriented parallel to the z axis so that we can write pi = (0, 0, pi). Then, on writing ri = (xi, yi, zi) in the formula for
E, we find that the z component of the field at the origin from the ith dipole is: Ei=pi(3z 2 i − ri2) 4π²0r5i .
We now sum over the cubic lattice of dipoles within the sphere. By sym-metry, the x and y components sum to zero, giving a resultant field along the z axis of magnitude:
Esphere= 1 4π²0 X i pi3z 2 i − r2i r5 i , as required.
(b) If all the dipoles have the same magnitude p, then the resultant field is given by: Esphere = p 4π²0 X i 2z2 i − x2i − yi2 r5 i .
The x, y and z axes are equivalent for the cubic lattice within the sphere, and so we must have:
X i x2 i r5 i =X i y2 i r5 i =X i z2 i r5 i . It is thus apparent that
X i 2z2 i − x2i − y2i r5 i = 0 . The net field is therefore zero.
(c) Consider a hollow sphere of radius a placed within a polarized dielectric medium as illustrated in Fig. 6. (cf. Fig 2.8.) We assume that the polar-ization is parallel to the z axis. The surface charge on the sphere must
balance the normal component of the polarization P . With P = (0, 0, P ), the normal component at polar angle θ is equal P cos θ, as shown in Fig. 6. Hence the surface charge density σ at angle θ is equal to −P cos θ. The charge contained in a circular element at angle θ subtending an incremen-tal angle dθ as defined in Fig. 6 is then given by:
dq = σ dA = −P cos θ × (2πa sin θ · a dθ) = −2πP a2cos θ sin θ dθ .
The x and y components of the field generated at the origin by this in-cremental charge sum to zero by symmetry, leaving just a z component, with a magnitude given by Coulomb’s law as:
dEz = − dq
4π²0a2
cos θ = +P cos2θ sin θ dθ 2²0
. On integrating over θ, we then obtain:
Ez= Z π θ=0 dEz= P 2²0 Z π 0 cos2θ sin θ dθ = P 3²0.
Since P is parallel to the z axis, and the x and y components of E are zero, we therefore have:
E = P 3²0,
as required for eqn 2.28.
+ + + + + ++
-
--
-P
q
dq
a
dqP
cosq
+ + + + + ++ + + + + + ++-
--
-P
q
dq
a
dqP
cosq
Figure 6: Definition of angles and charge increment as required for Exercise 2.9(c).
(2.10) If ²r− 1 is small, the left hand side of the Clausius–Mossotti relationship
becomes equal to (²r− 1)/3, and we then find:
²r= 1 + N χa≡ 1 + χ ,
where χ = N χa, as in eqn A.4. It is apparent that ²r− 1 will be small
if either N is small or χa is small. This means that we either have a low
density of absorbing atoms (as in a gas, for example), or we are working at frequencies far away from any resonances.
(2.11) We are working with a gas, and we can therefore forget about Clausius-Mossotti. At s.t.p. we have NA (Avogadro’s constant) molecules in a
volume of 22.4 litres. Hence N = 2.69 × 1025m−3. We then find χ afrom:
χa= (²r− 1)/N = 2.2 × 10−29m3.
The atomic dipole is worked out from p = ²0χaE .
The displacement of an electron by 1˚A produces a dipole of 1.6×10−29C m.
Hence we require a field of 0.8×1011 V/m. The field acting on an electron
at a distance r from a proton is given by Coulomb’s law as: E = e
4π²0r2
.
On substituting r = 1 ˚A, we find E = 1.4 × 1011 V/m. It is not surprising
that these two fields are of similar magnitude because the external field must work against the Coulomb forces in the molecule to induce a dipole. (2.12) We are given that α(E) = α0 for E2≥ E ≥ E1, where E is the photon
energy, and α(E) = 0 at all other energies. By using eqn 1.19 we then have:
κ(E) = c~α0/2E , E2≥ E ≥ E1,
κ(E) = 0 , elsewhere .
We can then find the refractive index from eqn 2.36 as follows: n(E) = 1 + 2 π Z E2/~ E1/~ ω0κ(ω0) ω02− ω2dω 0, = 1 + 2 π Z E2 E1 E0κ(E0) E02− E2dE 0.
Note that we do not have to worry about taking principal parts because κ(E) = 0 when E0= E if E < E
1. On substituting for κ(E0) we find:
n(E) = 1 +c~α0 π Z E2 E1 1 E02− E2dE 0. Now: Z 1 x2− a2dx = 1 2a Z · 1 x − a− 1 x + a ¸ dx , = 1 2aln µ x − a x + a ¶ . Hence: n(E) = 1 + c~α0 π 1 2E · ln µ E0− E E0+ E ¶¸E2 E1 , = 1 + c~α0 2πE · ln µ E2− E E2+ E ¶ − ln µ E1− E E1+ E ¶ ¸ , = 1 + c~α0 2πE ln µ (E2− E)(E1+ E) (E2+ E)(E1− E) ¶ , as required.
(2.13) (a) If we are far away from resonance frequencies, we can ignore the damping term, and write eqn 2.24 as:
²r≡ n2= 1 + N e2 ²0m0 X j fj ω2 0j− ω2 . On substituting ω = 2πc/λ, this becomes:
n2= 1 + N e2 ²0m0 1 (2πc)2 X j fjλ2jλ2 λ2− λ2 j , where λj = 2πc/ω0j. This is of the Sellmeier form if we take:
Aj = N e2fjλ2j/4π2²0m0c2.
(b) With the approximations stated in the exercise, we have: n2= 1 + A1λ2
λ2− λ2 1
= 1 + A1(1 − λ21/λ2)−1.
With x ≡ λ2
1/λ2¿ 1, we can expand this to:
n2= 1 + A 1(1 + x + x2+ · · · ) , which implies: n = [(1 + A1) + A1(x + x2+ · · · )]1/2 = (1 + A1)1/2[1 + A1/(1 + A1)(x + x2+ · · · )]1/2 = (1 + A1)1/2[1 + (1/2)A1/(1 + A1)(x + x2) − (1/8)(A1/(1 + A1))2(x + x2)2+ · · · ] = (1 + A1)1/2+ A1 2√1 + A1 x + µ A1 2√1 + A1 − A 2 1 8(1 + A1)3/2 ¶ x2+ · · · On re-substituting for x, we then find:
n = (1 + A1)1/2+ A1 2√1 + A1 λ2 1 λ2 + µ A1 2√1 + A1 − A21 8(1 + A1)3/2 ¶ λ4 1 λ4,
which shows that:
C1 = (1 + A1)1/2,
C2 = A1λ21/2(1 + A1)1/2,
C3 = A1(4 + 3A1)λ41/8(1 + A1)3/2.
Note that the Cauchy formula generally applies to transparent materials (eg glasses) in the visible spectral region. In this situation, the dispersion is dominated by the electronic absorption in the ultraviolet. We should then take λ1 as the wavelength of the band gap, and the approximation
λ2
1/λ2 ¿ 1 will be reasonable, as we are far away from the band gap
(2.14) (a) On neglecting the λ4 term in Cauchy’s formula, we have
n = C1+ C2/λ2.
On inserting the values of n at 402.6 nm and 706.5 nm and solving, we find C1= 1.5255 and C2= 4824.7 nm2, so that we have:
n = 1.5255 + 4824.7/λ2, where λ is measured in nm.
(b) The values are found by substituting into the result found in part (a) to obtain n = 1.5493 at 450 nm and n = 1.5369 at 650 nm.
(c) Referring to the angles defined in Fig. 7, we have sin θin
sin θ1 =
sin θout
sin θ2 = n ,
from Snell’s law. Furthermore, for a prism with apex angle α, we have θ1+ θ2= α. With θin= 45◦ and α = 60◦, we then find θout = 57.17◦ for
n = 1.5493 (450 nm) and θout = 55.91◦ for n = 1.5369 (650 nm). Hence
∆θout= 1.26◦. q in q out q 2 q 1 60° q in q out q 2 q 1 60°
Figure 7: Angles required for the solution of Exercise 2.14. (2.15) The transit time is given by:
τ = L vg = L
dk dω, which, with k = nω/c, becomes:
τ =L c µ n + ωdn dω ¶ .
We introduce the vacuum wavelength λ via ω = 2πc/λ, and write: dn dω = dn dλ· dλ dω = − λ2 2πc dn dλ. We then have (cf. eqn 2.26 with τ = L/vg):
τ = L c µ n − λdn dλ ¶ .
The difference in the transit time for two wavelengths separated by ∆λ, where ∆λ ¿ λ, is given by:
∆τ = dτ dλ∆λ .
On using the result for τ above, we find: dτ dλ = − L c λ d2n dλ2, which implies: ∆τ = L µ −λ c d2n dλ2 ¶ ∆λ = L D ∆λ ,
where D is the material dispersion parameter defined in eqn 2.40. We therefore finally obtain:
|∆τ | = L ¯ ¯ ¯ ¯−λc d 2n dλ2 ¯ ¯ ¯ ¯ ∆λ = L |D| ∆λ .
The time–bandwidth product of eqn 2.39 implies that a pulse of light contains a spread of frequencies and therefore a spread of wavelengths. In a dispersive medium, the different wavelengths will travel at different velocities, and this will cause pulse broadening. The precise amount of broadening depends on the numerical value of the time–bandwidth prod-uct assumed for the pulse. For a 10 ps pulse with ∆ν∆t = 1, we have ∆ν = 1011Hz. With λ = c/ν, we have |∆λ| = (λ2/c)|∆ν|, so that
|∆λ| = 0.8 nm in this case. We thus find:
|∆τ | = L|D|∆λ = 1 km × 17 ps km−1nm−1× 0.8nm = 14 ps . direction of propagation e-ray polarization vector index ellipsoid n e n o no n e n(q ) y z q direction of propagation e-ray polarization vector index ellipsoid n e n o no n e n(q ) y z q
Figure 8: Index ellipse for the e–ray of a wave propagating at an angle θ to the optic (z) axis of a uniaxial crystal, as required for Exercise 2.16.
(2.16) It is apparent from eqn 2.50 that ²11/²0= ²22/²0 = n2oand ²33/²0= n2e.
Hence we can re-cast the index ellipsoid in the form: x2 n2 o +y 2 n2 o +z 2 n2 e = 1 .
Owing to the x-y symmetry about the optic (z) axis, we can choose the axes of the index ellipsoid so that the x axis coincides with the polarization
vector of the o–ray as in Fig 2.13(a). The polarization of the e–ray will then lie in the y–z plane, as in Fig. 2.13(b). The projection of the index ellipsoid onto the plane that contains the direction of propagation and the polarization vector of the e–ray thus appears as the ellipse drawn in Fig. 8. The refractive index n(θ) that we require is the distance from the origin to the point of the ellipse where the E-vector cuts it. The co-ordinates of this point are x = 0, y = n(θ) cos θ and z = n(θ) sin θ. On substituting into the equation of the index ellipsoid, we then have:
0 + n(θ) 2cos2θ n2 o +n(θ) 2sin2θ n2 e = 1 , which implies: 1 n(θ)2 = cos2θ n2 o +sin 2θ n2 e , as required.
(2.17) The condition for total internal reflection for a medium–air interface is that the angle of incidence should exceed the critical angle θc given by:
sin θc = 1/n ,
where n is the refractive index of the medium. We see from Table 2.1 that no = 1.658 and ne = 1.486 for calcite. The critical angles for the
o- and e-rays are therefore 37.1◦ and 42.3◦respectively. For the polarizer
to work, we want the o-ray to suffer total internal reflection, but not the e-ray. With normal incidence as shown in Fig. 2.14, this will occur if the apex angle θ lies in the range 37.1◦− 42.3◦.
optic axis (z) Input E-vector front surface of wave plate y
q
E z E y (a) optic axis (z) Input E-vector direction yq
E z -Ey Output E-vector (b) optic axis (z) Input E-vector front surface of wave plate yq
E z E y (a) optic axis (z) Input E-vector direction yq
E z -Ey Output E-vector (b)Figure 9: (a) Input polarization vector for light propagating along the x direc-tion, relative to the crystal axes, as required for the solution of Exercise 2.18(a). (b) Output polarization of the half-wave plate.
(2.18) (a) We resolve the input polarization into two components, one along the optic axis, and one orthogonal to it, as shown in Fig. 2.15(b). We define our axes so that z lies along the optic axis, and y lies in the plane of the front surface of the waveplate, as shown in Fig. 9(a). The x axis is taken
to be the direction of propagation. If the input beam has amplitude E0,
the input polarization has components: Einz = E0cos θ ,
Einy = E0sin θ .
The component along z is the e-ray and the one along y is the o-ray. In a half-wave plate, the phase of the o-ray is shifted by π relative to the e-ray at the output of the retarder plate. This means that the output polarizations will be given by:
Eoutz = E0cos θ = Einz ,
Eouty = E0eiπ sin θ = −E0sin θ = −Einy ,
as shown in Fig. 9(b). The resultant E-vector points at an angle −θ with respect to the optic axis. Hence the output polarization is rotated by an angle 2θ with respect to the input polarization.
(b) If we set θ = 45◦, the input polarization will be given by:
Ein z = E0cos 45◦= E0/ √ 2 , Ein y = E0sin 45◦= E0/ √ 2 .
The quarter-wave plate introduces a phase difference of π/2 between the e-ray and o-rays. The output polarization will therefore be:
Eoutz = E0/ √ 2 , Eouty = (E0/ √ 2)eiπ/2= i E 0/ √ 2 .
The output therefore consists of two orthogonal linearly polarized com-ponents of equal amplitude with a phase difference of 90◦ between them.
This is circularly-polarized light.
(c) When θ 6= 45◦, the amplitudes of the e- and o-rays will be different.
The quarter-wave plate still introduces a phase shift of 90◦between them,
and so the output consists of two orthogonal linearly polarized components with unequal amplitude and with a 90◦ phase difference between them.
This is elliptically polarized light.
(2.19) We follow Example 2.4, but set the phase difference ∆φ to be π/2 because we have a quarter-wave plate instead of a half-wave plate. We therefore require:
∆φ = π 2 =
2π|∆n|d
λ ,
which implies d = λ/4|∆n|. In this case we have |∆n| = 0.0091 and λ = 500 nm, and so we require d = 1.4 × 10−5m.
In this exercise we have neglected the optical activity of the quartz plate because the thickness is very small. In a thicker multiple-order waveplate (i.e one with ∆φ = (2πm + π/2), where m is an integer) a small correction would have to be introduced to account for the optical activity.
(2.20) The crystal will be isotropic if the medium has high symmetry so that the x, y and z axes are equivalent. If not, it will be birefringent. For the crystals listed in the Exercise we have:
Crystal Structure x, y, z equivalent ? Birefringent ?
(a) NaCl cubic (fcc) yes no
(b) Diamond cubic yes no
(c) Graphite hexagonal no yes
(d) Wurtzite hexagonal no yes
(e) Zinc blende cubic yes no
(f) Solid argon cubic (fcc) yes no
(g) Sulphur orthorhombic no yes
The two hexagonal crystals are uniaxial, with the optic axis lying along the direction perpendicular to the hexagons. Sulphur has the lowest symmetry and is the only biaxial crystal included in the list.
(2.21) (a) The birefringent phase shift for a refractive difference of ∆n is given by eqn 2.46. For a half wavelength shift, this is equal to π. On substituting for the Kerr-induced birefringence from eqn 2.51, we obtain:
φ = π =2π|∆n|d
λ =
2π(λKE2)d
λ = 2πKE
2d .
On solving for the field, we then find: Eλ/2= 1/
√ 2Kd , as required.
(b) We substitute for K and d to find: Eλ/2 = 1/
p
2 × (8.7 × 10−14) × 0.02 = 1.7 × 107V/m .
The field is dropped across a distance of 5 mm, and so the voltage required is Vλ/2= Eλ/2× 0.005 = 85 kV.
As Table 2.2 shows, the value of K for the chalcogenide considered here is large for a glass, but the voltage required to make a Kerr cell is still very high. Practical Kerr cells use liquids with larger Kerr constants, for example, nitrobenzene.
(2.22) (a) Set up axes so that z corresponds to the direction of propagation and x to the input polarization. With these definitions the input polarization can be written as:
Ein= E0ˆx ,
where E0 is the amplitude of the light.
Circular polarization consists of two orthogonally polarized waves of equal amplitude with a 90◦ phase difference between them, and so we can write
left and right circular light as: σ+ = E 0(ˆx + iˆy)/ √ 2 , σ− = E 0(ˆx − iˆy)/ √ 2 ,
where the factor of i ≡ eiπ/2 represents the 90◦ phase shift. This allows
us to re-write the input polarization in terms of the left and right circular light as:
Ein= E0(σ++ σ−)/
√ 2 .
An optically-active medium will introduce a phase difference φ between the left- and right-circular components, where:
φ =2π λ ∆n d .
Here, λ is the vacuum wavelength, ∆n = (nR− nL) is the difference of the
refractive indices for right- and left-circular light, and d is the thickness of the medium. The output wave will therefore be of the form:
Eout= E0(σ++ eiφσ−)/
√ 2 .
We revert to the linear basis by substituting for σ+and σ−and re-writing
this as:
Eout= E0
¡
(ˆx + iˆy) + eiφ(ˆx − iˆy)¢/2 .
Collecting the terms in ˆx and ˆy, we find: Eout = E0
¡
(1 + eiφ)ˆx + i(1 − eiφ)ˆy)¢/2 , = E0eiφ/2
³
(e−iφ/2+ eiφ/2)ˆx + i(e−iφ/2− eiφ/2)ˆy)´/2 ,
= E0eiφ/2(2 cos(φ/2) ˆx + i(−2i sin(φ/2)) ˆy) /2 ,
= E0eiφ/2(cos(φ/2) ˆx + sin(φ/2) ˆy) .
This shows that the output polarization is linearly polarized at an an-gle φ/2 with respect to the x axis. The polarization rotation anan-gle θ is therefore equal to φ/2, and so we find:
θ = φ 2 = π λ∆n d = πd λ (nR− nL) .
(b) Using the result from part(a), we find that the rotatory power of an optically-active medium is given by:
RP = |θ/d| = π|∆n|/λ . For quartz we thus find
RP = π(7.1 × 10−5)/589 × 10−9= 379 rad/m = 21.7◦/mm .
(2.23) We apply a Lorentzian model according to the theory in Section 2.2. The Lorentzian lineshapes are shifted for σ+ and σ− light due to the Zeeman
effect, which shifts the energy of the transition from ~ω0 to ~ω0± µBB
according to the circular polarization.
(a) The Faraday rotation is caused by the difference of the real part of the refractive index for σ+ and σ−light. Figure 10(a) plots the refractive
0.6 0.8 1.0 1.2 1.4 0.96 0.98 1.00 1.02 1.04
R
e
fr
a
ct
iv
e
in
d
e
x
n
Angular frequency (w/w
0)
-0.04 -0.02 0.00 0.02 0.04D
n
0.6 0.8 1.0 1.2 1.4 -0.06 -0.04 -0.02 0.00 0.02 0.04 0.06E
x
ti
n
ct
io
n
co
e
ff
ici
e
n
tk
Angular frequency (w/w
0)
(a)
(b)
D
k
n
+n
-D
n
D
n
k
+k
-Dk
0.6 0.8 1.0 1.2 1.4 0.96 0.98 1.00 1.02 1.04R
e
fr
a
ct
iv
e
in
d
e
x
n
Angular frequency (w/w
0)
-0.04 -0.02 0.00 0.02 0.04D
n
0.6 0.8 1.0 1.2 1.4 -0.06 -0.04 -0.02 0.00 0.02 0.04 0.06E
x
ti
n
ct
io
n
co
e
ff
ici
e
n
tk
Angular frequency (w/w
0)
(a)
(b)
D
k
n
+n
-D
n
n
+n
-D
n
D
n
k
+k
-Dk
k
+k
-Dk
Figure 10: Lorentzian model calculation of Faraday effect and magnetic circular dichroism, as considered in Exercise 2.23. (a) Faraday rotation. (b) Magnetic circular dichroism.
difference. The lines are plotted against the normalized angular frequency (ω/ω0) for a Lorentzian line with γ = 0.05ω0. The value of µBB/~ is set
to be equal to γ. The value of ∆n = Re(˜n+− ˜n−), and hence the Faraday
rotation is negative above and below the line, and positive near ω0. The
rotation decays as the frequency is tuned away from resonance.
(b) The magnetic circular polarization is found by calculating the differ-ence ∆κ of the imaginary part of (˜n+− ˜n−). This is shown in Fig. 10 for
the same Lorentzian line as in part (a). We see that the magnetic circular dichroism follows a dispersive lineshape, with a negative signal below ω0
that peaks at ω0− µBB/~, and a positive signal above ω0 that peaks at
ω0+ µBB/~. The signal precisely at ω0 is zero.
(2.24) We require a Faraday rotation of π/4 at a magnetic field of 0.5 T. The Faraday rotation is given by eqn 2.53, and so the thickness of glass is given
by:
d = θ V b =
π/4
9.0 × 0.5 = 0.17 m = 17 cm .
This long length is impractical, and real Faraday isolators use specialist glasses or crystals with larger Verdet coefficients.
Chapter 3
Interband absorption
(3.1) The Born–von Karman boundary conditions are satisfied when: kxL = nxL ,
kyL = nyL ,
kzL = nzL ,
where nx, ny and nz are integers. The wave vector is therefore of the
form:
k = (2π/L)(nx, ny, nz) .
The allowed values of k form a grid as shown in Fig. 11. Each allowed k state occupies a volume of k space equal to (2π/L)3. This implies that
the number of states in a unit volume of k space is L3/(2π)3. Hence a
unit volume of the material would have 1/(2π)3states per unit volume of
k space.
In most calculations we actually require the density of states in k space as given in eqn 3.15. This is found by calculating the number of states within the incremental shell between k vectors of magnitude k and k + dk, as shown in Fig. 11. This volume of the incremental shell is equal to 4πk2dk,
and contains 4πk2dk × 1/(2π)3= k2dk/2π2states. Hence g(k) = k2/2π2.
kx
ky
2p/L k
dk
Figure 11: Grid of allowed values of k permitted by the Born–von Karman boundary conditions, as considered in Exercise 3.1. The points of the grid are separated from each other by distance 2π/L in all three directions, giving a volume per state of (2π/L)3. Note that the diagram only shows the x-y plane
of k space. The incremental shell considered for the derivation of eqn 3.15 is also shown.
(3.2) With E(k) = ~2k2/2m∗, we have:
dE dk =
~2k
m∗ .
On inserting into eqn 3.14–15 and substituting for k, we find: g(E) = 2k 2/2π2 ~2k/m∗ = m∗k π2~2 = m∗ π2~2 µ 2m∗E ~2 ¶1/2 = 1 2π2 µ 2m∗ ~2 ¶3/2 E1/2, as required.
(3.3) (a) The parity of a wave function is equal to ±1 depending on whether ψ(−r) = ±ψ(r). Atoms are spherically symmetric, and so measurable properties such as the probability amplitude must possess inversion sym-metry about the origin: i.e. |ψ(−r)|2 = |ψ(r)|2. This is satisfied if
ψ(−r) = ±ψ(r). In other words, the wave function must have a defi-nite parity.
(b) r is an odd function, and so the integral over all space will be zero unless the product ψ∗
fψiis also an odd function. This condition is satisfied
if the two wave functions have different parities (parity selection rule). Since the wave function parity is equal to (−1)l, the parity selection rule
implies that ∆l is an odd number.
(c) In spherical polar co-ordinates (r, θ, φ) we have: x = r sin θ cos φ = r sin θ (eiφ+ e−iφ)/2 , y = r sin θ sin φ = r sin θ (eiφ− e−iφ)/2i ,
z = r cos θ .
The selection rules on m can be derived by considering the integral over φ. For light polarized along the z axis we have:
M ∝ Z 2π
φ=0
e−im0φ
· 1 · eimφdφ ,
since z is independent of φ. The integral is zero unless m0 = m. The
selection rule for z-polarized light is therefore ∆m = 0. For x or y polarized light we have:
M ∝ Z 2π
φ=0
e−im0φ
(eiφ± e−iφ) eimφdφ ,
which is zero unless m0= m±1. We thus have the selection rule ∆m = ±1
for light linearly polarized along the x or y axis. (d) From eqn A.40 we have that:
E+ = E0(ˆx + iˆy)/ √ 2 , E− = E0(ˆx − iˆy)/ √ 2 ,
where E±are the electric fields for σ±polarizations. With x = r sin θ cos φ
and y = r sin θ sin φ, we therefore find:
E+ = E0r sin θ(cos φ + i sin φ)/
√
2 ∝ e+iφ, E− = E0r sin θ(cos φ − i sin φ)/
√
2 ∝ e−iφ. On inserting these into the matrix element, we have for σ+ light:
M+∝ Z 2π
φ=0
e−im0φe+iφeimφdφ ,
which is zero unless m0 = m + 1. Similarly, for σ− light we have:
M−∝
Z 2π φ=0
e−im0φ
e−iφeimφdφ ,
which is zero unless m0 = m − 1. With circularly polarized light we
there-fore have ∆m = +1 for σ+polarization and ∆m = −1 for σ−polarization.
(3.4) The apparatus required is basically the same as for Figs 3.14–15, but with modifications to take account of the fact that the required energy range of 0.3–0.6 eV corresponds to a wavelength range of 2–4 µm. This means that a detector with a band gap smaller than 0.3 eV must be used, e.g. InSb. (See Table 3.3.) As InSb array detectors are not available, a scanning monochromator with a single channel detector would normally be used. A thermal source would suffice as the light source.
Another point to consider is that standard glass lenses do not transmit in this wavelength range, and appropriate infrared lenses would have to be used, e.g. made from CdSe. (See Fig. 1.4(b).) Also, since the data is taken at room temperature, no cryostat is needed.
Figure 12 gives a diagram of a typical arrangement that could be used.
InAs sample scanning monochromator computer
:
infrared lenses white light source InSb detector InAs sample scanning monochromator computer:
infrared lenses white light source InSb detector InSb detectorFigure 12: Apparatus for measuring infrared absorption spectra in the range 2–4 µm, as discussed in Exercise 3.4.
(3.5) The type of band gap can be determined from an analysis of the variation of the absorption coefficient α with photon energy. The material is direct or indirect depending on whether a graph of α2 or α1/2 against ~ω is a
