Chapter 14
What are lasers?
A laser is a device that produces light by stimulated emission; the word laser being
an acronym for Light Amplification by Stimulated Emission of Radiation. When an atom in
an excited state makes a transition to the ground state, energy with a frequency  will be
emitted.
In 1917, Einstein suggested that there should be two possible types of emission
process rather than just one.
The most obvious is that an atom in an excited state can
randomly change to the ground state; a process called spontaneous emission.
Alternatively, a photon having energy equal to the energy difference between the two levels
can interact with the atom in the excited state causing it to fall to the lower state and emit a
photon at the same time; a process called stimulated emission.
Lasers make use of this
latter process to obtain light which is both coherent and polarised.
The key to laser action is to obtain atoms or molecules in a metastable excited state
and keep them there long enough for photons to pass and trigger stimulated emission.
This is achieved in ruby and neodymium lasers by optical pumping, in which an intense
beam of light excites large numbers of atoms into the appropriate energy state to allow for
stimulated emission to occur. In semiconductor lasers the metastable state is achieved by
using a high current across the device.
Why are thin films often brightly coloured?
Monochromatic light travelling through air falling upon a homogeneous thin film of an
insulator will be reflected from the top surface to give a reflected ray while the light
transmitted into the film will be repeatedly reflected from the bottom surface and the
underside of the top surface.
At each reflection, some of the light will escape to produce
additional reflected and transmitted rays.
Because of the difference in the paths taken by
the repeatedly reflected rays, the waves will interfere with each other. Depending upon the
thickness of the film, these will undergo constructive or destructive interference.
When a thin transparent film is viewed in white light the contributions of all of the
different wavelengths present must be added.
For any particular film thickness some of
the colours will be reinforced by constructive interference while others will be diminished by
destructive interference. The result of this is that the film will appear coloured. As the film
thickness increases or decreases, the sequence of colours seen varies in a cyclical fashion
as certain colours are either reinforced or cancelled. Each sequence of spectral colours is
called an order, which starts with the first order for the thinnest of films.
A new order
begins every 550 nm of retardation.
How do solar cells work?
Traditional solar cells exploit the electronic properties of semiconductors.
If
radiation of a suitable wavelength falls on a semiconductor, it will excite electrons across
the band gap giving rise to a voltage and a related increase in conductivity.
Solids that
behave in this way are called photovoltaic materials. The magnitude of this photovoltaic or
photoconductive effect is roughly proportional to the light intensity.
in a similar way to a single piece of semiconductor.
A p-n junction can act
However, the control afforded by the
junction makes the device, a photodiode, far more flexible and photodiodes form the basis
of solar cells.
Holes and electrons produced in the junction region by sunlight are swept
across the depletion region by the high built-in space charge present, the electrons going
from the p to n region and the holes from the n to p region.
This process produces a
photovoltage and causes a photocurrent to flow, thus producing energy that can be stored
or used to power devices.
Other types of solar cell rely on chemical reactions similar to that employed in natural
photosynthesis, where the central reactions are oxidation and reduction.
solar cells (DSSCs) are typical of this group.
Dye sensitized
The task of harvesting the light is left to a
sensitizer, which is a dye molecule, and the carrier transport task is allocated to a
semiconductor.
The generation of electrons and holes takes place in a dye which is
coated onto a semiconductor of large surface area, typically TiO 2.
Because the charge
separation takes place in the dye, the purity and defect structure of the semiconductor are
not crucial to satisfactory operation.
The electrons and holes so generated enter the
external circuit and can be used in the same way as n a traditional solar cell.
The dye is
restored to its initial state by a series of oxidation – reduction reactions taking place in a
liquid electrolyte which is in contact with the anode and cathode of the cell.
Quick quiz
1. a; 2. b; 3. b; 4. c; 5. c; 6. a; 7. a; 8. c; 9. b; 10. a; 11. c; 12. a; 13. b; 14. c;
15. b; 16. c; 17. c; 18. b; 19. a; 20. b; 21. a; 22. c; 23. c; 24. b; 25. c.
Calculations and questions
1.  425 nm,  = 7.05 x 1014 Hz, E = 4.67 x 10-19 J, violet;  575 nm,  = 5.21 x 1014 Hz, E
= 3.45 x 10-19 J, yellow-green;  630,  = 4.76 x 1014 Hz, E = 3.15 x 10-19 J, orange-red.
2. 299 kJ mol-1.
3. 7.97 kJ mol-1.
4. 242 nm.
5. 838 nm.
6. 6.9 x 10-5 m.
7.
(a) 3.57 x 10-19 J, 4.88 x 10-19 J; (b) 3.55 x 10-38, 6.42 x 10-52.
8. (a) 2.68 x 10-19 J; (b) 1.00 x 10-30.
9. (a) 2.17 x 10-19 J; (b) 1.87 x 10-19 J.
10. (a) GaAs, 918.4 nm, infrared, AlAs, 574 nm, yellow; (b) x = 0.075.
11. (a) GaN, 371.0 nm, ultraviolet, InN, 620 nm, orange-red; (b) x = 0.38.
12. (a) x = 0.4; (b) deep red,   700 nm.
13. 0.66.
14. (a) Zn, 44 cm, Cd, 12.5 cm; (b) transmittance, 0.1, absorbance, 1.
15. (a) Transmittance, 0.05, Absorbance, 1.3010; (b) 35 at. % Zn : 65 at. % Cd.
16. 62.9
17. (a) 19.2; (b) no critical angle; (c) 57.3.
18. (a) BaTiO3, 2.34; (b) PbTiO3, 2.78.
19. (a) spinel, 1.75; (b) akermanite, 1.61.
20. beryl, 1.57; garnet, 1.74.
21. 0.20.
22. (a) Na3AlF6, 0.022; (b) glass, 0.045; (c) Al2O3, 0.06; (d) ZrO2, 0.12; (e) Ta2O5, 0.13.
23. (a) Al, 0.916; (b) Ag, 0.919; (c) Au, 0.817; (e) Cr, 0.48, (f) Ni, 0.76.
24. (a), bright, reflecting; (b), dark, non-reflecting.
25. (a), dark, non-reflecting; (b), bright, reflecting.
26. reflecting in air, non-reflecting on oil.
27.
(a) 47.8 nm; (b) reflected colour yellow-white to straw yellow; (c) transmitted colour,
carmine red to deep violet.
29. (a) SiO, 0.21; (b) TiO2, 0.38.
30. (a) MgF2, /4, 0.015, /2, 0.041; (b) TiO2, /4, 0.04, /2, 0.23.
31. 6 layers.
32. Ta2O5, 75.6 nm; cryolite, 120.4 nm.
33. 2.20 m-1.
34. 4.6 x 10-4 m-1.
35. 1.7 x as much light is scattered by the dust.
36. limestone will cause about twice as much scattering as the water.
37. 10.5 km.
38. (a), 2; (b), 1.5; (c), 1; (d), 2.
39. (a) 5.36 x 10-6 m; (b) 4.28.
40. 14.4 mm.
41. (a) 638 nm; (b) orange-red; (c) no change.
42. 0.0218 dB km-1.
43. 3.01 mm.
44. 2060 km.
45. (a) yellow – green; (b) 1.83 nm.
46. 309 nm.
Solutions
1
Calculate the frequency and energy of photons associated with wavelengths 425 nm,
575 nm and 630 nm. What colours are attributed to these wavelengths?
E = h  = hc / ;  = c / 
hc = 6.626 x 10-34 x 2.998 x 108 = 1.986 x 10-25
E = 1.986 x 10-25 / 
 = 425 nm: E = 4.67 x 10-19 J,  = 7.05 x 1014 s-1, violet
 = 575 nm: E = 3.45 x 10-19 J,  = 5.21 x 1014 s-1, green-yellow
 = 630 nm: E = 3.15 x 10-19 J,  = 4.76 x 1014 s-1, red
2 Light of wavelength 400 nm is shone through a gas of absorbing molecules. Calculate
the energy absorbed by one mole of gas if each molecule absorbs one photon.
From Q1:
E = 1.986 x 10-25 /  = 1.986 x 10-25 / 400 x 10-9
1 mole of gas will absorb: (1.986 x 10-25 x 6.022 x 1023) / 400 x 10-9
= 299 kJ
3 Carbon dioxide, CO2, is a greenhouse gas that absorbs infrared radiation escaping from
the Earth. What is the energy per mole absorbed by the gas if each CO 2 molecule absorbs
one photon of wavelength 15 m.
From Q1:
E = 1.986 x 10-25 /  = 1.986 x 10-25 / 15 x 10-6
1 mole of gas will absorb: (1.986 x 10-25 x 6.022 x 1023) / 15 x 10-6
= 7.97 kJ
4 The energy required to break the bond linking the two oxygen atoms in a molecule of O 2
is 495 kJ mol-1. What is the longest wavelength light that could cause this decomposition
to occur?
From Q1:
E = 1.986 x 10-25 / 
To break the bonds in 1 mole of oxygen requires 495 kJ, hence:
495 x 103 = (1.986 x 10-25 x 6.022 x 1023) / 
 = 2.42 x 10-7 m = 242 nm
5 The energy required to dissociate ozone, O 3, into O2 plus O is 142.7 kJ mol-1. What is
the longest wavelength light that will dissociate ozone in the upper atmosphere?
From Q1:
E = 1.986 x 10-25 / 
To break the bonds in 1 mole of ozone requires 142.7 kJ, hence:
147.2 x 103 = (1.986 x 10-25 x 6.022 x 1023) / 
 = 8.38 x 10-7 m = 838 nm
6
Calculate the wavelength at which the rates of spontaneous and stimulated emission
become equal at 300 K.
The ratio of the rate of spontaneous to stimulated emission is:
R = exp [h  / kB T] – 1
When the rates are equal, R = 1, hence:
1 = exp [h  / kB T] – 1
2 = exp [h  / kB T]
[h  / kB T] = 0.69315
 = 0.69315 x 1.38 x 10-23 x 300 / 6.626 x 10-34
 = c /  = (2.9978 x 108 x 6.626 x 10-34 ) / (0.69315 x 1.38 x 10-23 x 300)
= 6.92 x 10-5 m
7 The optical transitions in ruby are: 4T2g  4A2g, 556 nm; 4T1g  4A2g, 407 nm. (a) What
is the energy gap of each of these transitions?
states relative to the ground state at 300 K?
(b) What are the populations of the upper
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
T2g  4A2g: E = 1.986 x 10-25 / 556 x 10-9 = 3.57 x 10-19 J
4
T1g  4A2g:
4
E = 1.986 x 10-25 /407 x 10-9 = 4.88 x 10-19 J
(b) N (4T1g) / N (4A2g) = exp[-E / kB T] = exp[-3.57 x 10-19 / (1.38066 x 10-23 x 300)
= 3.55 x 10-38
N (4T2g) / N (4A2g) = exp[-E / kB T] = exp[-4.88 x 10-19 / (1.38066 x 10-23 x 300)
= 6.42 x 10-52
8 The laser light from a ruby laser is at 694.3 nm. (a) What is the energy of the 2Eg lasing
state above the ground state, 4A2g? (b) Estimate the fraction of Cr3+ ions in this upper state
due to thermal equilibrium alone at 300 K.
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
Eg  4A2g: E = 1.986 x 10-25 / 694.3 x 10-9 = 2.86 x 10-19 J
2
(b) N (2Eg) / N (4A2g) = exp[-E / kB T] = exp[-2.86 x 10-19 / (1.38066 x 10-23 x 300)
= 1.00 x 10-30
9
What is the separation of the energy levels in neodymium ions that give rise to laser
lines at (a) 0.914 m and (b) 1.06 m.
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
E = 1.986 x 10-25 / 0.914 x 10-6 = 2.17 x 10-19 J
(b) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
E = 1.986 x 10-25 / 1.06 x 10-6 = 1.87 x 10-19 J
10 Gallium arsenide has a band gap of 1.35 eV, and aluminium arsenide has a band gap
of 2.16 eV. (a) What is the wavelength and colour of photons emitted by these solids? In
order to make an orange LED with an emission at a wavelength of 600 nm, it is proposed to
make a solid solution GaxAl1-xAs. (b) Taking the variation in band gap with composition as
linear, what value of x is required?
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
 = 1.986 x 10-25 / E (J)
GaAs:
Eg = 1.35 eV = 1.35 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 1.35 x 1.60218 x 10-19 = 918 nm, infrared
AlAs:
Eg = 2.16 eV = 2.16 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 2.16 x 1.60218 x 10-19 = 574 nm, yellow
(b) We need a band gap to give 600 nm emission;
Using similar triangles:
(918 – 574) / (1 – 0) = (600 – 574) / (x – 0)
x = 0.075
11 Gallium nitride has a band gap of 3.34 eV, and indium nitride has a band gap of 2.0 eV.
(a) What is the wavelength and colour of photons emitted by these solids?
In order to
make a green LED with an emission at a wavelength of 525 nm, it is proposed to make a
solid solution GaxIn1-xN.
(b) Taking the variation in band gap with composition as linear,
what value of x is required?
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
 = 1.986 x 10-25 / E (J)
GaN:
Eg = 3.34 eV = 3.34 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 3.34 x 1.60218 x 10-19 = 317 nm, ultraviolet
InN:
Eg = 2.00 eV = 2.00 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 2.00 x 1.60218 x 10-19 = 620 nm, orange-red
(b) We need a band gap to give 525 nm emission;
Using similar triangles:
(620 – 371) / (1 – 0) = (525 – 371) / (x – 0)
x = 0.38
12
A solid solution of indium phosphide and aluminium phosphide, In xAl1-xP is made up.
(a) At what value of x will the light emitted by an LED made from this compound just be
visible? (b) What colour will it be? The band gap of InP is 1.27 eV and that of AlP is 2.45
eV.
(a) From Q1: E = h  = hc /  = 1.986 x 10-25 / 
 = 1.986 x 10-25 / E (J)
InP:
Eg = 1.27 eV = 1.27 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 1.27 x 1.60218 x 10-19 = 976 nm, infrared
AlP:
Eg = 2.45 eV = 2.45 x 1.60218 x 10-19 J
 = 1.986 x 10-25 / 2.45 x 1.60218 x 10-19 = 506 nm, green-blue
As the solid solution evolves, light will just be visible as the emission passes from the
infrared into the visible, at  700 nm.
We need a band gap to give 700 nm emission;
Using similar triangles:
(976 – 506) / (1 – 0) = (700 – 506) / (x – 0)
x = 0.41
(b) Deep red,   700 nm
13 A solution is quoted as having a 22% transmittance. What is the absorbance?
Transmittance, T = I / I0; Absorbance, A = - log (transmittance)
A = - log (0.22) = 0.658
14 The linear absorption coefficient of zinc metal for X-rays from a nickel target is 5.187 m1
, and for cadmium metal is 18.418 m -1.
(a) What thickness of plates of these metals in
needed to reduce the irradiance of the radiation passing through the plate to 0.1 of the
incident irradiance. (b) What will be the transmittance and absorbance of the plates?
(a) Use:
I = I0 exp [- a L]
For Zn:
0.1 I0 = I0 exp [- 5.187 L]
0.1 = I0 exp [- 5.187 L]
-2.3026 = -5.187 L
L = 0.44 m = 44 cm
For Cd:
0.1 I0 = I0 exp [- 18.418 L]
0.1 = I0 exp [- 18.418 L]
-2.3026 = -18.418 L
L = 0.125 m = 12.5 cm
(b) Transmittance, T = I / I0; Absorbance, A = - log (transmittance)
T = 0.1 / 1 = 0.1
A = - log (0.1) = 1
15 A plate of a cadmium – zinc alloy 21.7 cm thick is used to reduce the X-radiation from a
nickel target to 0.05 of its incident value. (a) What is the transmittance and absorbance of
the plate?
(b) Assuming that the absorption coefficients, given in the previous question,
can be added, what is the composition of the alloy, in atom %?
(a) Transmittance, T = I / I0; Absorbance, A = - log (transmittance)
T = 0.05 / 1 = 0.05
A = - log(0.05) = 1.3010
(b) Use:
I = I0 exp (- a L)
Assume that the alloy composition is ZnxCd1-x:
0.05 I0 = I0 exp [- {5.187x + 18.418 (1 - x)} L]
0.05 = exp [- {5.187x + 18.418 (1 - x)} x 0.217]
-2.9957 = [- {5.187x + 18.418 (1 - x)} x 0.217]
-13.8051 = - 5.187x - 18.418 + 18.418 x
x = 0.35, i.e. Zn0.35Cd0.65
16
A lead glass fibre has a refractive index of 1.682.
What is the critical angle for total
internal reflection at the interface with an acrylic coating with refractive index 1.498?
sin c = n (low) / n (high) = 1.498 / 1.682 = 0.8906
c = 62.95 
17
A ray of light passing through water, (refractive index, n = 1.33), in a glass tank,
(refractive index 1.58), hits the water / glass surface at an angle of 23.
does it make with the surface as it continues through the glass?
(a) What angle
(b) What is the critical
angle for light passing through the water striking the water / glass interface?
(c) What is
the critical angle for light passing through the glass striking the glass / water interface?
(a) sin 1 / sin 2 =
sin 23 / sin 2 =
sin 2 =
n1 / n2
1.58 / 1.33
1.33 x sin 23 / 1.58 = 0.3289
2 = 19.2 
(b) There is no critical angle as we pass from n1 low to n2 high.
(c)
sin c = n (low) / n (high) = 1.33 / 1.58 = 0.8414
c = 57.3 
18
Estimate the refractive indices of the ceramics (a) barium titanate, BaTiO 3, (b) lead
titanate, PbTiO3, using the Gladstone – Dale formula.
PbTiO3, 8230 kg m-3.
Densities: BaTiO3, 6017 kg m-3;
n = 1 +  (p1kr1 + p2kr2 + p3kr3 ...)
Note: use densities in g cm-3 as in the original formula.
(a) BaTiO3 = BaO + TiO2
molar mass BaO = 153.3 g mol-1
molar mass TiO2 = 79.9 g mol-1
molar mass BaTiO3 = 233.2 g mol-1
weight fraction BaO = 153.3 / 233.2 = 0.6574
weight fraction TiO2 = 79.9 / 233.2 =
0.3426
n = 1 + 6.017[0.6574 x 0.13 + 0.3426 x 0.40]
= 1 + 6.017 [0.0855 + 0.1371]
(b)
= 2.34
PbTiO3 = PbO + TiO2
molar mass PbO = 223.3 g mol-1
molar mass TiO2 = 79.9 g mol-1
molar mass PbTiO3 = 303.2 g mol-1
weight fraction PbO = 223.3 / 303.2 = 0.7365
weight fraction TiO2 = 79.9 / 303.2 = 0.2635
n = 1 + 8.230[0.7365 x 0.13 + 0.2635 x 0.40]
= 1 + 8.230 [0.1105 + 0.1054]
= 1 + 6.017 [0.2226] = 2.78
19
Estimate the refractive index of the minerals (a) spinel, MgAl 2O4, (b) akermanite,
Ca2MgSi2O7, using the Gladstone – Dale formula. Densities: spinel, 3600 kg m-3;
akermanite, 2940 kg m-3.
n = 1 +  (p1kr1 + p2kr2 + p3kr3 ...)
Note: use densities in g cm-3 as in the original formula.
(a) MgAl2O4
= MgO + Al2O3
molar mass MgO = 40.3 g mol-1
molar mass Al2O3 = 102.0 g mol-1
molar mass MgAl2O4
= 142.3 g mol-1
weight fraction MgO = 40.3 / 142.3 = 0.2832
weight fraction Al2O3 = 102.0 / 142.3 =
0.7168
n = 1 + 3.6[0.2832 x 0.20 + 0.7168 x 0.21]
= 1 + 3.6 [0.0566 + 0.1505]
(b) Ca2MgSi2O7
= 1.75
= 2 CaO + MgO + 2 SiO2
molar mass CaO = 56.1 g mol-1
molar mass MgO = 40.3 g mol-1
molar mass SiO2
= 60.1 g mol-1
molar mass Ca2MgSi2O7
= 272.6 g mol-1
weight fraction CaO = 2 x 56.1 / 272.6 = 0.4116
weight fraction MgO = 40.3 / 272.6 = 0.1478
weight fraction SiO2 = 2 x 60.1 / 272.6 = 0.4410
n = 1 + 2.94[0.4116 x 0.21 + 0.1478 x 0.20 + 0.4410 x 0.21]
= 1 + 2.94 [0.0864 + 0.02956 + 0.0926]
20
= 1.61
Estimate the refractive index of the minerals (a) beryl, Be 3Al2(SiO3)6, (b) garnet,
Mg3Al2Si3O12, using the Gladstone – Dale formula. Densities: beryl, 2640 kg m-3; garnet,
3560 kg m-3.
n = 1 +  (p1kr1 + p2kr2 + p3kr3 ...)
Note: use densities in g cm-3 as in the original formula.
(a) Be3Al2(SiO3)6
= 3 BeO + Al2O3 + 6 SiO2
molar mass BeO = 25.01 g mol-1
molar mass Al2O3 = 102.0 g mol-1
molar mass SiO2
= 60.1 g mol-1
molar mass beryl
= 537.6 g mol-1
weight fraction BeO = 3x 25.01 / 537.6 = 0.1396
weight fraction Al2O3 = 102.0 / 537.6 =
0.1897
weight fraction SiO2 = 6 x 60.1 / 537.6 = 0.6708
n = 1 + 2.64[0.1396 x 0.24 + 0.0.1897 x 0.21 + 0.6708 x 0.21]
= 1 + 2.64 [0.0335 + 0.0398 + 0.1409]
(b) Mg3Al2Si3O12
= 1.57
= 3 MgO + Al2O3 + 3 SiO2
molar mass MgO = 40.3 g mol-1
molar mass Al2O3 = 102.0 g mol-1
molar mass SiO2
= 60.1 g mol-1
molar mass Ca2MgSi2O7
= 403.2 g mol-1
weight fraction MgO = 3 x 40.3 / 403.2 = 0.2999
weight fraction Al2O3 = 102.0 / 403.2 =
0.2530
weight fraction SiO2 = 3 x 60.1 / 403.2 = 0.4472
n = 1 + 3.56[0.2999 x 0.20 + 0.2530 x 0.21 + 0.4472 x 0.21]
= 1 + 3.56 [0.0600 + 0.0531 + 0.0939]
= 1.74
21
Estimate the refractive coefficient of Al2O3 using the information that the mineral
andalusite, Al2SiO5, has a density of 3150 kg m-3 and a refractive index of 1.639.
The
refractive coefficient of SiO2 is 0.21.
(a) Al2SiO5
= Al2O3 + SiO2
molar mass Al2O3 = 102.0 g mol-1
molar mass SiO2
= 60.1 g mol-1
molar mass Al2SiO5
= 162.1 g mol-1
weight fraction Al2O3 = 102.0 / 162.1 =
0.6292
weight fraction SiO2 = 60.1 / 162.1 = 0.3708
n = 1.639 = 1 + 3.15[0.6292 x kr + 0.0.3708 x 0.21]
0.639 = 3.15 [0.6292 x kr + 0.07787]
0.3937 = 1.9820 kr
kr = 0.20
22 Calculate the reflectivity of the surfaces of the following transparent materials in air; (a)
Na3AlF6, n = 1.35; (b) glass, n = 1.537; (c) Al2O3, n(average) = 1.63; (d) ZrO2, n = 2.05, (e)
Ta2O5, n = 2.15 (average).
R = [(n – 1) / (n + 1)]2
(a) Na3AlF6:
R = (0.35 / 2.35)2 = 0.022 (2.2 %)
(b) glass:
R = (0.537 / 2.537)2 = 0.045 (4.5 %)
(c) Al2O3:
R = (0.63 / 2.63)2 = 0.057 (5.7 %)
(a) ZrO2:
R = (1.05 / 3.05)2 = 0.119 (11.9 %)
R = (1.15 / 3.15)2 = 0.133 (13.3 %)
(c) Ta2O5:
23
Calculate the reflectivity of the surfaces of the following metals in air.
The optical
constants are for a wavelength of 550 nm: (a) aluminium, n = 0.82, k = 5.99; (b) silver, n =
0.255, k = 3.32; (c) gold, n = 0.33, k = 2.32; (d) chromium, n = 2.51, k = 2.66; (e) nickel, n =
1.85, k = 3.27.
R = [(n – 1)2 + k2 / (n + 1)2 + k2]
(a) Al:
R = [(0.82 – 1)2 + 5.992 / (1.82)2 + 5.992] = 0.916 (91.6 %)
(b) Ag:
R = [(0.255 – 1)2 + 3.322 / (1.225)2 + 3.322] = 0.924 (92.4 %)
(c) Au:
R = [(0.33 – 1)2 + 2.322 / (1.33)2 + 2.322] = 0.482 (48.2 %)
(d) Cr:
R = [(1.51)2 + 2.662 / (3.51)2 + 3.662] = 0.482 (48.2 %)
(e) Al:
R = [(0.85)2 + 3.272 / (2.85)2 + 3.272] = 0.607 (60.7 %)
24 A thin film on a substrate viewed in air has an optical thickness of /4. Will the film be
reflecting or not, (a), if the substrate has a lower refractive index than the film, and (b), if the
substrate has a higher refractive index than the film.
(a)
If the substrate has a lower refractive index than the film there is just one phase
change, at the top surface, so:
[p] = m
minimum (dark)
[p] = (m + ½ )
maximum (bright)
[p] = 2 ( / 4) = ( / 2)
(a)
the film will appear bright (reflecting)
If the substrate has a higher refractive index than the film there is a phase change at
both surfaces, so:
[p] = m
maximum (bright)
[p] = (m + ½ )
minimum (dark)
[p] = 2 ( / 4) = ( / 2)
the film will appear dark (non-reflecting)
25 A thin film on a substrate viewed in air has an optical thickness of /2. Will the film be
reflecting or not, (a), if the substrate has a lower refractive index than the film, and (b), if the
substrate has a higher refractive index than the film.
(a)
If the substrate has a lower refractive index than the film there is just one phase
change, at the top surface, so:
[p] = m
minimum (dark)
[p] = (m + ½ )
maximum (bright)
[p] = 2 ( / 2) = 
(a)
the film will appear dark (non-reflecting)
If the substrate has a higher refractive index than the film there is a phase change at
both surfaces, so:
[p] = m
maximum (bright)
[p] = (m + ½ )
minimum (dark)
[p] = 2 ( / 2) = 
26
the film will appear bright (non-reflecting)
Describe the differences between the reflectivity of a soap film, thickness /4, in air,
compared to the same film on an oil surface. Assume that the refractive index of the soap
film is equal to that of water, 1.33, and that of the oil is 1.44.
When the soap film is in air we have:
[p] =
m
minimum (dark)
[p] = (m + ½ )
maximum (bright)
[p] = 2 ( / 4) =  /2
the film will appear bright (reflecting)
When the soap film is on oil we have:
[p] = m
maximum (bright)
[p] = (m + ½ )
minimum (dark)
[p] = 2 ( / 4) =  /2
27
the film will appear dark (non-reflecting)
(a) What is the minimum physical thickness of a film of titanium dioxide in air that will
give rise to constructive interference of green light,  = 550 nm.
The refractive index of
TiO2 is 2.875 (average). Estimate the colour that the film would appear when viewed in
white light (b), by reflection and (c), by transmission.
When the film is in air we have:
[p] = (m + ½ )
maximum (bright)
The minimum optical path difference is when m = 0, i.e. [p] = ½  = 550 / 2 = 275
nm
[p] = 2 n t:
(b)
Using Table 14.5:
t = [p] / 2 n = 275 / 2 x 2.875 = 47.8 nm
optical path difference = 275 nm, reflected colour yellow-white to
straw yellow
(c) Using Table 14.5: optical path difference = 275 nm, transmitted colour carmine red to
deep violet.
28 Derive the relationship
nf = ns
for an antireflection coating, refractive index nf, on a substrate, refractive index ns, in air.
For values of [d] given by /4, 3/4 etc.
2
2
R = [(nf - n0ns)/(nf + n0ns)]2
The reflectance will be either a maximum or a minimum.
When the film has a higher
refractive index than the substrate the reflectivity will be a maximum. When the film has a
lower refractive index than the substrate the reflectivity will be a minimum.
We require R
= 0 i.e.
0 =
2
2
[(nf - n0ns)/(nf + n0ns)]2
In air, n0 = 1 hence ;
0 =
2
(nf - ns)
nf = ns
29
Determine the reflectivity of a / 4 film of (a), silicon oxide, n = 2.0, and (b), titanium
dioxide, n = 2.775, on glass, n = 1.504, in air.
2
2
R = [(nf - n0ns)/(nf + n0ns)]2
2
2
(a)
R = [(2.0 – 1.504)/(2.0 + 1.504)]2 = 0.206
(b)
R = [(2.504 – 1.504)/(2.504 + 1.504)]2 = 0.376
2
2
30 (a) Determine the reflectivity of a / 4 and a /2 film of magnesium fluoride, n = 1.384,
on glass, n = 1.504, in air.
(b) What changes would occur if the substrate was titanium
dioxide, n = 2.775 (average).
(a) For a / 4 film:
2
2
R = [(nf - n0ns)/(nf + n0ns)]2
2
2
= [(1.384 – 1.504)/(1.384 + 1.504)]2 = 0.015 (1.5 %)
For a / 2 film:
R = [(n0 - ns) / (n0 + ns)]2
R = [(1 – 1.504) / (1 + 1.504)]2 = 0.0405 (4 %)
(b)
Because nf < ns, for both glass and TiO2, no change in the pattern of reflectivity will
occur, but now:
For a / 4 film:
R
2
2
= [(1.384 – 2.875)/(1.384 + 2.875)]2 = 0.040 (4 %)
For a / 2 film:
R = [(1 – 2.7875) / (1 + 2.875)]2 = 0.23 (23 %)
31
(a) Plot a graph of reflectivity versus the number of pairs of layers for a quarter wave
stack on a glass substrate, n = 1.495, in air, using alternating layers of magnesium fluoride,
n = 1.384 and titanium dioxide, n = 2.775 (average).
(b) How many pairs are needed to
achieve a reflectivity of at least 99.9%.
R = [(ns - (nL/nH)2N) / (ns + (nL/nH)2N)]2
ns
= 1.495; nL = 1.384; nH
N
R
N
R
1
0.535
4
0.992
2
0.866
5
0.998
3
0.967
6
0.9996
7
0.9999
= 2.875
32 A quarter wave stack on a glass substrate, n = 1.545, in air, is required to reflect light of
650 nm from a laser.
The materials chosen have refractive indices of 2.15 (Ta 2O5) and
1.35, (Na3AlF6). What is the physical thickness of each layer?
The optical thickness of each layer is  / 4 = 650 / 4 = n t
For Ta2O5: t = 650 / 4 x 2.15 = 75.58 nm
For Na3AlF6:
33
t = 650 / 4 x 1.35 = 120.37 nm
The irradiance of a light beam traversing a solution placed into a cell of 10 cm path
length drops to 80.3% of the incident irradiance. Calculate the linear absorption coefficient
of the solution.
Use: I = I0 exp [- a L]
0.803I0 = I0 exp [-a x 0.1]
ln (0.803) = [-a x 0.1]
-a
34
= 2.19 m-1
The visibility of the atmosphere is reported as being 10 km.
Assuming that at this
distance an object can just be perceived, with 1% of the initial light falling on the object
reaching the observer, determine the linear absorption coefficient of the atmosphere.
Use: I = I0 exp [- a L]
0.1I0 = I0 exp [-a x 1 x 104]
ln (0.1) = [-a x 1 x 104]
-a
= 0.00046 m-1
= 4.6 x 10-4 m-1
35 Determine the relative amount of light scattered by dust particles with a refractive index
of 1.45 relative to that of water droplets with a refractive index 1.33, if Rayleigh scattering
occurs.
Assuming that the scattering centres have the same volume and scattering is viewed at the
same angle:
Is = k [(m2 – 1) / (m2 + 2)]2
m = n(particle) / n(medium) and for air n(medium) = 1 so:
Is(dust) / Is(water)
= [(1.452 – 1) / (1.452 + 2)]2 / [(1.332 – 1) / (1.332 + 2)]2
= 1.735
i.e. about 1.7 times as much light is scattered by the dust compared to water.
36 Will air visibility be improved if water droplets responsible for scattering are replaced by
similar sized limestone dust particles?
The refractive index of limestone is 1.53.
Assuming that the scattering centres have the same volume and scattering is viewed at the
same angle:
Is = k [(m2 – 1) / (m2 + 2)]2
m = n(particle) / n(medium) and for air n(medium) = 1 so:
Is(limestone) / Is(water)
= [(1.532 – 1) / (1.532 + 2)]2 / [(1.332 – 1) / (1.332 + 2)]2
= 2.29
limestone scatters about 2 ¼ times as much as water so visibility will not be improved.
37
The linear scattering coefficient of limestone dust in the air is 0.0002194 m -1.
Over
what distance will the irradiance of a light beam diminish to 10% of its initial value?
Use: I = I0 exp [- a L]
0.10 I0 = I0 exp [- 0.0002194 x L]
ln (0.1) = [- 0.0002194 x L]
-2.3026 = [- 0.0002194 x L]
L = 10.5 km
38 What is the relative amount of light scattered (a), in the incident direction, (b), at 45 to
the incident direction, (c), perpendicular to the incident direction, (d), in the reverse
direction, for Rayleigh scattering.
In Rayleigh scattering, the angular dependence is: (1 - cos2)
(a) incident direction,  = 0, Is = 2.0
(b) incident direction,  = 45, Is = 1.5
(c) incident direction,  = 90, Is = 1.0
(d) incident direction,  = 180, Is = 2.0
39 (a) What slit width is needed in a transmission diffraction grating to cause red light,  =
700 nm, to be deviated by 7.5 to the normal?
 = 400 nm?
The angle of deviation is given by:
(b) What will the deviation of the violet light,
sin m = m / d =  / d for 1st order diffraction;
(a)
sin 7.5 = 700 / d
d = 5363 nm
(b)
sin m = 400 / 5363
m = 4.28 
40
The first photonic crystal made was formed by drilling a block of material with a
refractive index of 3.6 so as to from a face-centred cubic array of holes with a unit cell
parameter of 2 mm. What wavelength radiation will not pass in the [100] direction?
Use Bragg’s law modified for transmission through a medium of refractive index ns
m = 2nsd sin B
For [100],  = 90, d100 = 2 mm for normal incidence and m = 1:
 = 2 x 3.6 x 2 = 14.4 mm
41 A photonic crystal is made by laying down close packed layers of polystyrene spheres
of refractive index 1.595, to form a hexagonal closest packed array. The sphere diameter
is 250 nm, and the layer separation normal to the layers can be taken as 0.8 of the sphere
diameter.
(a) What is the wavelength that will not be transmitted in light normal to the
close-packed layers?
(b) What colour does this correspond too?
(c) If the ordering is
imperfect and composed of random hexagonal (AB) and cubic (ABC) packing close-packed
layers how will the result change?
Use Bragg’s law modified for transmission through a medium of refractive index ns
m = 2nsd sin B
(a) For light normal to the layers  = 90 and m = 1:
d100 = 0.8 x 250 nm,
 = 2 x 1.595 x 0.8 x 250 = 638 nm
(b) Orange-red
(c) Changing the AB packing to ABC packing will not altered, so there is no change.
42 Estimate the attenuation of an optical fibre in which 0.5% of the initial power is lost in a
distance of 1 km.
attenuation = {-10 log10 [P(x) / P(0)] } / x
= -10 log [99.5 / 100] / 1 = 0.0218 dB km-1
43
The attenuation of ordinary window glass is of the order of 10 6 dB km-1.
What
thickness of glass would cause the incident light intensity to fall to 50%?
attenuation = {-10 log10 [P(x) / P(0)] } / x = 106
{-10 log10 [P(x) / P(0)] } = 106 x
x =
{-10 log10 [50 / 100] } / 106 km
x =
{-10 log10 [50 / 100] } / 103 m = 0.00301 m = 3.01 mm
44 An instantaneous pulse of white light is introduced into a silica optical fibre. What will
the pulse spread be, in km, after 1 second? The refractive indices of silica are, n(400 nm)
1.47000, n(average), 1.46265, n(700 nm) 1.45530.
Refractive index = velocity of light in vacuum / velocity of light in medium
In 1 second, red light will travel 2.99792 x 108 / n(red)
= 2.99792 x 108 / 1.45530 = 206 000 137.4m
In 1 second, violet light will travel 2.99792 x 108 / n(violet)
= 2.99792 x 108 / 1.47000 = 203 940 136.1m
The spread is:
45
2 060 001 m = 2060 km
A thin film of silver oxide forms on a silver surface.
The band gap for silver oxide is
2.25 eV. (a) What colour would be emitted by bulk silver oxide? (b) Treating the thin film
as a quantum well, what film thickness is needed to obtain an emission at a wavelength of
413 nm? The effective mass of both electrons and holes in Ag2O is 0.3 me.
(a)
E = Eg = h  = hc / ;
hc = 6.626 x 10-34 x 2.998 x 108 = 1.986 x 10-25
 = 1.986 x 10-25 / 2.25 x 1.60218 x 10-19 = 551 nm
This corresponds to yellow-green
(b) In a thin film E is modified to:
E = Eg + n2 h2 / 4 a2 m* = Eg + E(well) = hc / ;
Taking n = 1 and m* = 0.3me:
E(well) = 1 x (6.626 x 10-34)2 / 4 x a2 x 0.3 x 9.109 x 10-31
= 4.0165 x 10-37 / a2 J
To obtain emission at 413 nm:
Eg + E(well) = hc /  = 1.986 x 10-25 / 413 x 10-9 = 4.08087 x 10-19 J
4.08087 x 10-19 = 2.25 x 1.60218 x 10-19 + E(well)
E(well) = 1.2038 x 10-19
= 4.0165 x 10-37 / a2
a2 = 4.0165 x 10-37 / 1.2038 x 10-19
a = 1.827 nm
46 A film of zinc sulphide 4 nm in thickness forms on metallic zinc. Treating the film as a
quantum well, what is the wavelength of the transition n = 2. The bandgap of ZnS is 3.54
eV and the effective mass of both electrons and holes is 0.4 m e.
E = Eg + n2 h2 / 4 a2 m* = Eg + E(well) = hc / ;
Taking n = 2, Eg = 3.54 x 1.60218 x 10-19 J and m* = 0.4me:
Eg = 5.6717 x 10-19 J
E(well) = 4 x (6.626 x 10-34)2 / 4 x (4 x 10-9)2 x 0.4 x 9.109 x 10-31 J
= 7.531 x 10-20 J
E = 5.6717 x 10-19 + 7.531 x 10-20 J = 6.4248 x 10-19 J
 = hc / E
hc = 6.626 x 10-34 x 2.998 x 108 = 1.986 x 10-25
 = 1.986 x 10-25 / 6.4248 x 10-19 = 309 nm