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Kasap Solutions

This document is a solutions manual for the textbook 'Optoelectronics and Photonics: Principles and Practices, Second Edition' by Safa Kasap, providing solutions to problems and questions from Chapter 1. It covers topics such as Maxwell's wave equation, electromagnetic wave propagation in media, point light sources, Gaussian beams, and optical cavities with spherical mirrors. The manual includes detailed mathematical derivations and solutions for various optical phenomena and principles.

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Kasap Solutions

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Solutions Manual to

Optoelectronics and Photonics:

Principles and Practices, Second Edition
© 2013 Pearson Education

Safa Kasap

Revised: 31 January 2013

Check author's website for updates
http://optoelectronics.usask.ca

ISBN-10: 013308180X
ISBN-13: 9780133081800

NOTE TO INSTRUCTORS
If you are posting solutions on the internet, you must password the access and
download so that only your students can download the solutions, no one else. Word
format may be available from the author. Please check the above website. Report
errors and corrections directly to the author at safa.kasap@yahoo.com

S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright
and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a
retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or
likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education,
Inc., Upper Saddle River, NJ 07458.

Preliminary Solutions to Problems and Questions

Chapter 1

Note: Printing errors and corrections are indicated in dark red. See Question 1.47. These are
correct in the e-version of the textbook
1.1 Maxwell's wave equation and plane waves
(a) Consider a traveling sinusoidal wave of the form Ex = Eo cos(ωt − kz + φo). The latter can also be
written as Ex = Eo cos[k(vt − z) + φo], where v = ω/k is the velocity. Show that this wave satisfies
Maxwell's wave equation, and show that v = (µoεoεr)−1/2.
(b) Consider a traveling function of any shape, even a very short delta pulse, of the form Ex =
f[k(vt − z)], where f is any function, which can be written is Ex = f(φ), φ = k(vt − z). Show that this
traveling function satisfies Maxwell's wave equation. What is its velocity? What determines the form
of the function f?
Solution
(a)
Ex = Eo cos(ωt − kz + φo)
∂ 2 Ex
∴ =0
∂x 2
∂ 2 Ex
and =0
∂y 2
∂ 2 Ex
and = −k 2 E0 cos(ωt − kz + φ0 )
∂z 2

∂ 2 Ex
∴ = −ω 2 E0 cos(ωt − kz + φ0 )
∂t 2

∂2E ∂2E ∂2E ∂2E

Substitute these into the wave equation + + − ε ε µ = 0 to find
∂x 2 ∂y 2 ∂z 2 ∂t 2
o r o

−k 2 E0 cos(ωt − kz + φ0 ) + ε oε r µo + ω 2 E0 cos(ωt − kz + φ0 ) =0
ω2 1
∴ =
k 2
ε oε r µ o
ω −1
∴ = (ε oε r µo ) 2
k
−1
∴ v = (ε oε r µo ) 2

(b) Let
E=
x )] f (φ )
f [k (v t − z=
Take first and second derivatives with respect to x, y, z and t.

∂ 2 Ex
=0
∂x 2
∂ 2 Ex
=0
∂y 2
∂Ex df
= −k
∂z dφ
∂ 2 Ex 2
2 d f
= k
∂z 2 dφ 2
∂Ex df
= kv
∂t dφ
∂ 2 Ex 2
2 2 d f
= k v
∂t 2 dφ 2

∂2 E ∂2 E ∂2 E ∂2 E
Substitute these into the wave equation + + − ε ε µ =
0 to find
∂x 2 ∂y 2 ∂z 2 ∂t 2
o r o

d2 f 2
2 2 d f
k2 − ε ε µ k v =
0
dφ 2 dφ 2
o r o

1
∴ v2 =
ε oε r µ o
−1
∴ v = (ε oε r µo ) 2

1.2 Propagation in a medium of finite small conductivity An electromagnetic wave in an

isotropic medium with a dielectric constant εr and a finite conductivity σ and traveling along z obeys
the following equation for the variation of the electric field E perpendicular to z,
d 2E ∂2E ∂E
− ε ε µ = µ oσ (1)
∂t ∂t
2 o r o 2
dz
Show that one possible solution is a plane wave whose amplitude decays exponentially with
propagation along z, that is E = Eoexp(−αz)exp[j(ωt – kz)]. Here exp(−αz) causes the envelope of the
amplitude to decay with z (attenuation) and exp[j(ωt – kz)] is the traveling wave portion. Show that in
a medium in which α is small, the wave velocity and the attenuation coefficient are given by
ω 1 σ
v= = and α=
k µoε oε r 2ε o cn

where n is the refractive index (n = εr1/2). (Metals with high conductivities are excluded.)

Solution

We can write E = Eoexp(−αz)exp[j(ωt – kz)] as E = Eoexp[jωt – j(k – jα)z]. Substitute this into the
wave resonance condition
[– j(k – jα)]2Eoexp[jωt – j(k – jα)z] − (jω)2εoεrµoEoexp[jωt – j(k – jα)z] =
jωµoσ Eoexp[jωt – j(k – jα)z]
∴ −(k – jα)2 + ω2εoεrµo = jωµoσ
∴ −k2 + 2jkα – α2 + ω2εoεrµo = jωµoσ
Rearrange into real and imaginary parts and then equating the real parts and imaginary parts
∴ −k2 – α2 + ω2εoεrµo + 2jkα = jωµoσ
Real parts
−k2 – α2 + ω2εoεrµo = 0
Imaginary parts
2kα = ωµoσ
ωµ σ ω µ σ µ cσ σ
Thus, α= o = ⋅ o = o =
2k k 2 2n 2ε o n
where we have assumed ω/k = velocity = c/n (see below).
From the imaginary part
k 2 = ω 2 µ oε oε r − α 2
Consider the small α case (otherwise the wave is totally attenuated with very little propagation). Then
k 2 = ω 2 µ oε oε r
and the velocity is
ω 1
v= =
k µ oε oε r

1.3 Point light source What is the irradiance measured at a distance of 1 m and 2 m from a 1 W
light point source?

Solution
Then the irradiance I at a distance r from O is
P 1W
I = o2 = = 8.0 µW cm-2
4πr 4π (1 m) 2
which drops by a factor of 4 at r = 2 m to become 2.0 µW cm-2

1.4 Gaussian beam A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a
Gaussian beam, what is the divergence of the beam? What are its Rayleigh range and beam width at 10
m?
Solution
Using Eq. (1.1.7), we find,
4λ 4(633 ×10−9 m)
= 2θ = = 1.01×10-3 rad = 0.058°
π (2wo ) π (0.8 ×10 m) −3

The Rayliegh range is

π wo2 π [ 12 (0.8 ×10−3 m)]2
= zo = = 0.79 m
λ (633 ×10−9 m)

The beam width at a distance of 10 m is

2w = 2wo[1 + (z/zo)2]1/2 = (0.8×10-3 m){1 + [(10 m)/(0.79 m)]2}1/2

= 0.01016 m or 10.16 mm.

1.5 Gaussian beam in a cavity with spherical mirrors Consider an optical cavity formed by two
aligned spherical mirrors facing each other as shown in Figure 1.54. Such an optical cavity is called a
spherical mirror resonator, and is most commonly used in gas lasers. Sometimes, one of the reflectors
is a plane mirror. The two spherical mirrors and the space between them form an optical resonator
because only certain light waves with certain frequencies can exist in this optical cavity. The radiation
inside a spherical mirror cavity is a Gaussian beam. The actual or particular Gaussian beam that fits
into the cavity is that beam whose wavefronts at the mirrors match the curvature of the mirrors.
Consider the symmetric resonator shown in Figure 1.54 in which the mirrors have the same radius of
curvature R. When a wave starts at A, its wavefront is the same as the curvature of A. In the middle of
the cavity it has the minimum width and at B the wave again has the same curvature as B. Such a wave
in the cavity can replicate itself (and hence exist in the cavity) as it travels between the mirrors
provided that it has right beam characteristics, that is the right curvature at the mirrors. The radius of
curvature R of a Gaussian beam wavefront at a distance z along its axis is given by
R(z) = z[1 + (zo/z)2] ; zo = πwo2/λ
is the Rayleigh range
Consider a confocal symmetric optical cavity in which the mirrors are separated by L = R.
(a) Show that the cavity length L is 2zo, that is, it is the same as the Rayleigh range, which is the reason
the latter is called the confocal length.
(b) Show that the waist of the beam 2wo is fully determined only by the radius of curvature R of the
mirrors, and given by
2wo = (2λR/π)1/2
(c) If the cavity length L = R = 50 cm, and λ = 633 nm, what is the waist of the beam at the center
and also at the mirrors?

Figure 1.54 Two spherical mirrors reflect waves to and from each other. The optical cavity contains a
Gaussian beam. This particular optical cavity is symmetric and confocal; the two focal points coincide at F.

Solution
(a) At z = R / 2 we have R ( z ) = R . Substitute these into R(z) = z[1 + (zo/z)2] to find
R = (R/2)[1 + (2zo/R)2]

2
 2z 
∴ 2 = 1+  o 
 R 
 2 zo 
∴   =1
 R 
∴ L = 2 zo
(b) R = (R/2)[1 + (2zo/R)2]
2
 2z 
∴ 2 = 1+  o 
 R 
 2 zo 
∴   =1
 R 
Now use zo = πwo2/λ,
 2πwo2 
∴   = 1
 Rλ 
2 Rλ
∴ 2 wo =
π

(c) Substitute λ = 633 nm, L = R = 50 cm into the above equation to find 2wo = 449 µm or 0.449 mm.
At the mirror, z = R/2, and also zo = R/2 so that
1/ 2
  z 2 
1/ 2
  R / 2 2 
2 w = 2 wo 1 +    = 2 wo 1 +    = 2 wo ( 21 / 2 ) = 0.635 mm
  zo     R / 2  

1.6 Cauchy dispersion equation Using the Cauchy coefficients and the general Cauchy equation,
calculate refractive index of a silicon crystal at 200 µm and at 2 µm, over two orders of magnitude
wavelength change. What is your conclusion?

Solution
At λ = 200µm, the photon energy is
hc (6.62 ×10−34 J s)(3 × 108 m s -1 ) 1
hυ = = −6 × 6.2062 ×10−3 eV
=
λ (200 ×10 m) −19
1.6. ×10 J eV -1

Using the Cauchy dispersion relation for silicon with coefficients from Table 9.2,
n = n-2(hυ)−2 + n0 + n2(hυ)2 + n4(hυ)4
= (−2.04×10-8)( 6.2062 ×10−3 )−2 + 3.4189+ (8.15×10-2)( 6.2062 ×10−3 )2
+ (1.25×10-2)( 6.2062 ×10−3 )4
= 3.4184

At λ = 2µm, the photon energy is

hc (6.62 ×10−34 J s)(3 × 108 m s -1 ) 1
hυ = = −6 × =
0.6206eV
λ (2 ×10 m) −19
1.6 ×10 J eV -1
Using the Cauchy dispersion relation for silicon with coefficients from Table 9.2,

n = n-2(hυ)−2 + n0 + n2(hυ)2 + n4(hυ)4

= (−2.04×10-8)( 0.6206 )−2 + 3.4189+ (8.15×10-2)( 0.6206 )2
+ (1.25×10-2) ( 0.6206 )4
= 3.4521

1.7 Sellmeier dispersion equation Using the Sellmeier equation and the coefficients, calculate the
refractive index of fused silica (SiO2) and germania GeO2 at 1550 nm. Which is larger, and why?

Solution
The Sellmeier dispersion relation for fused silica is
0.696749λ 2 0.408218λ 2 0.890815λ 2
n2 =1+ 2 + +
λ − 0.06906602 μm 2 λ 2 − 0.1156622μm 2 λ 2 − 9.9005592μm 2

0.696749 (1550 nm ) 2 0.408218(1550 nm) 2 0.890815(1550 nm) 2

n2 = 1+ + +
(1550 nm) 2 − (69.0660 nm) 2 (1550 nm) 2 − (115.662 nm) 2 (1550 nm) 2 − (9900.559 nm) 2
so that

n = 1.4443

The Sellmeier dispersion relation for germania is

0.8068664λ 2 0.7181585λ 2 0.8541683λ 2
n2 =1+ 2 + +
λ − (0.0689726μm) 2 λ 2 − (0.1539661μm) 2 λ 2 − (11.841931μm) 2

0.8068664 (1550 nm) 2 0.7181585(1550 nm) 2 0.8541683(1550 nm) 2

n2 =
1+ + +
(1550 nm) 2 − (68.9726 nm) 2 (1550 nm) 2 − (153.9661nm) 2 (1550 nm) 2 − (11841.931nm) 2

so that n = 1.5871

1.8 Sellmeier dispersion equation The Sellmeier dispersion coefficient for pure silica (SiO2) and
86.5%SiO2-13.5 mol.% GeO2 re given in Table 1.2 Write a program on your computer or calculator,
or use a math software package or even a spread sheet program (e.g. Excel) to obtain the refractive
index n as a function of λ from 0.5 µm to 1.8 µm for both pure silica and 86.5%SiO2-13.5%GeO2.
Obtain the group index, Ng, vs. wavelength for both materials and plot it on the same graph. Find the
wavelength at which the material dispersion becomes zero in each material.

Solution

Excel program to plot n and differentiate and find Ng

Figure 1Q8-1 Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength (Excel).
The minimum in Ng is around 1.3 µm. Note that the smooth line option used in Excel to pass a continuous smooth line
through the data points. Data points are exactly on the line and are not shown for clarity.

Figure 1Q8-2 Refractive index n and the group index Ng of 86.5%SiO213.5%GeO as a function of wavelength (Excel).
The minimum in Ng is around 1.4 µm. Note that the smooth line option used in Excel to pass a continuous smooth line
through the data points. Data points are exactly on the line and are not shown for clarity.

Material dispersion is proportional to derivative of group velocity over wavelength. The corresponding
values are close to 1.3 and 1.4 µm.

1.9 The Cauchy dispersion relation for zinc selenide ZnSe is a II-VI semiconductor and a very
useful optical material used in various applications such as optical windows (especially high power
laser windows), lenses, prisms etc. It transmits over 0.50 to 19 µm. n in the 1 – 11 µm range described
by a Cauchy expression of the form
0.0485 0.0061
n = 2.4365 + + − 0.0003λ 2 ZnSe dispersion relation
λ2 λ4
in which λ in µm. What are the n-2, n0, n2 and n4 coefficients? What is ZnSe's refractive index n and
group index Ng at 5 µm?

Solution
hc
hυ =
λ
1
(6.62 ×10−34 J s ×
hc = −19
1.24 ×10−6 eVm
)(3 ×108 m s -1 ) =
1.6 ×10 J eV -1

so that
0.0485 0.0061
2
(hν ) 2 +
n=
2.4365 + 4
(hν ) 4 − 0.0003(hc) 2 (hν ) −2
(hc) (hc)
Comparing with Cauchy dispersion equation in photon energy: n = n-2(hυ)−2 + n0 + n2(hυ)2 + n4(hυ)4 ,
we have

n0 = 2.4365

0.0485 0.0485
=
n2 = = 3.15 ×1010 eV -2
−6 2
(hc) 2
(1.24 × 10 )

n−2 = 0.0003(hc) 2 = 0.0003 × (1.24 ×10−6 ) 2 = 4.62 ×10−16 eV 2

0.0061 0.0061
and =
n4 = = 2.58 ×1021 eV -4
−6 4
(hc) 4
(1.24 ×10 )

At λ = 5 µm
0.0485 0.0061
n =2.4365 + + − 0.0003(5µ m) 2
(5µ m) (5µ m)
2 4

0.0485 0.0061
= 2.4365 + + − 0.0003(25) = 2.43
25 625

dn
Group index Ng = n − λ
dλ
0.0485 0.0061
and n = 2.4365 + + − 0.0003λ 2
λ2 λ4
dn −2λ × 0.0485 −4λ 3 × 0.0061
∴ = + − 2 × 0.0003 λ
dλ λ4 λ8
dn −0.097 −0.0244
∴ = 3 + − 0.0006 λ
dλ λ λ5
At λ = 5 µm
dn −0.097 −0.0244
= + − 0.0006 × (5µm)
d λ (5µm)3 (5µm)5
dn
∴ = −0.003783µm −1
dλ
dn
∴ Ng = n − λ = 2.43 − 5µm × (−0.003783µm −1 ) = 2.45
dλ

1.10 Refractive index, reflection and the Brewster angle

(a) Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the
phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the
phase velocity?
(b) What is the Brewster angle (the polarization angle θp) and the critical angle (θc) for total internal
reflection when the light wave traveling in this silica medium is incident on a silica/air interface. What
happens at the polarization angle?
(c) What is the reflection coefficient and reflectance at normal incidence when the light beam
traveling in the silica medium is incident on a silica/air interface?

(d) What is the reflection coefficient and reflectance at normal incidence when a light beam
traveling in air is incident on an air/silica interface? How do these compare with part (c) and what is
your conclusion?

Solution

Figure 1.8 Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength.

(a) From Figure 1.8, at λ = 1300 nm, n = 1.447, Ng = 1.462, so that

The phase velocity is given by
v = c/n = (3×108 m s-1)/(1.447) = 2.073×108 m s-1.
The group velocity is given by
vg = c/Ng = (3×108 ms-1)/(1.462) = 2.052×108 m s-1.
The group velocity is about ~1% smaller than the phase velocity.

(b)

The Brewster angle θp is given by

n2 1
tan θ=
p = = 0.691
n1 1.447

∴ =θ p tan
= −1
0.691 34.64
At the Brewster angle of incidence θi = θp, the reflected light contains only field oscillations normal to
the plane of incidence (paper).

The critical angle is

n2 1
sin θ=
c = = 0.691
n1 1.447
∴ =θ c sin
= −1
(0.691) 43.7

R= R= r⊥= r/ =
2 2
and ⊥ // / 0.0333

(d) Given
n1 = 1
n2 = 1.447
n − n 1 − 1.447
∴ r/ / = r⊥ = 1 2 = = −0.1827
n1 + n2 1.447 + 1
R= R= r⊥= r/ =
2 2
and ⊥ // / 0.0333

Reflection coefficients are negative, which means that in external reflection at normal incidence there
is a phase shift of 180°.

1.11 Snell's law and lateral beam displacement What is the displacement of a laser beam passing
through a glass plate of thickness 2 mm and refractive index 1.570 if the angle of incidence is 40°? (See
Figure 1.14)

Figure 1.14 Lateral displacement of light passing obliquely through a transparent plate

Solution
The problem is sketched in Figure 1Q12-1

Figure 1Q12-1 Light beam deflection through a glass plate of thickness L = 2 mm. The angle of
incidence is 40° and the glass has a refractive index of 1.570

d  cos θi 
= sin θi 1 − 
L
 ( n / no ) 2 − sin 2 θi 

d  cos 40 
∴ = sin 40 1 − 
L  (1.570 /1) 2 − sin 2 40 
 0.7660 
=
0.6428 1 − =0.2986
 2.46 − 0.4132 
d
∴ = 0.2986
2 mm
∴ d = 0.60 mm
This is a significant displacement that can be easily measured by using a photodiode array.

1.12 Snell's law and lateral beam displacement An engineer wants to design a refractometer (an
instrument for measuring the refractive index) using the lateral displacement of light through a glass
plate. His initial experiments involve using a plate of thickness L, and measuring the displacement of
a laser beam when the angle of incidence θi is changed, for example, by rotating (tilting) the sample.
For θi = 40°, he measures a displacement of 0.60 mm, and when θi = 80° he measures 1.69 mm. Find
the refractive index of the plate and its thickness. (Note: You need to solve a nonlinear equation for n
numerically.)

Solution

Figure 1.14 shows the lateral beam deflection through a transparent plate.

Figure 1.14 Lateral displacement of light passing obliquely through a transparent plate

 cos θi 
Apply d = L sin θi 1 − 
 n 2 − sin 2 θi 
 cos 40°   cos 80° 
0.60 mm = L sin 40°1 −  and 1.69 mm = L sin 80°1 − 
 n 2 − sin 2 40°   n 2 − sin 2 80° 
Divide one by the other
 cos 40°   cos 40° 
1 −  1 − 
0.60  sin 40°   n 2 − sin 2 40°  0.60  sin 40°   n 2 − sin 2 40° 
=  ∴ 0= − 
1.69  sin 80°   cos 80°  1.69  sin 80°   cos 80° 
1 −  1 − 
 n − sin 80° 
2 2
 n − sin 80° 
2 2

 cos 40° 
1 − 
0.60  sin 40°   x − sin 40° 
2 2
Define y = − 
1.69  sin 80°   cos 80° 
1 − 
 x 2 − sin 2 80° 
 0.76606 
1 − 
x 2 − 0.41318 
∴ y = 0.35503 − 0.6527  = f(x)
 0.17365 
1 − 
 x 2 − 0.96985 

We can plot y = f(x) vs. x, and find where f(x) cross the x-axis, which will give x = n
0 .1

0 .0 5

0
y 1.4 x 1.5 1.6 1.7 1.8
0 .0 5

The above graph was generated in LiveMath (Theorist) (http://livemath.com)

Clearly, the x-axis is cut at n ≈ 1.575
Substitute n = 1.575 into one of the equations i.e.
 cos 40° 
0.60 mm = L sin 40°1 − 
 1.5752 − sin 2 40° 
Solving for L we find L ≈ 2.0 mm.

1.13 Snell's law and prisms Consider the quartz prism shown in Figure 1.55 that has an apex angle
α = 60°. The prism has a refractive index of n and it is in air.
(a) What are Snell's law at interfaces at A (incidence and transmittance angles of θi and θt ) and B
(incidence and transmittance angles of θi′ and θt′)?
(b) Total deflection δ = δ1 + δ2 where δ1 = θi − θt and δ2 = θt′ − θi′. Now, β + θi′ + θt = 180° and α
+ β = 180°. Find the deflection of the beam for an incidence angle of 45° for the following three colors
at which n is known: Blue, n = 1.4634 at λ = 486.1 nm; yellow, n = 1.4587 at λ = 589.2 nm; red, n =
1.4567 at λ = 656.3 nm. What is the separation in distance between the rays if the rays are projected on
a screen 1 m away.

Figure 1.55 A light beam is deflected by a prism through an angle δ. The angle of incidence is υi. The
apex angle of the prism is α.

Solution
(a) Snell's law at interfaces at A:
sin θi n
=
sin θt 1
Snell's law at interfaces at B:

sin θi′ 1
=
sin θt′ n
(b) Consider the deflection angle δ,
δ = δ1 + δ2 where δ1 = θi − θt and δ2 = θt′ − θi′, i.e. δ = θi − θt + θt′ − θi′. Now,
 sin θ i 
θ t = arcsin  
 n 
and from Figure 1.55
 sin θi 
θi′ = 180 − β − θt = α − θt = α − arcsin 
 n 
so that
   sin θi  
θt′ = arcsin[n sin θi′] = arcsin n sin α − arcsin  
   n  
and the deflection is,
 sin θi     sin θi     sin θi  
δ = θi − arcsin   + arcsin n sin α − arcsin    − α − arcsin  
 n     n     n  
so that finally,
   sin θi  
δ = θi − α + arcsin n sin α − arcsin  
   n  
Substituting the values, and keeping n as a variable, the deflection δ(n) as a function of n is
   1  
δ ( n ) = ( 45 − 60 ) + arcsin n sin 60 − arcsin   
   2n  
where sin(45°) = 1/√2.
The separation δL for two wavelengths λ1 and λ2 corresponding to n1 and n2 at the screen at a distance
L away is therefore
δL = L[δ(n1) − δ(n2)]
where the deflections must be in radians.
Consider the deflection of blue light
   1  
δ blue = ( 45 − 60 ) + arcsin (1.4634) sin 60 − arcsin   
   2 (1 . 4634 )  
∴ δblue = 34.115°
Similarly, δyellow = 33.709°
The separation of blue and yellow beams at the screes is
δLblue-yellow = L(δblue − δblue) = (1m)(π/180)( 34.115° − 33.709°) = 7.08 mm
Table 1Q13-1 summarizes the results of the calculations for blue, yellow and red light.

Table 1Q13-1 Deflection of blue, yellow and red light through a prism with apex angle 60°. The angle
of incidence is 45°.

Blue Yellow Red

486.1 nm 589.2 nm 656.3 nm

n 1.4634 1.4587 1.4567

0.5954 rad 0.5883 rad 5853 rad
δ (Deflection angle)
34.115° 33.709° 33.537°
δL between colors 7.08 mm 3.00 mm
δL between blue and
10.1 mm
rede

1.14 Fermat's principle of least time Fermat's principle of least time in simple terms states that
when light travels from one point to another it takes a path that has the shortest time. In going from a
point A in some medium with a refractive index n1 to a point B in a neighboring medium with
refractive index n2 as in Figure 1.56 the light path is AOB that involves refraction at O and satisfies
Snell's law. The time it takes to travel from A to B is minimum only for the path AOB such that the
incidence and refraction angles θi and θt satisfy Snell's law. Let's draw a straight line from A to B
cutting the x-axes at O′. The line AO′B will be our reference line and we will place the origin of x and
y coordinates at O′. Without invoking Snell's law, we will vary point O along the x-axis (hence OO′ is
a variable labeled x), until the time it takes to travel AOB is minimum, and thereby derive Snell's law.
The time t it takes for light to travel from A to B through O is
AO OB [( x1 − x ) 2 + y12 ]1 / 2 [( x 2 + x ) 2 + y 22 ]1 / 2
t= + = + (1)
c / n1 c / n 2 c / n1 c / n2
The incidence and transmittance angles are given by
x1 − x ( x2 + x)
sin θ i = and sin θt = (2)
[( x1 − x ) + y1 ]
2 2 1/ 2
[( x2 + x) 2 + y22 ]1/2
Differentiate Eq. (1) with respect to x to find the condition for the "least time" and then use Eq. (2) in
this condition to derive Snell's law.

Figure 1.56 Consider a light wave traveling from point A (x1, y2) to B (x1, y2) through an arbitrary point O at a distance x
from O′. The principle of least time from A to B requires that O is such that the incidence and refraction angles obey Snell's
law.

Solution
Differentiate t with respect to x
dt −1/ 2 × 2( x1 − x)[( x1 − x) 2 + y12 ]−1/2 1/ 2 × 2( x2 + x)[( x2 + x) 2 + y22 ]−1/2
= +
dx c / n1 c / n2
The time should be minimum so
dt
=0 condition for the "least time"
dx
−( x1 − x)[( x1 − x) 2 + y12 ]−1/2 ( x2 + x)[( x2 + x) 2 + y22 ]−1/2
∴ + = 0
c / n1 c / n2
( x1 − x) ( x2 + x)
∴ =
c / n1[( x1 − x) + y1 ]
2 2 1/2
c / n2 [( x1 − x) 2 + y12 ]1/2

Use Eq. (2) in the above expression to find

n1 sin θi = n2 sin θt

sin θi n2
∴ = Snell's law
sin θt n1

1.15 Antireflection (AR) coating

(a) Consider three dielectric media with flat and parallel boundaries with refractive indices n1, n2,
and n3. Show that for normal incidence the reflection coefficient between layers 1 and 2 is the same as
that between layers 2 and 3 if n2 = √[n1n3]. What is the significance of this result?
(b) Consider a Si photodiode that is designed for operation at 900 nm. Given a choice of two
possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of
2.3, which would you use and what would be the thickness of the antireflection coating? The
refractive index of Si is 3.5. Explain your decision.
(c) Consider a Ge photodiode that is designed for operation around 1200 nm. What are the best
AR refractive index and coating thickness if the refractive index of Ge is about 4.0?

Solution

(a) Start with the reflection coefficient r12 between n1 and n2,
n1 − n2 n1 − n1n3 n1 − n1n3 n n + n3
= r12 = = × 1 3
n1 + n2 n1 + n1n3 n1 + n1n3 n1n3 + n3
n1 n1n3 + n1n3 − n3 n1n3 − n1n3 n1 n1n3 − n3 n1n3
= =
n1 n1n3 + n1n3 + n1n3 + n3 n1n3 n1 n1n3 + 2n1n3 + n3 n1n3

Now consider r23 between n2 and n3,

n2 − n3 n1n3 − n3 n1n3 − n3 n1 + n1n3
= r23 = = ×
n2 + n3 n1n3 + n3 n1n3 + n3 n1 + n1n3
n1 n1n3 − n1n3 + n1n3 − n3 n1n3 n1 n1n3 − n3 n1n3
= =
n1 n1n3 + n1n3 + n1n3 + n3 n1n3 n1 n1n3 + 2n1n3 + n3 n1n3
∴ r12 = r23

To reduce the reflected light, waves A and B must interfere destructively. To obtain a good degree of
destructive interference between waves A and B, the amplitudes of reflection coefficients must be
comparable. When n2 = (n1n3)1/2, then the reflection coefficient between the air and coating is equal to
that between the coating and the semiconductor. So the reflection is minimum.
(b) We use n1 = 1 for air, n2 for the antireflection coefficient and n3 =3.5 for Si photodiode,
2
 n 2 − n1n3 
Rmin =  22 
 n2 + n1n3 
For SiO2 n2 = 1.5
2
 1.52 − 1× 3.5 
= Rmin (SiO 2 ) =  0.047
 1.5 + 1× 3.5 
2

For TiO2 n2 = 2.3

2
 2.32 − 1× 3.5 
=
Rmin (TiO 2 ) =  0.041
 2.3 + 1× 3.5 
2

∴ Rmin (TiO 2 ) < Rmin (SiO 2 )

So, TiO2 is a better choice

The thickness of the AR layer should be

d = λ/(4n2) = (900 nm)/[4(2.3)] = 97.8 nm
(c) Consider the Ge photodiode. Ge has n = 4.0. We use n1 = 1 for air, n2 for the antireflection
coefficient and n3 = 4.0 for Ge photodiode,
The ideal AR coating would have n2 = (n1n3)1/2 = 2.0
The thickness of the AR layer should be
d = λ/(4n2) = (1200 mm)/[4(2)] = 150 nm

1.16 Single and double layer antireflection V-coating For a single layer AR coating of index n2 on
a material with index n3 (> n2 > n1), as shown in Figure 1.57(a), the minimum reflectance at normal
incidence is given by
2
 n 2 − n1n3 
R min =  22  Single layer AR coating
 n 2 + n1n3 
when the reflections A, B, … all interfere as destructively as possible. Rmin = 0 when n2 = (n1n3)1/2. The
choice of materials may not always be the best for a single layer antireflection coating. Double layer
AR coatings, as shown in Figure 1.57(b) can achieve lower and sharper reflectance at a specified
wavelength as in Figure 1.57(c). To reduce the reflection of light at the n1/n4, interface, two layers n2
and n3, each quarter wavelength in the layer (λ/n2 and λ/n3) are interfaced between n1 and n4. The
reflections A, B and C for normal incidence result in a minimum reflectance given by
2
 n 2 n − n 4 n 22 
R min =  32 1 2
Double layer AR coating
 n3 n1 + n 4 n 2 
Double layer reflectance vs. wavelength behavior usually has V-shape, and they are called V-coatings.
(a) Show that double layer reflectance vanishes when
(n2/n3)2 = n1/n4 Best double layer AR coating
(b) Consider an InGaAs, a semiconductor crystal with an index 3.8, for use in a photodetector.
What is the reflectance without any AR coating?
(c) What is the reflectance when InGaAs is coated with a thin AR layer of Si3N4? Which material
in the table would be ideal as an AR coating?
(d) What two materials would you choose to obtain a V-coating? Note: The choice of an AR
coating also depends on the technology involved in depositing the AR coating and its effects on the
interface states between the AR layer and the semiconductor. Si1-xNx is a common AR coating on
devices inasmuch as it is a good passive dielectric layer, its deposition technology is well established
and changing its composition (x) changes its index.
A B CBA
n1 Reflectance
n1 n2
Double layer
n2 n3
Single
n3 n4 layer
λ

(a) (b) (c)

Figure 1.57(a) A single layer AR coating. (b) A double layer AR coating and its V-shaped reflectance
spectrum over a wavelength range.

Solution
(a) The minimum reflectance,
2
 n32 n1 − n4 n22 
=Rmin = 2
0
n 2
 3 1 n + n4 2 
n
∴ n32 n1 − n4 n22 =
0
∴ n32 n1 = n4 n22
n1 n22
∴ =
n4 n32
n1 n
∴ = ( 2 )2 Best double layer AR coating
n4 n3

(b) Without an AR coating, the reflectance is

R = [(n1 − n3)/ (n1 + n3)]2 = [(1 − 3.8)/ 1 + 3.8)]2 = 0.34 or 34%

(c) Take n3 = 3.8, n2 = 1.95, n1 = 1, and find the minimum reflectance from
2
 n 2 − n1n3 
R min =  22 
 n 2 + n1n3 
2
1.952 − 3.8  −7
∴ R=  = 1.08 ×10
1.95 + 3.8 
min 2

For ideal an AR coating:

n2 = n1n3
∴ n1n3 = 1× 3.8 =1.9493
∴ n2 = 1.9493
Looking at table, Si3N4 (n2 = 1.95) would be ideal.

(d) Two find 2 materials for a V-coating, consider first,

(n2 / n3 ) 2 = n1 / n4 Best double layer AR coating

(n2 / n3 ) = n1 / n4

(n=
2 / n3 ) =1/ 3.8 0.51
This is the ratio we need. From the table MgF2 (n2 = 1.38) and CdS (n3 = 2.60) are the best two
1.38
materials for V-coating: (n2 /=
n3 ) = 0.53 .
2.60
The minimum reflectance would be
2 2
 n32 n1 − n4 n22   2.602 − 3.8 ×1.382 
=Rmin = 2  = 2
0.001
n 2
 3 1 n + n4 2 
n  2.60 2
+ 3.8 × 1.38 

1.17 Single, double and triple layer antireflection coatings

Figure shows the reflectance of an uncoated glass, and glass that has a single (1), double (2) and triple
(3) layer antireflection coatings? The coating details are in the figure caption. Each layer in single and
double layer AR coatings has a thickness of λ/4, where λ is the wavelength in the layer. The triple
layer AR layer has three coatings with thicknesses λ/4, λ/2 and λ/4. Can you qualitatively explain the
results by using interference? What applications would need single, double and triple layer coatings?

Solution
Instructor’s choice of answers. Can be given out as a short project to students.

1.18 Reflection at glass-glass and air-glass interfaces

A ray of light that is traveling in a glass medium of refractive index n1 = 1.460 becomes incident on a
less dense glass medium of refractive index n2 = 1.430. Suppose that the free space wavelength of the
light ray is 850 nm.
(a) What should the minimum incidence angle for TIR be?
(b) What is the phase change in the reflected wave when the angle of incidence θi = 85° and when
θi = 90°?

(c) What is the penetration depth of the evanescent wave into medium 2 when θi = 85° and when θi
= 90°?
(d) What is the reflection coefficient and reflectance at normal incidence (θi = 0°) when the light
beam traveling in the silica medium (n = 1.460) is incident on a silica/air interface?
(e) What is the reflection coefficient and reflectance at normal incidence when a light beam
traveling in air is incident on an air/silica (n = 1.460) interface? How do these compare with part (d)
and what is your conclusion?

Solution
(a) The critical angle θc for TIR is given by
−1 n2 1.430
= θ c sin
= sin −1= 78.36
n1 1.460
(b) Since the incidence angle θi > θc, there is a phase shift in the reflected wave. The phase change
in Er,⊥ is given by φ⊥. With n1 = 1.460, n2 = 1.430, and θi = 85°,
1/2
 2  1.430  
2

1/2 sin (85°) −   

sin 2 θi − n 2    1.460  
tan 2 ϕ⊥
= 1
= ( cos θi
) 
cos(85°)
= 2.08675
(
tan 2 ϕ⊥ = 2.08675
1
)
ϕ⊥ = 2 tan −1 (2.08673)=128.79
so that the phase change is 128.79 . For the Er,// component, the phase change is
1/2
sin 2 θi − n 2 
tan 2=
1
(
ϕ/ / + 2 π
1
) 1
= 2 tan 12 ϕ⊥
n cos θi
2
n
( )
so that tan(1/2φ// + 12 π ) = (n1/n2)2tan(φ⊥/2) = (1.460/1.430)2tan(1/2(128.79°)) = 2.17522
which gives ϕ / / = 2 tan −1 (2.17522)-π = −49.38
We can repeat the calculation with θi = 90°.
The phase change in Er,⊥ is given by φ⊥. With n1 = 1.460, n2 = 1.430, and θi = 90°,
1/2
 2  1.430  
2

1/2  sin (90 °) −   

sin 2 θi − n 2    1.460  
tan 2 ϕ ⊥
= 1
= ( cos θi
) 
cos(90°)
∴ tan ( 1
2
ϕ⊥ )= ∞
∴ =ϕ⊥ 2 tan −1 (∞)=180
so that the phase change is 180 . For the Er,// component, the phase change is
1/2
sin 2 θi − n 2 
tan 2=
1
(ϕ/ / + 2 π
1
) = 2 tan
n cos θi
2
1
n
( 1
2
ϕ⊥ )

so that tan( /2φ// + 2 π )= ∞

1 1

which gives φ// = 2tan-1(∞) − π = 0°

(c) The amplitude of the evanescent wave as it penetrates into medium 2 is

Et,⊥(y,t) = Eto,⊥exp(–α2y)
We ignore the z-dependence, expj(ωt − kzz), as this represents propagation along z. The field strength
drops to e-1 when y = 1/α2 = δ, the penetration depth. The attenuation constant α2 for θi = 85° is
1/ 2
2πn2  n1  
2

α2 =   sin 2 θ i − 1
λo  n2  

1/ 2
2π (1.430)  1.460 
2

∴ α2 = − 9   sin (85 ) − 1
2 
= 1.96×106 m .
(850 × 10 )  1.430  
so the penetration depth δ = 1/α2 = 1/(1.96×106 m) = 5.1×10-7 m, or 0.51µm.
For 90°, repeating the calculation above,
1/ 2
2π (1.430)  1.460 
2

α2 =    sin 2
( 90
) − 1 = 2.175×106 m
(850 × 10−9 )  1.430  

so the penetration depth δ = 1/α2 = 1/(2.175×106 m) = 4.5×10-7 m, or 0.45 µm.

Conclusion: The penetration depth increases as the angle of incidence decreases

(d)
n1 − n2 1.460 − 1
r// = r⊥ = = = 0.187
n1 + n2 1.460 + 1
and R// = R⊥ = r//2 = 0.035 or 3.5%

(e)
n1 − n2 1 − 1.460
r// = r⊥ =
= = −0.187
n1 + n2 1 + 1.460
and R// = R⊥ = r//2 = 0.035 or 3.5%
When r⊥ is a negative number, then there is a phase shift of 180° (or π) which is in agreement with
part (b).

1.19 Dielectric mirror Consider a dielectric mirror that is made up of quarter wave layers of GaAs
with nH = 3.382 and AlAs with nL = 2.912, both around 1500 nm. The GaAs-AlAs dielectric mirror is
inside a vertical cavity surface emitting laser diode operating at 1.5 µm. The substrate is GaAs with n3
= nsubstrate = 3.382 . The light is incident on the mirror from another semiconductor that is GaAlAs with
an index n0 = 3.40. Calculate how many pairs of layers N would be needed to get a reflectance above
95%. What would be the bandwidth?

Solution

2
 2 N  n0  2 N 
2
  n1  2 N  n0  
 1n −   n
n3  2   n  −  n  
 1 + RN
2N
  2  3  n  n
RN =   =  ∴  1
 = 0n 
 n12 N +  n0 n22 N    n1 
2N
  n 2  3 1−
+  0  
n RN
  n 3    
  n 2  n3 

Insert RN = 0.95 and solve,

 n 1 + RN 
ln 0 
 n 1 − R 
1  3 N 
N= =14.58 i.e. 15 pairs are needed
2 
ln
n1 

 n2 
∆λ 4  n − n2 
≈ arcsin 1  =0.095167
λ0 π  n1 + n2 
But, λ0 = 1500 nm, ∴ ∆λ = 142 nm

1.20 TIR and polarization at water-air interface

(a) Given that the refractive index of water is about 1.33, what is the polarization angle for light
traveling in air and reflected from the surface of the water?
(b) Consider a diver in sea pointing a flashlight towards the surface of the water. What is the
critical angle for the light beam to be reflected from the water surface?

Solution
n2
(a) Apply tan θ p =
n1
n1 = 1
with
n2 = 1.33
∴ =θ p tan
= −1
=
(n2 / n1 ) tan −1
(1.33 /1) 53.06

(b) Given
n1 = 1.33
and
n2 = 1
The critical angle is
−1 n2 −1 1
=θ c sin
= sin= 48.75
n1 1.33

1.21 Reflection and transmission at a semiconductor-semiconductor interface A light wave with

a free space wavelength of 890 nm (free space wavelength) that is propagating in GaAs becomes
incident on AlGaAs. The refractive index of GaAs is 3.60, that of AlGaAs is 3.30.
(a) Consider normal incidence. What are the reflection and transmission coefficients and the
reflectance and transmittance? (From GaAs into AlGaAs)

(b) What is the Brewster angle (the polarization angle θp) and the critical angle (θc) for total internal
reflection for the wave in (a); the wave that is traveling in GaAs and incident on the GaAs/AlGaAs
interface.
(c) What is the reflection coefficient and the phase change in the reflected wave when the angle of
incidence θi = 79°?
(d) What is the penetration depth of the evanescent wave into medium 2 when θi = 79° and when θi
= 89°? What is your conclusion?

Solution
(a) Given,
n1 = 3.60
n2 = 3.30
n1 − n2 3.60 − 3.30
we have r/= r⊥= = = 0.043
n1 + n2 3.60 + 3.30
/

∴ R= R= r⊥= r/ =
2 2
⊥ // / 0.0018
2n1 3.60
and t// =t⊥ = = 2× = 1.043
n1 + n2 3.60 + 3.30
4n1n2 4 × 3.60 × 3.30
∴ T = T=⊥ T= = = 0.998
(n1 + n2 ) (3.60 + 3.30) 2
// 2

(b) Apply
n2
tan θ p =
n1
n
and sin θ c = 2
n1
n1 = 3.60
Take
n2 = 3.30
∴ =θ p tan
= −1
(n2 / n1 ) tan −1 (3.30
= / 3.60) 42.51
−1 n2 −1 3.30
∴ =θ c sin
= sin= 66.44
n1 3.60
(c) Take
n2 3.30
=
n = = 0.9166
n1 3.60
cos θi − [n 2 − sin 2 θi ]1/2
r⊥ =
cos θi + [n 2 − sin 2 θi ]1/2
cos(79 ) − [(0.9166) 2 − sin 2 (79 )]1/2
then, =
cos(79 ) + [(0.9166) 2 − sin 2 (79 )]1/2
0.1908 − 0.3513 j
=
0.1908 + 0.3513 j

0.1908 − 0.3513 j 0.1908 − 0.3513 j

=r⊥ ×
0.1908 + 0.3513 j 0.1908 − 0.3513 j

and 0.19082 − 0.35132 − 2 j 0.1908 × 0.3513

=
0.1598
−0.087 − 0.1340 j
=
0.1598
= −0.5444 − 0.8385 j
∴ r⊥ =−0.5444 + j (−0.8385)
∴ r⊥ = 0.9997exp( j 237 )
=
We know that r⊥ | r⊥ | exp(− jφ⊥ ) , thus
=
0.9997exp( j 57 ) 0.9997exp(− jφ⊥ )
∴ exp(− jφ⊥ ) =
exp( j 237 )
φ⊥ = −237
∴
φ⊥ = 123
1/2 1/2
sin 2 θi − n 2  sin 2 (79°) − (0.9166) 2 
or = tan ϕ⊥ =
cos θi
( 1
2 ) cos(79°)
∴ ϕ⊥ = 122.98 = 123°
Now the parallel component, r//,

[n 2 − sin 2 θi ]1/2 − n 2 cos θi

r/ / =
[n 2 − sin 2 θi ]1/2 + n 2 cos θi
[(0.9166) 2 − sin 2 (79 )]1/2 − (0.9166) 2 cos(79 )
=
[(0.9166) 2 − sin 2 (79 )]1/2 + (0.9166) 2 cos(79 )
0.3513 j − 0.1603
=
0.3513 j + 0.1603

0.3513 j − 0.1603 0.3513 j − 0.1603

=r/ / ×
0.3513 j + 0.1603 0.3513 j − 0.1603
−0.0977 − j 0.1126
∴ =
−0.1491
= 0.6552 + j 0.7552

=r/ / 0.6552 + j (0.7552)

∴
=
r/ / 1.088 × exp( j 49.05 )
We know that =r/ / | r // | exp(− jφ // ) , thus
exp(− jφ // ) =
exp( j 49.05 )

∴ φ/ / = −49.05

or,
tan ( 1
2
+ 12 π )
ϕ / /=
1
n 2
tan=
1
ϕ
2 ⊥
1
0.91662
(
tan ) ( 1
2
=
122.98
)
2.1913

∴ = (ϕ / / + π ) tan −1=
1
2
(2.1913) 65.47
∴ ϕ/ / + π ) =
( 130.94
∴ ϕ/ / =
130.94 − 180 =
−49.05

(d)
The attenuation coefficient α2 for θi =79° is
1/ 2
2πn2  n1 
2

α2 =   sin 2 θ i − 1
λo  n2  

1/2
2π (3.30)  3.60  
2

i.e. α= −9    sin 2
(79 °) − 1 = 8.92 ×106 m −1 .
(890 ×10 m)  3.30 
2

so the penetration depth is δ = 1/α2 = 1/(8.92×106 m-1) = 1.12×10-7 m, or 0.112 µm
For 89°, repeating the calculation

1/2
2π (3.30)  3.60  
2

α= −9   sin (89°) − 1 = 1.01×10 m

2 7 −1

(890 ×10 m)  3.30 

2


So, the penetration depth is δ = 1/α2 = 1/( 1.01×107 m −1 ) = 9.9×10-8 m, or 990nm.

The conclusion is that the penetration depth decreases as the incidence angle increases

1.22 Phase changes on TIR Consider a light wave of wavelength 870 nm traveling in a
semiconductor medium (GaAs) of refractive index 3.60. It is incident on a different semiconductor
medium (AlGaAs) of refractive index 3.40, and the angle of incidence is 80°. Will this result in total
internal reflection? Calculate the phase change in the parallel and perpendicular components of the
reflected electric field.

Solution
−1 n2 −1 3.40
=θ c sin
= sin= 70.81
n1 3.60
Clearly, =θi 80 > θ c
So, this results in total internal reflection.

n2 3.40
=
n = = 0.9444
n1 3.60

1/2 1/2
sin 2 θi − n 2  sin 2 (80°) − (0.9444) 2 
tan ϕ⊥
= = (
cos θi
1
2

) = 1.6078
cos(80°)
∴ =ϕ⊥ 2=
tan −1 (1.6078) 116.24

tan ( 1
2
+ 12 π )
ϕ / /=
n
1
2
tan=
1
ϕ
2 ⊥
1
0.94442
(
tan ) ( 1
2
=
116.24 
)
1.8027

∴ = (ϕ / / + π ) tan −1=
1
2
(1.8027) 60.98
∴ ϕ/ / + π ) =
( 121.96
∴ ϕ/ / =
121.96 − 180 =
−58.04

1.23 Fresnel’s equation Fresnel's equations are sometimes given as follows:

E n cosθ i − n 2 cosθ t
r ⊥ = r 0,⊥ = 1
Ei 0,⊥ n1 cosθ i + n 2 cosθ t
E r 0,// n1 cosθ t − n 2 cosθ i
r // = =
Ei 0,// n1 cosθ t + n 2 cosθ i
E t 0,⊥ 2n1 cosθ i
t⊥ = =
E i 0,⊥ n1 cosθ i + n 2 cosθ t
2n1 cosθ iE t 0,//
and t // = =
Ei 0,// n1 cosθ t + n 2 cosθ i
Show that these reduce to Fresnel's equation given in Section 1.6.
Using Fresnel's equations, find the reflection and transmission coefficients for normal
incidence and show that
r⊥ + 1 = t ⊥ and r // + nt // = 1
where n = n2/n1.

Solution
n2 / n1 = n
From Snell’s law
sin θi n2
= = n
sin θt n1
sin 2 θi
∴ sin 2 θt =
n2
Perpendicular component

Er 0,⊥ n1 cos θi − n2 cos θt cos θi − n2 / n1 cos θt

=
r⊥ = =
Ei 0,⊥ n1 cos θi + n2 cos θt cos θi + n2 / n1 cos θt
cos θi − n cos θt cos θi − n[1 − sin 2 θt ]1/2
∴ =r⊥ =
cos θi + n cos θt cos θi + n[1 − sin 2 θt ]1/2

sin 2 θi 1/2
cos θi − n[1 − 2
]
∴ r⊥ = n
sin 2 θi 1/2
cos θi + n[1 − ]
n2
cos θi − [n 2 − sin 2 θi ]1/2
∴ r⊥ =
cos θi + [n 2 − sin 2 θi ]1/2
Parallel component
Er 0,/ / n1 cos θt − n2 cos θi cos θt − n2 / n1 cos θi cos θt − n cos θi
=
r/ / = = =
Ei 0,/ / n1 cos θt + n2 cos θi cos θt + n2 / n1 cos θi cos θt + n cos θi
sin 2 θi 1/2
] − n cos θi [1 −
[1 − sin θt ] − n cos θi n2
2 1/2
∴ =r/ / =
[1 − sin 2 θt ]1/2 + n cos θi sin 2 θi 1/2
[1 − ] + n cos θi
n2
[n 2 − sin 2 θi ]1/2 − n 2 cos θi
∴ r/ / = 2
[n − sin 2 θi ]1/2 + n 2 cos θi

Et 0,⊥ 2n1 cos θi 2 cos θi 2 cos θi

= =
t⊥= =
Ei 0,⊥ n1 cos θi + n2 cos θt cos θi + n2 / n1 cos θt cos θi + n cos θt
2 cos θi 2 cos θi
∴ =t⊥ =
cos θi + n[1 − sin θt ]
2 1/2
sin 2 θi 1/2
cos θi + n[1 − ]
n2
2 cos θi
∴ t⊥ =
cos θi + [n 2 − sin 2 θi ]1/2
Et 0,/ / 2n1 cos θi 2 cos θi 2 cos θi
=
t// = = =
Ei 0,/ / n1 cos θt + n2 cos θi cos θt + n2 / n1 cos θi cos θt + n cos θi

2 cos θi 2 cos θi
∴ =t// =
cos θt + n cos θi sin θi 1/2
2
[1 − ] + n cos θi
n2
2n cos θi
∴ t// =
[n − sin 2 θi ]1/2 + n 2 cos θi
2

cos θi − [n 2 − sin 2 θi ]1/2

=r⊥ + 1 +1
cos θi + [n 2 − sin 2 θi ]1/2
cos θi − [n 2 − sin 2 θi ]1/2 + cos θi + [n 2 − sin 2 θi ]1/2
=
cos θi + [n 2 − sin 2 θi ]1/2

2 cos θi
∴ r⊥ + 1 =
cos θi + [n 2 − sin 2 θi ]1/2
∴ r⊥ + 1 =t⊥

[n 2 − sin 2 θi ]1/2 − n 2 cos θi 2n cos θi

=r/ / + nt / / +n 2
[n − sin θi ] + n cos θi
2 2 1/2 2
[n − sin 2 θi ]1/2 + n 2 cos θi
[n 2 − sin 2 θi ]1/2 − n 2 cos θi + 2n 2 cos θi
∴ r/ / + nt / / =
[n 2 − sin 2 θi ]1/2 + n 2 cos θi
[n 2 − sin 2 θi ]1/2 + n 2 cos θi
∴ r/ / + nt / / =
[n 2 − sin 2 θi ]1/2 + n 2 cos θi
∴ r/ / + nt / / =
1

1.24 Fresnel's equations Consider a light wave traveling in a glass medium with an index n1 = 1.440
and it is incident on the glass-air interface. Using Fresnel equations only i.e. Eqs (6a) and 6(b) in §1.6,
calculate the reflection coefficients r⊥ and r// and hence reflectances R⊥ and R// for (a) θi = 25° and (b)
θi = 50°. In the case of θi = 50°, find the phase change φ⊥ and φ// from the reflection coefficients by
writing r = |r|exp(−jφ). Compare φ⊥ and φ// from r⊥ and r// calculations with those calculated from Eqs
(11) and (12).

Solution

The above problem is solved using LiveMath and reproduced below. It should be relatively
straightforward to follow.

na = n1; nb = n2; rπ = r//; rσ = r⊥; subscript σ means ⊥; π means //.

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For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
1.25 Goos-Haenchen phase shift A ray of light which is traveling in a glass medium (1) of
refractive index n1 = 1.460 becomes incident on a less dense glass medium (2) of refractive index n2 =
1.430. Suppose that the free space wavelength of the light ray is 850 nm. The angle of incidence θi =
85°. Estimate the lateral Goos-Haenchen shift in the reflected wave for the perpendicular field
component. Recalculate the Goos-Haenchen shift if the second medium has n2 = 1 (air). What is your
conclusion? Assume that the virtual reflection occurs from a virtual plane in medium B at a distance d
that is the same as the penetration depth. Note that d actually depends on the polarization, the direction
of the field, but we will ignore this dependence.

Solution

Figure 1.20 The reflected light beam in total internal reflection appears to have been laterally shifted by an amount ∆z at the
interface. It appears as though it is reflected from a virtual plane at a depth d in the second medium from the interface.

The problem is shown in Figure 1.20. When θi = 85°,

1/ 2
2π n2  n1  
2
α2 =  sin θ i − 1
2
λ o  n 2  
−7
The penetration depth is δ = 1/α2 = 5.09×10 m.
As an estimate, we can assume that d ~ δ so that the Goose-Haenchen shift is
∆z ≈ 2dtanθ = 2(5.09×10-7 m)(tan85°) = 11.6×10-6 m = 11.6 µm

We can repeat the calculation using n2 = 1 (air), then we find δ = 1/α2 = 1.28×10-7 m, and ∆z ≈
2dtanθ = 2(1.28×10-7 m)(tan85°) = 2.93×10-6 m = 2.93 µm. The shift is small when the refractive index
difference is large. The wave penetrates more into the second medium when the refractive index
difference is smaller. Note: The use of d ≈ δ is a rough approximation to estimate ∆z.

1.26 Evanescent wave Total internal reflection (TIR) of a plane wave from a boundary between a more
dense medium (1) n1 and a less dense medium (2) n2 is accompanied by an evanescent wave propagating
in medium 2 near the boundary. Find the functional form of this wave and discuss how its magnitude
varies with the distance into medium 2.

Solution

The transmitted wave has the general form

Et,⊥ = t⊥Eio,⊥expj(ωt − kt⋅r)

in which t⊥ is the transmission coefficient. The dot product, examining

kt⋅r = yktcosθt + zkt sinθt.

However, from Snell's law, when θi > θc, sinθt = (n1/n2)sinθi > 1 and cosθt = √[1 − sin2θt] = ±jA2
is a purely imaginary number. Thus, taking cosθt = −jA2

Et,⊥ = t⊥Eio,⊥expj(ωt – zktsinθt + jyktA2) = t⊥Eio,⊥exp(−yktA2)expj(ωt – zktsinθt)

which has an amplitude that decays along y as exp(–α2y) where α2 = ktA2. Note that +jA2 is ignored
because it implies a wave in medium 2 whose amplitude and hence intensity grows. Consider the
traveling wave part expj(ωt – zktsinθt). Here, ktsinθt = kisinθi (by virtue of Snell's law). But kisinθi = kiz,
which is the wave vector along z, that is, along the boundary. Thus the evanescent wave propagates along
z at the same speed as the incident and reflected waves along z. Furthermore, for TIR we need sinθi >
n2/n1. This means that the transmission coefficient,

ni cosθ i
t⊥ = 12 = t⊥ 0 exp (j ψ ⊥ )
 n  2 
cosθ i +   2  − sin2 θ i 
n 
 1 

must be a complex number as indicated by t⊥0exp(jψ⊥) in which t⊥0 is a real number and ψ⊥ is a phase
change. Note that t⊥ does not, however, change the general behavior of propagation along z and the
penetration along y.

1.27 TIR and FTIR

(a) By considering the electric field component in medium in Figure 1.22(b), explain how you can
adjust the amount of transmitted light.
(b) What is the critical angle at the hypotenuse face of a beam splitter cube (Figure 1.22 (b)) made
of glass with n1 = 1.6 and having a thin film of liquid with n2 =1.3. Can you use 45° prisms with
normal incidence?
(c) Explain how a light beam can propagate along a layer of material between two different media
as shown in Figure 1.59 (a). Explain what the requirements are for the indices n1, n2, n3. Will there be
any losses at the reflections?
(d) Consider the prism coupler arrangement in Figure 1.59(b). Explain how this arrangement
works for coupling an external light beam from a laser into a thin layer on the surface of a glass
substrate. Light is then propagated inside the thin layer along the surface of the substrate. What is the
purpose of the adjustable coupling gap?

Figure 1.22 (a) A light incident at the long face of a glass prism suffers TIR; the prism deflects the light. (b) Two prisms
separated by a thin low refractive index film forming a beam-splitter cube. The incident beam is split into two beams by FTIR.

Figure 1.59 (a) Light propagation along an optical guide. (b) Coupling of laser light into a thin layer - optical guide - using a
prism. The light propagates along the thin layer.

Solution
(a) Consider the prism A when the neighboring prism C in Figure 1.22 (b) in far away. When the
light beam in prism A is incident on the A/B interface, hypotenuse face, it suffers TIR as θi > θc. There
is however an evanescent wave whose field decays exponentially with distance in medium B. When
we bring prism C close to A, the field in B will reach C and consequently penetrates C. (The tangential
field must be continuous from B to C). One cannot just use the field expression for the evanescent
wave because this was derived for a light beam incident at an interface between two media only; no
third medium. The transmitted light intensity from A to C depends on the thickness of B.

(b) For the prism A in Figure 1.22 (b), n1 = 1.6 and n2 = 1.3 so that the critical angle for TIR at the
hypotenuse face is
θc = arcsin(n2/n1) = arcsin(1.3/1.6) = 54.3°

In this case, one cannot use a 45 ° prism.

(d) If the angle of incidence θi at the n1/n2 layer is more than the critical angle θc12 and if angle of
incidence θi at the n1/n3 layer is more than the critical angle θ13 then the light ray will travel by TIR,
zigzagging between the boundaries as sketched in Figure 1.59(a). For example, suppose that n1 = 2
(thin layer); n2 = 1 (air) and n3 = 1.6 (glass),
θc12 = arcsin(n2/n1) = arcsin(1/2) = 38.8°,

and θc13 = arcsin(n3/n1) = arcsin(1.6/2) = 53.1°,

so that θi > 53.1° will satisfy TIR. There is no loss in TIR as the magnitude of the amplitude of the
reflected way is the same as that of the incident wave.
Note: There is an additional requirement that the waves entering the thin film interfere constructively,
otherwise the waves will interfere destructively to cancel each other. Thus there will be an additional
requirement, called the waveguide condition, which is discussed in Chapter 2.

(e) The light ray entering the prism is deflected towards the base of the prism as shown in Figure
1.59 (b). There is a small gap between the prism and the thin layer. Although the light arriving at the
prism base/gap interface is reflected, because of the close proximity of the thin layer, some light is
coupled into the thin layer per discussion in Part (a) due to frustrated TIR. This arrangement is a much
more efficient way to couple the light into the thin layer because the incident light is received by the
large hypotenuse face compared with coupling the light directly into the thin layer.

1.28 Complex refractive index and dielectric constant The complex refractive index Ν = n − jK can
be defined in terms of the complex relative permittivity εr = εr1 − jεr2 as
Ν = n − jK = ε r1/ 2 = (εr1 − jεr 2)1/2
where εr1 and εr2 are the real and imaginary parts of εr. Show that
1/ 2 1/ 2
 (ε 2 + ε r22 )1 / 2 + ε r1   (ε 2 + ε 2 )1/ 2 − ε r1 
n =  r1  and K =  r1 r 2 
 2   2 

Solution
Given Ν = n − jK = ε r = ε r1 − jε r 2

we have n 2 − 2 jnK − K 2 = ε r1 − jε r 2
∴ 2nK = ε r 2 (1)
and n 2 − K 2 = ε r1 (2)
∴ K = ε r 2 / 2n
and substituting into the second equation above,
2
ε 
n −  r 2  = ε r1
2

 2n 
∴ n − n 2ε r1 − 14 ε r22 = 0
4

ε r1 ± ε r21 − 4( − 14 ε r22 ) ε r1 ± ε r21 + ε r22

∴ n2 = =
2 2
It is apparent that n2 has two solutions. The negative sign has to be excluded because this would make
the numerator negative and lead to a complex number for n. By definition, n is a real number, and not
imaginary. Thus,
ε r1 + ε r21 + ε r22
n2 =
2

1/ 2
 (ε 2 + ε 2 )1 / 2 + ε r1 
∴ n =  r1 r 2 
 2 
From Eq. (1), n =εr2/2K, Substitute this into Eq. (2),
2
 ε r2 
  − K 2 = ε r1
 2K 
∴ K 4 + K 2ε r1 − 14 ε r22 = 0
− ε r1 ± ε r21 − 4( − 14 ε r22 ) − ε r1 + ε r21 + ε r22
∴ K =2
=
2 2
Where the negative sign is excluded as K cannot imaginary. Thus,
1/ 2
 (ε 2 + ε 2 )1 / 2 − ε r1 
∴ K =  r1 r 2 
 2 

1.29 Complex refractive index Spectroscopic ellipsometry measurements on a germanium crystal at

a photon energy of 1.5 eV show that the real and imaginary parts of the complex relative permittivity
are 21.56 and 2.772 respectively. Find the complex refractive index. What is the reflectance and
absorption coefficient at this wavelength? How do your calculations match with the experimental
values of n = 4.653 and K = 0.298, R = 0.419 and α = 4.53 × 106 m-1?
Solution
From problem 1.28 we have
1/ 2
 ( 21.562 + 2.7722 )1 / 2 + 21.56 
1/ 2
 (ε r21 + ε r22 )1 / 2 + ε r1 
n =  =  = 4.653
 2   2 
Similarly
1/ 2
 ( 21.562 + 2.7722 )1 / 2 − 21.56 
1/ 2
 (ε 2 + ε 2 )1 / 2 − ε r1 
K =  r1 r 2  =  = 0.298
 2   2 
Almost an exact agreement (not surprisingly).

The reflectance R is given by

(n − 1) 2 + K 2 (4.653 − 1) 2 + 0.2982
= R = = 0.42 or 42%
(n + 1) 2 + K 2 (4.653 + 1) 2 + 0.2982
The absorption coefficient α is 2k″ as in Eq. (1.8.67) so that
α = 2k″ = 2koK = 2(2π/λo)K = 2(2πυ/c)K
2( 2π )hυK 2( 2π )(1.5 eV )(0.298)
∴ α= = −15
= 4.53×10-6 m-1
hc ( 4.136 × 10 eV s)(3 × 10 m s )
8 -1

which agrees with the measurements.

1.30 Complex refractive index Figure 1.26 shows the infrared extinction coefficient K of CdTe .
Calculate the absorption coefficient α and the reflectance R of CdTe at 60 μm and 80 μm.

Solution

At 60 μm:

K ≈ 1, and n ≈ 0.6, so that the corresponding free-space wave vector is

ko = 2π/λ = 2π/(60×10-6 m) = 1.05×105 m-1.
The absorption coefficient α is 2k″ as in Eq. (1.8.6) so that
α = 2k″ = 2koK = 2(1.05×105 m-1)(1) = 2.1×105 m-1
which corresponds to an absorption depth 1/α of about 4.8 micron. The reflectance is
( n − 1) 2 + K 2 (0.6 − 1) 2 + 12
R= = = 0.32 or 32%
( n + 1) 2 + K 2 (0.6 + 1) 2 + 12
At 80 μm:
K ≈ 0.27, and n ≈ 4.5, so that the corresponding free-space wave vector is
ko = 2π/λ = 2π/(80×10-6 m) = 7.85×104 m-1.
The absorption coefficient α is 2k″ so that
α = 2k″ = 2koK = 2(7.85×104 m-1)(0.27) = 4.2×104 m-1
which corresponds to an absorption depth 1/α of about 24 micron. The reflectance is
( n − 1) 2 + K 2 (0.6 − 1) 2 + 12
R= = = 0.41 or 41%
( n + 1) 2 + K 2 (0.6 + 1) 2 + 12

1.31 Refractive index and attenuation in the infrared region - Reststrahlen absorption Figure
1.26 shows the refractive index n and the attenuation (absorption) coefficient K as a function of
wavelength λ in the infrared for a CdTe crystal due to lattice absorption, called Reststrahlen
absorption. It results from the ionic polarization of the crystal induced by the optical field in the light
wave. The relative permittivity εr due to positive (Cd2+) and negative (Te2−) ions being made to
oscillate by the optical field about their equilibrium positions is given in its simplest form by

ε rH − ε rL
ε r = ε r′ − jε r′′ = ε rH + 2
(1)
ω  γ ω 
  − 1 + j  
 ωT  ωT  ωT 
where εrL and εrΗ are the relative permittivity at low (L) and high (H) frequencies, well below and
above the infrared peak, γ is a loss coefficient characterizing the rate of energy transfer from the EM
wave to lattice vibrations (phonons), and ωT is a transverse optical lattice vibration frequency that is
related to the nature of bonding between the ions in the crystal. Table 1.3 provides some typical values
for CdTe and GaAs. Eq. (1) can be used to obtain a reasonable approximation to the infrared refractive
index n and absorption K due to Reststrahlen absorption.
(a) Consider CdTe, and plot n and K vs. λ from 40 µm to 90 µm and compare with the experimental
results in Figure 1.26 in terms of the peak positions and the width of the absorption peak.
(b) Consider GaAs, and plot n and K vs. λ from 30 µm to 50 µm.
(c) Calculate n and K for GaAs at λ = 38.02 µm and compare with the experimental values n = 7.55
and K = 0.629.

Figure 1.26 Optical properties of CdTe as a function of wavelength in the infrared region.

Solution
From Question 1.28

1
 ε r′ + (ε r′ ) + (ε r′′) 
n= K = n 2 − ε r′
2 2

2 
which means that we can substitute the values of εr′ and εr′′ from Eq. (1) into the above two equations and
plot n and K as a function of wavelength.
(a) CdTe

(b) GaAs

More points can also be used

(c) For GaAs at 38.02 µm, the calculated values are n = 7.44 and K = 0.586, which compare
reasonable well with experimental values of n = 7.55 and K = 0.629.

1.32 Coherence length A particular laser is operating in single mode and emitting a continuous
wave lasing emission whose spectral width is 1 MHz. What is the coherence time and coherence
length?

Solution
The spectral width in frequency ∆υ and the coherence time ∆t are related by
1
∆υ ≈
∆t

Thus, the coherence time is

∆t ≈ 1/ ∆υ = 1/(1×106 Hz) =10-6 s or 1µs.
The coherence length is
lc = c∆t = (3 ×108 m s −1 ) ×10-6 s = 300m

1.33 Spectral widths and coherence

(a) Suppose that frequency spectrum of a radiation emitted from a source has a central frequency
υo and a spectral width ∆υ. The spectrum of this radiation in terms of wavelength will have a central
wavelength λo and a spectral width ∆λ. Clearly, λo = c/υo. Since ∆λ << λo and ∆υ << υo, using λ =
c/υ, show that the line width ∆λ and hence the coherence length lc are
λo λ2 λo2
∆λ = ∆υ = ∆υ o and l c = c ∆t =
υo c ∆λ
(b) Calculate ∆λ for a lasing emission from a He-Ne laser that has λo = 632.8 nm and ∆υ ≈ 1.5
GHz. Find its coherence time and length.

Solution
(a) See Example 1.9.3
(b) Consider the width in wavelength,
∆λ = ∆υ( λ2/c) = (1.5×109 s-1 (632/8×10-9 m)2/(3×108 m s-1) = 3.16×10-6 m.
The coherence time is
∆t ≈ 1/ ∆υ = 1/(1.5×109 Hz) = 0.666×10-9 s
The coherence length is
lc = c∆t = (3×108 m s-1)(0.666×10-9 s) = 0.20 m = 20 cm

1.34 Coherence lengths Find the coherence length of the following light sources
(a) An LED emitting at 1550 nm with a spectral width 150 nm
(b) A semiconductor laser diode emitting at 1550 nm with a spectral width 3 nm
(c) A quantum well semiconductor laser diode emitting at 1550 nm with a spectral with of 0.1 nm
(d) A multimode HeNe laser with a spectral frequency with of 1.5 GHz
(e) A specially designed single mode and stabilized HeNe laser with a spectral width of 100 MHz
Solution
∆υ dυ c
(a) ≈ = − 2
∆λ dλ λ
so that ∆υ = ∆λ(c/λ2) = (150×10-9 m)(3×108 m s-1)/(1550×10-9 m)2 = 1.873×1013 Hz
Thus, the coherence time is
∆t ≈ 1/ ∆υ = 1/(1.873×1013 Hz) = 5.34×10-14 s
The coherence length is
lc = c∆t = 1.6×10-5 m or 16 µm
(b)

∆υ = ∆λ(c/λ2) = (3×10-9 m)(3×108 m s-1)/(1550×10-9 m)2 = 3.746×1011 Hz

The coherence time is

∆t ≈ 1/ ∆υ = 1/(3.746×1011Hz) = 2.67×10-12 s or 2.67 fs
The coherence length is
lc = c∆t = 8.01×10-4 m or 0.8 mm

(c) Apply ∆υ = ∆λ(c/λ2), that is,

∆υ = ∆λ(c/λ2) = (0.1×10-9 m)(3×108 m s-1)/(1550×10-9 m)2 = 1.248×1010 Hz

The coherence time is

∆t ≈ 1/ ∆υ = 1/(1.248×1010 Hz) = 8.01×10-11 s or 80.1 fs
The coherence length is
lc = c∆t = 2.4×10-2 m or 24 mm

(d) Apply
1
∆υ ≈
∆t
Thus, the coherence time is
∆t ≈ 1/ ∆υ = 1/(1.5×109 Hz) =6.66×10-10 s or 666 fs.
The coherence length is
lc = c∆t = (3 ×108 m s −1 ) ×6.66×10-10 s = 0.2 m = 20 cm
(e)
1
∆υ ≈
∆t
Thus, the coherence time is
∆t ≈ 1/ ∆υ = 1/(100×106 Hz) =10-8 s
The coherence length is
lc = c∆t = (3 ×108 m s −1 ) ×10-8 s = 3 m

1.35 Fabry-Perot optical cavity Consider an optical cavity formed between two identical mirrors.
The cavity length is 50 cm and the refractive index of the medium is 1. The mirror reflectances are
0.97 each. What is the nearest mode number that corresponds to a radiation of wavelength 632.8 nm?
What is the actual wavelength of the mode closest to 632.8 nm? What is the mode separation in
frequency and wavelength? What are the finesse F and Q-factors for the cavity?

Solution
For λ = λo = 632.8 nm, the corresponding mode number mo is,
mo = 2L / λo = (2×0.5 m) / (632.8×10-9 m) = 1580278.1
and actual mo has to be the closest integer value to 1580278.1, that is 1580278
The actual wavelength of the mode closest to 632.8 nm is λo = 2L / mo =(2×0.5 m) / (1580278) =
632.80005 nm
The frequency separation ∆υm of two consecutive modes is
c c c c c
∆ υm = υ m +1 – υm = − = 2L − =
λ m +1 λ m 2L 2L
(m + 1) m

c 3 × 10 8
or ∆ υm =
= = 3×108 Hz.
2L 2(0.5)
The wavelength separation of two consecutive modes is
λ2m (632.8 × 10 −9 ) 2
∆ λm = = = 4.004×10-13 m or 0.400 pm.
2L 2(0.5)
Finesse is
πR 1 / 2
F= =103.1
1− R
Quality factor is
Q = m0F =1580278×103.14 = 1.63×108

1.36 Fabry-Perot optical cavity from a ruby crystal Consider a ruby crystal of diameter 1 cm and
length 10 cm. The refractive index is 1.78. The ends have been silvered and the reflectances are 0.99
and 0.95 each. What is the nearest mode number that corresponds to a radiation of wavelength 694.3
nm? What is the actual wavelength of the mode closest to 694.3 nm? What is the mode separation in
frequency and wavelength? What are the finesse F and Q-factors for the cavity?

Solution
Number mode nearest to the emission wavelength is
2L
m= =(2×10 cm)×(1.78)/(694.3 nm) = 512746.65 i.e. i.e. m0 = 512746.
λ
n
The actual wavelength of the mode closest to 694.3 nm is
2 Ln
λ0 = =(2×10 cm) ×(1.78)/ (512746) = 694.3008819 nm
m0
The frequency separation ∆υm of two consecutive modes is
c c c c c
∆υ m = υ m +1 − υ m = − = − = = 8.43×109 Hz.
λm +1 λm 2 Ln 2 Ln 2 Ln
(m + 1) m
The wavelength separation of two consecutive modes is
λ2m
∆λ = = 0.00135408 nm = 1.35 pm
2 Ln
Average geometric reflectance is R = (R1R2)1/2=0.96979
πR 1 / 2
Finesse, F = = 102.42
1− R
Quality factor, Q = m0F =1580278×103.14= 5.25×107

1.37 Fabry-Perot optical cavity spectral width Consider an optical cavity of length 40 cm. Assume
the refractive index is 1, and use Eq. (1.11.3) to plot the peak closest to 632.8 nm for 4 values of R =
0.99, 0.90, 0.75 and 0.6. For each case find the spectral width δλm,, the finesse F and Q. How accurate
is Eq.(1.11.5) in predicting δλm? (You may want to use a graphing software for this problem.)

Solution

 c 
υ m = m  = mυ f ; υf = c/(2L) Cavity resonant frequencies (2)
 2L 
Io
I cavity = Cavity intensity (3)
(1 − R ) + 4R sin 2 (kL)
2

Io
I max = ; kmL = mπ Maximum cavity intensity (4)
(1 − R ) 2
υf πR 1 / 2
δυ m = ; F= Spectral width (5)
F 1− R
Resonant frequency υm
Quality factor, Q = = = mF (6)
Spectral width δυm
Cavity fundamental mode is υf = c/(2L) = 3.75×108 Hz. The Graph below shows that the peak closest
to 632.8 nm is 632.80025 nm which corresponds to υm = 4.7408325×1014 Hz.

AU: Arbitrary Units

R 0.99 0.9 0.75 0.6 Source

δλ, nm 1.5000E-06 1.6750E-05 4.6000E-05 8.3000E-05 From Graph
δνexp, MHz 1.12 12.55 34.46 62.18 From Graph
υf
δνcalc, MHz 1.20 12.58 34.46 61.64 δυ m =
F
πR 1 / 2
Fcalculation 312.58 29.80 10.88 6.08 F=
1− R
υf
Fexperiment 333.70 29.88 10.88 6.03 F=
δυ m
υm
Q 4.219E+14 3.778E+13 1.376E+13 7.624E+12 Q=
δυ m

1.38 Diffraction Suppose that a collimated beam of light of wavelength 600 nm is incident on a
circular aperture of diameter of 200 µm. What is the divergence of the transmitted beam? What is the
diameter at a distance 10 m? What would be the divergence if the aperture were a single slit of width
200 µm?

Solution
(a)
λ 600 ×10−9
sin=θ o 1.22= 1.22 =−6
3.66 ×10−3
D 200 ×10
θ o = 0.209 

The divergence angle is

2θ o = 0.418
If R is the distance of the screen from the aperture, then the radius of the Airy disk, approximately b,
can be calculated from = b / R tanθ o ≈ θ o
b ==R tanθ o 10 × tan(0.209 ) =
0.036m =
3.6cm
Thus, the diameter is
2b = 0.072 m or 7.2 cm

(b)
Divergence from a single slit of width a is
2λ 2 × 600 ×10−9 360
∆θ = 2θ o ≈ = = 6 ×10−3 rad = 6 ×10−3 × = 0.34
a 200 ×10 −6
2π

1.39 Diffraction Consider diffraction from a uniformly illuminated circular aperture of diameter D.
The far field diffraction pattern is given by a Bessel function of the first kind and first order, J1, and
the intensity at a point P on the angle θ with respect to the central axis through aperture is
2
 2 J (γ ) 
I (γ ) = I o  1 
 γ 

where Io is the maximum intensity, γ = (1/2)kDsinθ is a variable quantity that represents the angular
position θ on the screen as well as the wavelength (k = 2π/λ) and the aperture diameter D. J1(γ) can be
calculated from
1 π
J 1 (γ ) = ∫ cos(α − γ sin α )dα
π 0

Using numerical integration (or a suitable math software package), plot [J1(γ)/γ ] vs. γ for γ = 0 to 10
using suitable number of points, and then find the zeros. What are the first two γ that lead to dark
rings?

Solution

6.000E-01
5.000E-01 ×1
4.000E-01
3.000E-01
2.000E-01
×5
J
1.000E-01
0.000E+00
-1.000E-01
-2.000E-01
-3.000E-01

γ
-4.000E-01
0.00 2.00 4.00 6.00 8.00 10.00

There are two zeros at γ = 3.80 and 7.00

0 .4

0 .2

Y
0
0 γ 2 4 6 8

The above is from Livemath (Theorist)

The ratio of the intensity of first bright ring to the intensity at the center of the Airy disk
2
  J (γ )  
 1 
  γ  
γ →0 
= = 0.017
 J 1 (5.14) 
 
 5.14 

Additional Problem
Consider an aperture that is 50 µm in diameter and illuminated by a 550 nm green laser light beam. If
the screen is 2 m away, what are the radius of the first dark ring?The first dark ring occurs when
γ = 3.83 so that γ = (1/2)kDsinθ = (1/2)(2π/λ)Dsinθ

γ = 3.83 = (1/2)kDsinθ = (1/2)[2π/(0.550 µm)](50 µm) sinθ

gives θ = 1.16o.
Let R be the distance from the aperture to the screen. If the radius of the dark ring is r then r/R = tanθ.
Thus substituting R = 2 m, and θ = 1.16o we find r = 0.024 m and 2r = 0.048 m.

1.40 Bragg diffraction Suppose that parallel grooves are etched on the surface of a semiconductor to
act as a reflection grating and that the periodicity (separation) of the grooves is 1 micron. If light of
wavelength 1.3 µm is incident at an angle 89° to the normal, find the diffracted beams.

Solution
When the incident beam is not normal to the diffraction grating, then the diffraction angle θm for the
m-th mode is given by,
d(sinθm − sinθi) = mλ ; m = 0, ±1, ±2, …

so that for first order

(1 µm)(sinθm − sin(89°) = (+1)(1.3 µm)
and (1 µm)(sinθm − sin(89°) = (−1)(1.3 µm)
Solving these two equations, we find θm = complex number for m = 1, and θm = 17.5° for m = −1. θm =
17.5° for m = −1, in fact, is the only solution.

Figure 1Q40-1 There is only one diffracted beam, which corresponds to m = −1.

1.41 Diffraction grating for WDM Consider a transmission diffraction grating. Suppose that we
wish to use this grating to separate out different wavelengths of information in a WDM signal at 1550
nm. (WDM stands of wavelength division multiplexing.) Suppose that the diffraction grating has a
periodicity of 2 µm. The angle of incidence is 0° with respect to the normal to the diffraction grating.

What is the angular separation of the two wavelength component s at 1.550 µm and 1.540 µm? How
would you increase this separation?
Solution
Consider the transmission grating shown in Figure 1Q41-1 with normal incidence, θi = 0

Figure 1Q41-1 Transmission gratings.

The grating equation for normal incidence with the grating in air is given by
dsinθ = mλ ; m = 0, ±1, ±2, …
in which we need to set
d = 2μm
For λ = 1.550μm
2μm × sinθ =× m 1.550μm
∴ sinθ = 0.775m
For m = 1
= θ sin= −1
(0.775) 50.08

For λ = 1.540μm
2μm × sinθ =×m 1.540μm
∴ sinθ = 0.770 m
For m = 1
=θ sin
= −1
(0.770) 50.35

∴ ∆θ 50.35 − 50.08
= = 
0.28

1.42 A monochromator Consider an incident beam on a reflection diffraction grating as in Figure

1.60. Each incident wavelength will result in a diffracted wave with a different diffraction angle. We
can place a small slit and allow only one diffracted wave λm to pass through to the photodetector. The
diffracted beam would consist of wavelengths in the incident beam separated (or fanned) out after
diffraction. Only one wavelength λm will be diffracted favourably to pass through the slit and reach the
photodetector. Suppose that the slit width is s = 0.1 mm, and the slit is at a distance R = 5 cm from the
grating. Suppose that the slit is placed so that it is at right angles to the incident beam: θi + θm = π/2.
The grating has a corrugation periodicity of 1 µm.
(a) What is the range of wavelengths that can be captured by the photodetector when we rotate the
grating from θi = 1° to 40°?
(b) Suppose that θi = 15°. What is the wavelength that will be detected? What is the resolution,
that is, the range of wavelengths that will pass through the slit? How can you improve the resolution?
What would be the advantage and disadvantage in decreasing the slit width s?

Figure 1.60 A monochromator based on using a diffraction grating

Solution
Grating equation is d (sin(θ m ) − sin(θ i ) ) = mλ where m = 0, ±1, ±2, … Moreover, in this particular
case θ m + θi = 90° . Therefore the grating equation may be transformed as
 1 
d (sin(90° − θi ) − sin(θi ) ) = 2d sin (90° − θi − θi ) cos (90° − θi + θi ) = 2d sin (45° − θi ) = mλ or
1
 2 2 

sin (45° − θi )
2d
∴ λ=
m
(a) The equation shows that the largest λ may be achieved for m = 1 which is usually done in
monochromator. Substituting 40° and 1° into above formula we get the values of 123.3 nm and 982.4
nm that can be captured by the photodetector when we rotate the grating from θi = 1° to 40°.
(b) At θi=15° and m = 1 the above formula gives λ = 707.1 nm
Spectral resolution may be found by differentiating the above formula
δλ = 2d cos(45° − θi )δθi ≈ 2d cos(45° − θi )
s
R
which gives δλ = 2.45 nm

1.43 Thin film optics Consider light incident on a thin film on a substrate, and assume normal
incidence for simplicity.

(a) Consider a thin soap film in air, n1 = n2 = 1, n2 = 1.40. If the soap thickness d = 1 µm, plot the
reflectance vs. wavelengt from 0.35 to 0.75 µm, which includes the visible range. What is your
conclusion?
(b) MgF2 thin films are used on glass plates for the reduction of glare. Given that n1 = 1, n2 = 1.38
and n3 = 1.70. (n for glass depends on the type of glass but 1.6 is a reasonable value), plot the
reflectance as a function of wavelength from 0.35 to 0.75 µm for a thin film of thickness 0.10 µm.
What is your conclusion?

Solution

(a) Substitute φ = 2dn2(2π/λ) in

r + r e − jφ r + r e − j ( 4πn2 d / λ )
r = 1 2 − jφ = 1 2 − j ( 4πn2 d / λ )
1 + r1r2 e 1 + r1r2 e
and plot R = | r | as a function of wavelength from 0.35 to 0.75 µm as in the figure. Clearly, certain
2

wavelengths, in this case, violet, green, orange-red are reflected more than others (blue and yellow).

Figure: Reflectance vs wavelength in the visible range for a soap film

(b)

Figure: Reflectance vs wavelength in the visible range for a MgF2 tin film coating on glass

The reflectance is lowered substantially by the thin film coating, and remains low over the visible
spectrum. Without the coating, the reflectance is 6.0%. With the coating, it is below 1%
Authors comment: The above are for normal incidence. Obviously reflections at other angles will have
R vs λ shifted in wavelength. This will affect the spectrum of the reflected light from the soap film but

the MgF2 coating will still result in a relatively low reflectance over the visible because the minimum
in the reflectance is very broad over the visible range.

1.44 Thin film optics Consider a glass substrate with n3 = 165 that has been coated with a
transparent optical film (a dielectric film) with n2 = 2.50, n1 = 1 (air). If the film thickness is 500 nm,
find the minimum and maximum reflectances and transmittances and their corresponding wavelengths
in the visible range for normal incidence. (Assume normal incidence.) Note that the thin n2-film is not
an AR coating, and for n1 < n3 < n2,
2 2
 n2 − n n  n −n 
Rmax =  22 1 3  and Rmin =  3 1 
 n2 + n1n3   n3 + n1 
Solution
Minimum reflectance Rmin occurs at φ = 2π or multiples of 2π , and maximum reflectance Rmax occurs
at φ = π or an odd integer multiple of 2π . The corresponding equations to Eq. (1.11.8) are
2
 n − n   1.65 − 1 
2

Rmin =  3 1  =   = 0.060or 6.0 %

 n3 + n1   1.65 + 1 
2 2
 n 2 − n n   2.52 − (1)(1.65) 
and Rmax =  22 1 3  =  2  = 0.34 or 34%
 n2 + n1n3   2.5 + (1)(1.65) 
Corresponding transmittances are,
Tmax = 1 – Rmin = 0.94 or 94%
and Tmin = 1 – Rmax = 0.66 or 66%.
Since n2 is not an intermediate index between n1 and n3, the n2-film does not reduce the reflection that
would have occurred at the n1-n3 interface had there been no n2-layer. Indeed R13 in the absence of n2,
is the same as Rmin and the n2-layer increases reflection
Since φ = 2dn2(2π/λ), and d = 500 nm, and the wavelengths for maximum reflectance are given
by the condition φ = (2m+1)π, m = 0, 1,2… we can calculate the maximum reflectance wavelengths
λmax = 4dn2/φ = 4dn2/[(2m+1)π]

φ/π 1 3 5 7 9 11 13
λmax 5000 1,667 1000 714 555 455 385
(nm)

1.45 Thin film optics Consider light incident on a thin film on a substrate, and assume normal
incidence for simplicity. Plot the reflectance R and transmittance Τ as a function of the phase change φ
from φ = −4π to +4π for the following cases
(a) Thin soap film in air, n1 = n2 = 1, n2 = 1.40. If the soap thickness d = 1 µm, what are the
maxima and minima in the reflectance in the visible range?
(b) A thin film of MgF2 on a glass plate for the reduction of glare, where that n1 = 1, n2 = 1.38 and
n3 = 1.70. (n for glass depends on the type of glass but 1.7 is a reasonable value). What should be the
thickness od MgF2 to for minimum reflection at 550 nm?
(c) A thin film of semiconductor on glass where n1 = 1, n2 = 3.5 and n3 = 1.55.

Solution

(a) R and T vs φ for a thin soap film in air, n1 = (b) R and T vs φ for a thin film of MgF2 on
n3 = 1, n2 = 1.4 (LiveMath) glass n1 = 1, n2 = 1.38, n3 = 1.70 (LiveMath)

R and T vs φ for a thin film of semiconductor on

glass n1 = 1, n2 = 3.5, n3 = 1.55 (LiveMath)

1.46 Transmission through a plate Consider the transmittance of light through a partially
transparent glass plate of index n2 in which light experiences attenuation (either by absorption or
scattering). Suppose that the plate is in a medium of index n1,the reflectance at each n1-n2 interface is
R and the attenuation coefficient is α.
(a) Show that
(1 − R ) 2 e −αd
Tplate =
(1 − R 2 )e −2αd
(b) If T is transmittance of a glass plate of refractive index n in a medium of index no show that, in
the absence of any absorption in the glass plate,
n/no= T-1 + (T-2 – 1)1/2
if we neglect any losses in the glass plate.
(c) If the transmittance of a glass plate in air has been measured to be 89.96%. What is its refractive
index? Do you think this is a good way to measure the refractive index?

Solution

Figure 1.47 Transmitted and reflected light through a slab of material in which there is no
interference.

(a) Consider a light beam of unit intensity that is passed through a thick plate of partially transparent
material of index n2 in a medium of index n1 as in Figure 1.47. The first transmitted light intensity into
the plate is (1−R), and the first transmitted light out is (1−R) ×(1−R) e −α d = (1−R)2 e −α d . However,
there are internal reflections as shown, so that the second transmitted light is (1−R) × e −α d ×R× e −α d
×R × e −α d × (1−R) = R2(1−R)2 e −3α d so that the transmitted intensity through the plate is
−3α d −5α d
Tplate = (1−R)2 e −α d + R2(1−R)2 e + R4(1−R)2 e + … = (1−R)2 e −α d [1 +
−2α d
R2 e + R4 e−4α d + …]
(1 − R ) 2 e −α d
or Tplate =
1 − R 2 e −2α d
(b) For the transparent plate α=0 and the transmittance of plate becomes
(1 − R ) 2 e −α d (1 − R ) 1 − R
2
Tplate = = =
1 − R 2 e −2α d 1− R2 1+ R
2
 n − n0 
Assuming that R =   the equation above becomes
 n + n0 
2
 n − n0 
1 −  
 n + n0  (n + n0 )2 − (n − n0 )2 = (n / n0 + 1)2 − (n / n0 − 1)2 = 2(n / n0 )
T plate = =
 n − n0 
2
(n + n0 )2 + (n − n0 )2 (n / n0 + 1)2 + (n / n0 − 1)2 (n / n0 ) 2 + 1
1 +  
 n + n0 
which leads to quadratic equation (n / n0 ) 2 − 2 (n / n0 ) + 1 = 0 with the solution n/no= T-1 + (T-2 – 1)1/2
T
.
(c) The transmittance of 89.96% leads to refractive index of 1.5967. In practice, this is not very good
method because it does not give sufficient precision.

1.47 Scattering Consider Rayleigh scattering. If the incident light is unpolarized, the intensity Is of
the scattered light a point at a distance r at an angle θ to the original light beam is given by

1 + cos2 θ
Is ∝
r2
Plot a polar plot of the intensity Is at at a fixed distance r from the scatter as we change the angle θ
around the scattered. In a polar plot, the radial coordinate (OP in Figure 1.48 b)) is Is. Construct a
contour plot in the xy plane in which a contour represents a constant intensity. You need to solve vary
r and θ or x and y such that Is remains constant. Note x = rcosθ and y = rsinθ ; θ = arctan(y/x), r = (x2 +
y2)1/2.
Author's Note: There is a printing error. The minus sign should have been plus as in the above
expression. This should have been obvious from Figure 1.48(b). The error will be corrected in the next
reprint. The e-version of the book is correct.

Solution

(a) Polar plot

Take I s = 1 + cos2 θ as we are interested in the angular dependence only (set r = 1).
In the polar plot, the distance from the origin is the intensity. Do not confuse this with r. r is contant
but in the polar plot the coordinate r is now the intensity.

θ
1 2
Intensity 3
6

Polar plot on LiveMath (Theorist)

(b) Contour plot

We can set the proportionality constant to 1, and write
1 + cos2 θ
Is =
r2
1 + cos2 θ
∴ r=
Is
We can now plot the above on a polar plot in which the distance from the center is r, and r and θ pairs
of coordinates are such that they always yield a constant Is because we have set Is = constant. We can
arbitrarily set I = 1, 2 or 3 to get 3 contour lines.

Blue, Is = 1, black, Is = 2 and red, Is = 3 in AU (arbitrary units)

θ
0.5 r 1 1.5
6

Contour plots on LiveMath (Theorist)

Another interesting plot is the density plot in which the density represents the intensity Is. The
brightness (density) represents the intensity at a point r,θ.

0
-2 x 0 2
y
-2

Density plot of Rayleigh scattering. Brightness represents more light intensity at the point r,θ

1.48 One dimensional photonic crystal (a Bragg mirror) The 1D photonic crystal in Figure 1.50(a),
which is essentially a Bragg reflector, has the dispersion behavior shown in Figure 1.51. The stop-
band ∆ω for normal incidence and for all polarizations of light is given by (R.H. Lipson and C. Lu,
Eur. J. Phys. 30, S33, 2009)
∆ω n −n 
= 2(2 / π ) arcsin 2 1 
ωo  n2 + n1 
where ∆ω is the stop-band, ωo is the center frequency defined in Figure 1.50(a) and n2 and n1 are the
high and low refractive indices. Calculate the lowest stop band in terms of photon energy in eV, and
wavelength (nm) for a structure in which n2 = 4 and n1 = 1.5, and n1d1 = n2d2 = λ/4 and d1 = 2 µm.

Solution

 n −n 
∆ω = ωo 2(2 / π ) arcsin  2 1 
 n2 + n1 
n2 = 4 and n1 = 1.5, d1 = 2 µm
n=
1d1 n=
2d2 λ/4
1.5 × 2 µ m =4d 2 =λ / 4
d 2 = 0.75µ m
λ0 = 12 µ m
c
ω0 2π = 1.57 ×1014
=
λ0
 4 − 1.5 
∆ω= 1.57 ×1014 × 2(2 / π ) arcsin  = 5.41×10
15

 4 + 1.5 
2π × 3 ×10 8
∆λ= 2π c / ∆ω= = 3.48 ×10−7 m= 348nm
5.41×10 15

1 5.41×1015
∆E = h∆ω / 2π = 6.62 ×10−34 J s × ) = 3.56eV
1.6 ×10−19 J eV -1 2π

1.49 Photonic crystals Concepts have been borrowed from crystallography, such as a unit cell, to
define a photonic crystal. What is the difference between a unit cell used in a photonic crystal and that
used in a real crystal? What is the size limit on the unit cell of a photonic crystal? Is the refractive
index a microscopic or a macroscopic concept? What is the assumption on the refractive index?

Solution
The size limit on the unit cell of a photonic crystal is that it must be longer than the wavelength scale.
Refractive index is a macroscopic concept.

NOTES FROM THE AUTHOR

Some of the problems have been solved by using LiveMath (previously Mathview and Theorist).
http://livemath.com

LiveMath interpretation

This is a comment, and is not used in calculations

A square represents a mathematical statement

The first line with a square is a mathematical statement.

The second line with a triangle isolates θc. It is a mathematical conclusion from the first line. The dot at
the center of the triangle represents a working conclusion, something that will be used elsewhere in
calculations.
The third line with a triangle is a mathematical conclusion from the second line. It calculates θc
Italic text next to mathematical conclusions explains the mathematical operation e.g. isolate

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