Contents
Page
Chapter 08 Wave and Geometric Optics ........................ 171
8.0 Introduction ............................................................................171
8.1 Transverse and Longitudinal Waves....................................172
8.1.1 Parameters of Wave .................................................................... 173
8.1.2 The speed of a Travelling Wave ................................................. 175
8.1.3 Wave Speed on a Stretched String ............................................. 176
8.1.4 Interference of Waves ................................................................. 176
8.1.5 Energy and Power of a Traveling String Wave ........................ 177
8.1.6 Phasors .......................................................................................... 179
8.2 Sound Wave ............................................................................179
8.2.1 Traveling Sound Wave ................................................................ 181
8.2.2 Interference of Sound Wave ....................................................... 182
8.3 Doppler Effect .........................................................................184
8.4 Supersonic Speeds ..................................................................186
8.5 Geometric Optics ....................................................................188
8.5.1 Mirror ........................................................................................... 188
8.5.2 Spherical Refracting Surfaces .................................................... 193
8.5.3 Thin Lenses................................................................................... 195
8.5.3.1 Two Lens Systems ............................................................................ 197
8.5.4 Optical Instruments ..................................................................... 198
8.5.4.1 Simple Magnifying Lens .................................................................. 198
8.5.4.2 Compound Microscope .................................................................... 201
8.5.4.3 Refracting Telescope ........................................................................ 202
Tutorials ........................................................................................203
-i-
List of Figures
Page
Figure 8.1:
Figure 8.2:
Figure 8.3:
Figure 8.4:
Figure 8.5:
Figure 8.6:
Figure 8.7:
Figure 8.8:
Figure 8.9:
Figure 8.10:
Figure 8.11:
Figure 8.12:
Figure 8.13:
Figure 8.14:
Figure 8.15:
Figure 8.16:
Figure 8.17:
Figure 8.18:
Figure 8.19:
Figure 8.20:
Figure 8.21:
Figure 8.22:
Figure 8.23:
Figure 8.24:
Figure 8.25:
Figure 8.26:
Figure 8.27:
Figure 8.28:
Illustration of transverse wave ....................................................................... 172
Illustration of longitudinal wave .................................................................... 172
A moving beetle creates both transverse and longitudinal waves ................. 173
Waveform of the wave function y(x, t )  y m sin(kx  t ) ........................... 174
Illustrations of the interference displacement of two waves of different phases
........................................................................................................................ 177
Phasor representation of two waves of small frequency and the displacement
result of interference ...................................................................................... 179
Illustration of a sound wave travels from a point source S ............................ 180
Plots of (a) displacement and (b) air pressure at t = 0 ................................... 181
Two point sources S1 and S2 emit spherical sound waves in phase ............... 182
Illustration of Doppler Effect ......................................................................... 185
Illustration of a source moving at speed v equal to speed vs of sound showing it
moving as fast as the wavefronts ................................................................... 187
Illustration of a source moving at speed v faster than speed vs of sound
showing Mach cone of wave.......................................................................... 187
An extended object O and its virtual image l in a plane mirror ..................... 189
An illustration of image formed by a concave mirror.................................... 190
An illustration of image formed by a convex mirror ..................................... 190
Real image formed by concave mirror........................................................... 191
Focal point F and length of a concave mirror ................................................ 192
Focal point F and focal length of a convex mirror......................................... 192
Images formed by spherical refracting surfaces ............................................ 194
(a) Convergence of light by convex lens and (b) divergence of light by
concave lens ................................................................................................... 197
Image formed by convex lens and concave lens ............................................ 197
Object placed on near point of human eye..................................................... 199
Object placed closer to the human eye, a distance shorted than near point ... 199
Correction for fussy image formed by object placed shorter than Pn closer to
eye and closed to the focal point .................................................................... 200
Correction for fussy image formed by object placed shorter than Pn closer to
eye and is inside the focal point ..................................................................... 201
The structure of a compound microscope ...................................................... 201
Structure of a refracting telescope ................................................................. 202
The height of image and angle made by the parallel rays and observer ........ 203
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Chapter 08
Wave and Geometric Optics
_____________________________________________
8.0 Introduction
Wave and geometric optics is a study of behavior and properties of wave
transmitting in the medium or vacuum and certainly we will study the
applications of light wave for mankind.
Wave can be classified into three types, which are mechanical wave,
electromagnetic wave, and matter wave. Mechanical wave is mechanically
generated wave such as water wave, sound wave, and seismic wave. All these
waves are governed by Newton’s laws and they can exist only with material
medium like water, air, and rock etc.
Electromagnetic wave is wave that does not require medium for
transmission. Wave such as visible light, x-ray, and radar wave are examples of
this type of wave. All electromagnetic waves travel in vacuum with the speed of
2.99792x108m/s.
Matter wave is commonly used in modern technology. The wave is
associated with electron, proton, and other fundamental particles including
atoms and molecules. These particles and molecules constitute matter. This is
the reason the wave is termed as matter wave.
Light wave is commonly known as light. Geometric optics, a study of how
light is transmitted through medium such as lens and uses its properties through
these mediums to design instruments such as microscope and telescope to
extend human visual capability.
In this chapter, we begin to discuss the types of mechanical waves like
transverse, longitudinal wave, and sound wave, in the aspects of their properties
and interference.
We will also discuss how is form image, reflection due to mirror, refraction
and refraction from lenses, and basic design of optical instruments.
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08 Wave and Geometric Optics
8.1 Transverse and Longitudinal Waves
Transverse wave has its displacement perpendicular to the direction of traveling
wave. An example of generating transverse wave is by moving the tied string up
and down continuously as shown in Fig. 8.1.
Figure 8.1: Illustration of transverse wave
Longitudinal wave is the wave type that has its displacement parallel to the
direction of the moving wave. Moving a piston back and forth in the air filled
pipe will create longitudinal wave as shown in Fig. 8.2.
Figure 8.2: Illustration of longitudinal wave
A moving beetle on the surface of sand will both create transverse and
longitudinal waves as illustrated in Fig. 8.3. Each of these waves has different
speed. It allows its predator like scorpion to catch.
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08 Wave and Geometric Optics
Figure 8.3: A moving beetle creates both transverse and longitudinal waves
Example 8.1
The movement of a beetle sends out two pulses, which are longitudinal and
transverse pulses. The speed of longitudinal pulse is 150ms -1, while the speed of
transverse pulse is 50m/s. A scorpion has eight legs spread roughly in circle
about 5cm in diameter intercepts the faster longitudinal pulse first and learns the
direction of beetle and then sense the time difference t between first intercept
and second intercept of slow transverse pulse to determine the distance d to the
beetle. Find the distance d if the time difference is 5.0ms.
Solution
The time difference t is equal to t 
d
d
.

vT vL
 1
1 
1 
 1
  = 5.0x103 / 
 = 37.5cm.
 50 150 
 vT vL 
Thus, the distance d is equal to d  t /
8.1.1 Parameters of Wave
A description of a wave on a string and the motion of any element along the
length require a function that provides the shape of the wave. This means that it
needs a relation in the form y = h(x, t). It is a function of transverse replacement,
which depends on time t and position x. Mathematically, it can be expressed by
equation (8.1).
y(x, t )  y m sin(kx  t )
(8.1)
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08 Wave and Geometric Optics
where ym is the amplitude, sin(kx  t ) is oscillation term, k is the angular wave
number, x is the position, and  is the angular frequency. Based on equation
(8.1), the waveforms are shown in Fig. 8.4.
(a)
(b)
Figure 8.4: Waveform of the wave function y(x, t )  y m sin(kx  t )
The phase of the wave is the argument (kx  t ) . The wave sweeps through a
string element at a particular position x. The phase changes linearly with time t.
This shall mean that the sine function also oscillates between +1 and -1, which
is corresponded to a peak and a valley. The peak is called amplitude ym.
The wavelength  of a wave is the distance parallel to the direction of the
wave’s travel, between repetitions of the shape of the wave. The illustration is
shown in Fig. 8.4(a).
At time t = 0, the wave function from equation (8.1) is y(x,0)  y m sin(kx) .
Based on this equation, the displacement is the same at position x = x1 and x =
x1+ as shown in Fig. 8.4(a), which is y m sin(kx1 ) = y m sinkx1   . This implies
that k = 2. Thus, the angular wave number k is equal to
k
2

(8.2)
Period of the oscillation T of a wave is the time taken at any string element to
move through one complete oscillation. If one fixes the position say x = 0, the
wave equation is equal to y(0, t )  y m sin(t ) . It is also equal to y(0, t )  y m sin(t ) .
At time t = t1 and t = t+T, the displacements are the same as shown in Fig.
8.4(b). Thus, function y(0, t1 )  y m sin(t1 ) =  y m sin(t1  T) . This implies that
T = 2. This shall mean that the angular frequency  of the wave is equal to

2
T
(8.3)
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08 Wave and Geometric Optics
The frequency of the oscillation f is equal to 1/T. Thus, the angular frequency 
is also equal to 2f.
8.1.2 The speed of a Travelling Wave
If the wave is travelling in the position x-direction, after a small time t, it
travels a distance x. The ratio of x/t is the wave speed . As the wave
moves at a fixed point, the displacement is retained. This shall mean the phase
(kx-t) is constant. Thus, wave speed can be determined from (kx-t) =
constant. Thus,
k
dx
0
dt
(8.4)
This implies that the wave speed  is

dx 

dt k
(8.5)
Example 8.2
A wave traveling along a string is described by y(x, t)  0.00327 sin(72.1x  2.72t) .
(a) What is the amplitude of this wave?
(b) What are the wavelength, period, and frequency of this wave?
(c) What is the velocity of this wave?
(d) What is the displacement at x = 22.5cm and t = 18.9s?
Solution
The amplitude of this wave is 0.00327m.
The angular wave number is 72.1/m. The wavelength of the wave is 2/72.1 =
87.15mm.
The angular frequency of the wave is 2.72Rad/s. Thus, the frequency of the
wave is 2.72/2 = 0.432Hz.
The period of this wave is 1/0.432 = 2.31s.
 2.72

=3.77cm/s.
k 72.1
The displacement is y(x, t)  0.00327 sin(72.1x  2.72t)
The velocity of this wave is  
= 0.00327 sin(72.1x 22.5x102  2.72x18.9) = 0.00327 sin(35.18rad)
= 0.00327 sin(35.18Rad) = 0.00327x0.583 = 1.91mm.
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08 Wave and Geometric Optics
8.1.3 Wave Speed on a Stretched String
The speed of wave defined by equation (8.5) is set by the properties of the
medium it travels. If the wave is traveled through a medium such as water, air,
steel, or stretched string, it causes that medium to oscillate. For this to happen,
the medium must possess both mass and elasticity. For a string that has mass m
and length l, we define the linear density  as equal to the ratio of m and l,
which is

m
l
(8.6)
We cannot send a wave along a string unless it is under tension. Thus, the
tension  in the string is equal to the common magnitude of the two forces tying
the ends of the string. The speed of the wave traveling in the stretched string can
then be defined as



(8.7)
8.1.4 Interference of Waves
Suppose we send two sinusoidal waves of same wavelength and amplitude in
the same direction along a stretched string. The superposition principles applied,
which states that the displacement of the string when the waves overlap is the
algebraic sum of the displacement of the individual wave. The waves can
combine which we call interference. Supposing one wave is traveling along a
stretched string is given by y1 (x, t)  ym sin(kx  t) and the second one is given by
y2 (x, t )  ym sin(kx  t  ) , where  is the phase constant. When these two waves
interference, the resultant displacement is given by
y' (x, t )  ym sin(kx  t )  ym sin(kx  t  )
(8.8)
1  
1 

y ' ( x, t )  2 y m cos  sin kx  t    . Note that
2  
2 

1
1
sin   sin   2 sin    cos    . From this equation, the amplitude at
2
2
1
1
interference is 2 y m cos  and oscillation term is sin kx  t    . If  is equal
2 
2 


which
is
equal
to
to zero then at interference, the magnitude is 2ym. This is a case of fully
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08 Wave and Geometric Optics
constructive interference. If  is equal to 1800 then at interference, the
magnitude is equal to zero. This is a case of fully destructive interference. If  is
equal to 600 then the amplitude is ym. This is a case of intermediate interference.
The illustrations are shown in Fig. 8.5.
(a)  = 0
(b)  =  rad
(c)  = 2/3 rad
Figure 8.5: Illustrations of the interference displacement of two waves of different phases
8.1.5 Energy and Power of a Traveling String Wave
As the wave travels in the stretched string, it transports both kinetic and elastic
potential energies.
An element of string has mass dm oscillating transversely in simple
harmonic motion as the wave pass through it. It has kinetic energy associated
with its transverse velocity u . When the element is rushing through its y = 0
position, its transverse velocity is maximum. When the element is at y = ym, its
transverse velocity is zero. Thus, at y = 0, the kinetic energy of the element is
maximum, while at y = ym, the kinetic energy is zero.
The string must be stretched in order to send the sinusoidal wave. As the
string element of length dx oscillates transversely, its length must increase and
decrease in a periodic way if the string element is to fit sinusoidal waveform.
When the string element is at y = ym, its potential energy is equal to zero, while
at y = 0, its potential energy is maximum because it stretches to maximum. The
string element has maximum elastic and kinetic energies at y = 0 and zero
energy at y = ym.
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08 Wave and Geometric Optics
The kinetic energy dK associated with a string element of mass dm is given
by
dK 
1 2
u dm
2
(8.9)
The transverse speed of the oscillating string can be determined by
differentiating equation (8.1) with respect time t. while keep to position x
constant, which is
u
dy( x, t ) dym sin(kx  t 
=  ym cos(kx  t)

dt
dt
(8.10)
Equation (8.9) becomes equation (8.11) after substituting equation (8.10) and
dm = dx from equation (8.6).
dK 


 2 2
 y m cos 2 kx  t  dx
2
(8.11)
Diving equation (8.11) with dt, it yields the rate of energy transmission equation,
which is shown in equation (8.12).


dK  2 2
dx
  y m cos 2 kx  t 
dt 2
dt
Note that
(8.12)
dx
is equal to , the traveling speed of the wave. The average rate at
dt
which kinetic energy is transported is the average over an integral number of
wavelength and using the fact that the average value of the cosine square
function over an integral number of period is 1/2. Hence,
1
 dK 
2 2

   y m
dt
4

avg
(8.13)
The average rate of elastic potential energy carried along with the wave is same
as the average rate of kinetic energy carried along with the wave. Thus, the
average power Pavg transmitted by the wave is the sum of average rate of kinetic
energy and average rate of elastic potential energy transmitted. i.e.
1
Pavg  2 y 2m
2
(8.14)
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08 Wave and Geometric Optics
8.1.6 Phasors
Supposing two sinusoidal waves of functions y1 (x, t)  ym1 sin(kx  t) and
y2 (x, t )  ym2 sin(kx  t  ) , when they interfere each other, the displacement of
interference is equation (8.15).
y' (x, t )  y m1 sin(kx  t ) + y m2 sin(kx  t  )
(8.15)
The displacement of the interference is also equal to y' (x, t )  y'm sin(kx  t  ) .
The amplitude y 'm and phase constant  can be calculated from phasor
representation of the waves, which are illustrated in Fig. 8.6. The amplitude
y 'm can be calculated using equation (8.16).
y'm 
ym2 cos   ym1 2  ym2 sin 2
(8.16)
The phase constant  can be calculated using equation (8.17).
 y m 2 sin  
  tan 1 

 y m 2 cos   y m1 
(8.17)
Figure 8.6: Phasor representation of two waves of small frequency and the displacement
result of interference
8.2 Sound Wave
Sound wave is defined roughly as longitudinal wave. Seismic engineer uses
such wave to probe Earth’s crust for fossil oil. Submarine uses sound wave to
stalk other submarine by listening for the characteristic noise produced by the
propulsion system.
Point source sound represents a tiny sound source that emits sound in all
directions. Figure 8.7 illustrates a sound wave travels from a point source S. The
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08 Wave and Geometric Optics
wavefronts and rays indicate the directions of travel and the spread of the sound
waves. Wavefronts are surfaces over which the oscillation of the air due to the
sound wave has the same value. This surface is represented by whole or partial
circles in two dimensional drawing for point source. Rays are directed lines
perpendicular to the wavefronts that indicate the direction of travel of the
wavefronts. The double short arrows indicate longitudinal oscillations of the air
are parallel to the rays.
Near point source the wavefronts are spherical and spread out in three
dimensions. The wave is termed as spherical wave. As the wavefronts move
outward and their radii become larger and their curvature decreases. Far from
the source, we can approximate the wavefronts as planes or lines in two
dimensional drawing. Thus, this wave is called planar wave.
Figure 8.7: Illustration of a sound wave travels from a point source S
In stretched string, potential energy is associated with the periodic stretching of
the string element as the wave passes through them. Sound wave passes through
air, potential energy is associated with periodic compression and expansion of
small volume of air. Thus, the pressure on the medium is changed to bulk
modulus B, which is defined as
B
p
V / V
(8.18)
Here V / V is the fractional change in volume due to change in pressure p.
Bulk modulus has dimension pascal, which has unit kgm-1s-2. In order to the
dimensional unit for the speed of sound wave traveling in medium, linear
density  has to be changed to density of the medium . Thus, the speed of
sound in a medium is
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08 Wave and Geometric Optics

B

(8.19)
8.2.1 Traveling Sound Wave
The air element of the sound wave oscillates longitudinally. The displacement
s(x, t) of the sound wave is
s(x, t )  s m cos(kx  t )
(8.20)
or it can be s(x, t )  s m sin(kx  t ) . As the wave moves, the air pressure at
position x varies sinusoidally. Thus, the description of air pressure variation is
p(x, t )  pm cos(kx  t )
(8.21)
where pm is the pressure amplitude. The pressure amplitude is normally very
much smaller than the pressure p when there is no wave. The relationship
between displacement s(x, t) and pressure amplitude is shown in equation (8.22).
p m  ()s m
(8.22)
Figure 8.8 shows plots of displacement and air pressure variation at t = 0. The
result shows the displacement s and pressure variation p are 900 out of phase.
The pressure variation p at any point along the wave is zero when the
displacement is at maximum.
(a)
(b)
Figure 8.8: Plots of (a) displacement and (b) air pressure at t = 0
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08 Wave and Geometric Optics
8.2.2 Interference of Sound Wave
It is like the transverse wave, sound wave can undergo interference. Let’s
consider two identical point sources as shown in Fig. 8.9 emitting two identical
sound waves that are in phase and of identical wavelength . The waves will be
in phase if they travel with identical paths to reach point P. If they are in phase
at point P, like the transverse waves, they will undergo constructive interference
at point P. However, if the path traveled by wave from S 2 is longer than the
wave from S1 or vice versa then the waves may not be in phase when they reach
point P.
Figure 8.9: Two point sources S1 and S2 emit spherical sound waves in phase
The difference in path length is L =|L2 –L1|. To relate phase difference  with
different path length L, their relationship is shown in equation (8.23).

2L

(8.23)
For constructive interference, the phase difference  should be either 0, 2,
4,….. It would mean that the phase difference  is integral multiple of 2. i.e.
  m2 , where m = 0, 1, 2, 3, …….
This is also meant that
(8.24)
L
= 0, 1, 2, 3, ……..

For destructive interference, the phase difference  should be , 3 ,
5 , …... It will mean that the phase difference  is odd multiple of . i.e.
  2m  1 , where m = 0, 1, 2, 3, …….
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(8.25)
08 Wave and Geometric Optics
This is also meant that
L
= 0.5, 1.5, 2.5, 3.5, ……..

Two waves at point P may neither constructive interference nor destructive
interference. It can be intermediate interference. This shall mean that
L
has

value not specified by either constructive or destructive interference.
8.2.3 Intensity and Sound Level
The intensity l of a sound wave at a surface is the average rate per unit area at
which energy is transferred by wave through or onto the surface. Thus,
l
P
A
(8.26)
P is the time rate of transfer of energy of the sound wave, which is power. A is
area of surface intercepting sound. The intensity is related with displacement s
illustrated by equation (8.27).
1
l  2s 2m
2
(8.27)
The intensity varies with distance from a real sound source is often complex.
Like loudspeaker may transmit sound only in a particular direction and
environment usually produces echo that overlap the direct sound wave. If we
assume that the source is point source, which produces isotropically sound then
the energy emitted from the source must pass through the surface of the sphere
that has surface are 4r2. Thus, the time rate at which energy is transferred
through the surface by the sound waves must be equal the time rate at which
energy is emitted from the source. From equation (8.26), the intensity of the
sound wave is
l
PS
4r 2
(8.28)
PS is power of the source. From equation (8.28), we notice that the intensity l
decreases with the square of the distance r.
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08 Wave and Geometric Optics
Instead of deal with intensity of sound, it is better and convenient to use
sound level , which is defined as
  10dB log
l
l0
(8.29)
Here dB is denoted as decibel, which is unit for sound. l0 is the standard
reference intensity that has value 10-12W/m2.
8.3 Doppler Effect
The Doppler Effect is observed whenever the source of waves is moving with
respect to an observer. The Doppler Effect can be described as the effect
produced by a moving source of waves in which there is an apparent upward
shift in frequency for observers towards whom the source is approaching and an
apparent downward shift in frequency for observers from whom the source is
receding. It is important to note that the effect does not result because of an
actual change in the frequency of the source. Doppler Effect was proposed y in
1842 by Austrian physicist Johann Christian Doppler.
Astronomers who use the shift in frequency of electromagnetic waves
produced by moving stars in our galaxy and beyond in order to derive
information about these stars and galaxies. The belief that the universe is
expanding is based in part upon observations of electromagnetic waves emitted
by stars in distant galaxies. Furthermore, specific information about stars within
galaxies can be determined by application of the Doppler Effect. Galaxies are
clusters of stars that typically rotate about some center of mass point.
Electromagnetic radiation emitted by such stars in a distant galaxy would appear
to be shifted downward in frequency (a red shift) if the star is rotating in its
cluster in a direction that is away from the Earth. On the other hand, there is an
upward shift in frequency (a blue shift) of such observed radiation if the star is
rotating in a direction that is towards the Earth.
The Doppler Effect can be observed for any type of wave - water wave,
sound wave, light wave, radio wave etc. We are most familiar with the Doppler
Effect because of our experiences with sound waves. As illustrated in Fig. 8.10,
when a police car or emergency vehicle is traveling towards you on the highway,
as the car approached with its siren blasting, the pitch of the siren sound is high
and after the car pass by the pitch of the siren sound is low. This is a typically
example illustrating the Doppler Effect. There is an apparent shift in frequency
for a sound wave produced by a moving source.
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08 Wave and Geometric Optics
Figure 8.10: Illustration of Doppler Effect
If either the detector or source is moving, or both are moving, the emitted
frequency f and the detected frequency f’ are related by equation (8.30).
f'  f
  D
  S
(8.30)
where  is the speed of sound through air, D is the detector’s speed relative to
air and S is the source’s speed relative to the air. The choice of plus or minus
signs is set by the rule, which is stated here. When motion of detector or source
is moved toward other, the sign on its speed must give an upward shift in
frequency. When the motion of detector or source is way from the other, the
sign on its speed must give a downward shift in frequency.
With the above understanding, let’s apply equation (8.30) for the case with
stationery source and moving detector. For a stationery source, S is equal to
zero. Thus, equation (8.30) becomes equation (8.31).
f'  f
  D

(8.31)
The plus sign denotes the detector is moving toward the source, while minus
sign denotes the detector is moving away from the source.
For the case of moving source and stationery detector, D is equal to zero.
Thus, equation (8.30) becomes
f' f

  S
(8.32)
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08 Wave and Geometric Optics
The minus sign denotes the source is moving toward the detector, while plus
sign denotes the source is moving away from the detector.
Example 8.3
A rocket moves at a speed of 242m/s. directly toward a stationery pole while
emitting sound wave at frequency f = 1,250Hz.
(a). What is the frequency f’ measured by a detector that is mounted on the pole?
(b). Some of the sounds reach the pole reflected back to rocket as an echo. What
is the frequency detected by the detected mounted on the rocket?
Let’s look at the general equation f '  f
  D
and apply it to this scenario.
  S
To the detector mounted on the pole, the rocket (source) is approaching it,
thus, the relative speed is v+vD.
To the rocket, the detector is closer as time lapsed. Thus, the relative speed
is v-vS. Thus, the Doppler frequency equation applying to this scenario is
f' f
  D
  S
Since vD is equal to zero, the f’ measured by the detector mounted on the
pole is f '  f

343
= 4,250Hz.
 1,250x
  S
343  242
vS is equal to zero, the frequency f” measured by the detector mounted on
the rocket is f "  f
  D
343  242
= 7,240Hz.
 4250x

343
8.4 Supersonic Speeds
If a source is moving toward stationery detector at the speed of sound, which is
speed of sound v equal to speed of source vs, then from equation (8.32) the
detected frequency will be infinitely large. This means that the source is moving
so fast that its keeps pace with its own spherical wavefronts as shown in Fig.
8.11.
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08 Wave and Geometric Optics
Figure 8.11: Illustration of a source moving at speed v equal to speed vs of sound showing it
moving as fast as the wavefronts
If the speed of sound v is exceeded the speed of sound vs, which is at supersonic
speed then all the spherical wavefronts bunch along a V-shaped cone envelope
as illustrated in Fig. 8.12. This V-shaped cone is called Mach cone. A shock
wave is said to exist along the surface of this cone because the bunching of
wavefornts causes an abrupt rise and fall of air pressure as the surface passes
through any point. W1 is the wavefront when the source is at S1. Similarly W6 is
the wavefront when the source is at S6.
Figure 8.12: Illustration of a source moving at speed v faster than speed vs of sound showing
Mach cone of wave
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08 Wave and Geometric Optics
The radius of any wavefront in Fig. 8.12 is vt, where v is the speed of sound
after and t is the time that has lapsed since the source emitted that wavefront.
From Fig. 8.12, the half angle  of the cone is called Mach cone angle and it is
given by equation (8.33).
sin  
vt v

vs t vs
(8.33)
where vst is the distance travelled by the source in time t. The ratio of vs/v is
called Mach number.
8.5 Geometric Optics
We shall begin with the study of image formed by mirror and lens. There are
two types of image, which are real image and virtual image. Real image can be
formed on a surface such as on a card or movie screen. Virtual image is only
exist within the brain but nevertheless is said to be in perceived location. When
a person stands in front of a mirror, virtual image is formed behind the mirror.
This image of course is not real image. We shall explore several ways in which
real and virtual images are formed by reflection with mirror and refraction with
lenses. Finally we shall discuss the use of lenses for designing optical
instruments.
8.5.1 Mirror
We shall discuss three types of mirror namely plane mirror, concave mirror, and
convex mirror pertaining their reflection, formation of image, central axis, focal
point, and formulae to calculate their lateral magnifications.
The image formed by a plane mirror from an extended object as illustrated
in Fig. 8.13 is a virtual image. The distance between the object and the plane
mirror p is equal to the distance between the virtual image and the mirror is the
same. Thus, p is equal to i. If the object has height h and the height of virtual
image is h’ then the lateral magnification m is defined as
m
h'
h
(8.34)
For the case of plane mirror, the lateral magnification is equal to -1 since the
height of the object and image is the same. It can also be shown that the lateral
magnification is also equal to
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08 Wave and Geometric Optics
m
i
p
(8.35)
For plane mirror i = -p since i is at the other side of mirror. This implies that m
is equal to +1 for plane mirror. The “+” means that the object and image have
same orientations.
Figure 8.13: An extended object O and its virtual image l in a plane mirror
Concave and convex mirrors are spherical mirrors. Unlike the plane mirror, they
have a finite radius of curvature. Concave mirror has the following
characteristics.
1. The center of curvature C is closer and is in front of the mirror.
2. The field view is the extent of the scene that is reflected to the observer. It
is decreased.
3. The image of the object is farther behind the concave mirror.
4. The image is large. This is the feature used for making makeup mirror
and shaving mirror.
The illustration of the image formed by concave mirror is shown in Fig. 8.14.
The image is a virtual image formed at the opposite of the mirror.
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08 Wave and Geometric Optics
Figure 8.14: An illustration of image formed by a concave mirror
Convex mirror has the following characteristics.
1. The center of curvature C is closer and is behind the mirror.
2. The field view is the extent of the scene that is reflected to the observer. It
is increased.
3. The image of the object is closer behind the convex mirror.
4. The image is small. This is the feature used for mirror placed at road
junction and mirror placed at store to have more view.
The illustration of the image formed by convex mirror is shown in Fig. 8.15.
The image is a virtual formed at the opposite of the mirror.
Figure 8.15: An illustration of image formed by a convex mirror
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08 Wave and Geometric Optics
For concave mirror, a real image would be formed if the object is placed larger
than the focal length from the center point c as illustrated in Fig. 8.16. The
lateral magnification m has a negative value since the object and image are in
same orientation. Based on equation (8.34) both p and i have are positive value.
This is true because the image is a real image. Thus, the distance i between real
image and central point c has a positive value.
Figure 8.16: Real image formed by concave mirror
When the parallel rays reach a concave mirror as illustrated in Fig. 8.17, the ray
near the central axis reflected through a common point F called focal point F or
focus of the mirror. If one placed a card board at this point, an image of distance
point object would be formed. The distance from focal point F to the center of
mirror c is called focal length f. This focal point is a real focal point that has a
positive value since real image can be formed.
When the parallel rays reach a convex mirror as illustrated in Fig. 8.18, the
rays near the central axis diverged away. If one’s eyes intercept some of these
rays, one would perceive the rays as originated from point source. This
perceived point source is the focal point F. The distance from focal point F to
the center of mirror c is called focal length f. This focal point is a virtual focal
point that has a negative value since virtual image can only be formed.
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08 Wave and Geometric Optics
Figure 8.17: Focal point F and length of a concave mirror
Figure 8.18: Focal point F and focal length of a convex mirror
For mirror of both types, it can be shown that the focal length f is equal half of
radius of the mirror, which is
f 
1
r
2
(8.36)
A simple equation relating the object distance p, image distance i to the center
of spherical mirror, and focal length f of the mirror is shown in equation (8.37).
1 1 1
 
f
p i
(8.37)
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08 Wave and Geometric Optics
Example 8.4
A tarantula (spider) of height h sits cautiously before a spherical mirror whose
focal length has absolute value |f| = 40cm. The image form the tarantula
produced by the mirror has same orientation as the tarantula and has height h’ =
0.20h.
(a). Is the image real or virtual, and is it on the same side or opposite side?
(b). Is the mirror concave or convex and what is the focal length f?
Solution
The image is a virtual image since it has same orientation. It must be at the
opposite side of the mirror.
The image and object have same orientation. It means that the lateral
magnification has positive value. Thus, from equation (8.34), i = - 0.20p.
The focal length f follows equation (8.36), which is
1 1 1
1 1
1
.
  .i.e.
 
f
p 0.2p
f
p i
This implies that focal length f is equal to –p/4. In another word the focal length
has negative value, which is -40cm. Negative focal length means the mirror is a
convex mirror.
8.5.2 Spherical Refracting Surfaces
Let’s look at the images formed by reflections to images formed by refraction
through surfaces of transparent materials. Spherical surface with radius of
curvature r and center of curvature C shall be considered in this study. The light
emitted through a point object O in a medium with refractive index n1 will
refract through a spherical surface into a medium of refractive index n2. After
refracting through the surface, the question is whether a real image or virtual
image would be formed. The answer depends on the relative value of n1, n2 and
the geometry of the situation.
Let’s consider six situations as illustrated in Fig. 8.19. Figure 8.19(a) and
Fig. 8.19(b) show that real images are formed. These situations would happen if
the refracted light diverges toward the central axis. Situations shown in Fig.
8.19(c), Fig. 81.9(d), Fig. 8.19(e), and Fig. 8.19(f) would form virtual images
since the refracted light diverges away from central axis.
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08 Wave and Geometric Optics
Based on above scenario, real images form on the side of refracting surface
that is opposite the object and virtual images form on the same side as object.
(a)
(b)
(c)
(d)
(e)
(f)
Figure 8.19: Images formed by spherical refracting surfaces
The equation governing the spherical refracting surface is shown in equation
(8.38).
n1 n2 n2  n1
 
p i
r
(8.38)
Like in the case of mirror, p is positive for object. i is positive for real image,
whereas i is negative for virtual image. r is the radius of curvature. However,
rule must be applied for radius of curvature. For object facing convex surface
the radius of curvature is positive. For object facing concave surface, the radius
of curvature is negative.
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08 Wave and Geometric Optics
Example 8.5
A Jurassic mosquito is discovered embedded in a chunk of amber which has
refractive index of 1.6. One surface of the amber is spherical convex with radius
of curvature 3.0mm. The mosquito head happened to be on the central of axis of
the surface viewing along axis appears to be buried 5.0mm into amber. What is
the actual depth of the mosquito head?
Solution
The virtual image is formed from this scenario. The illustration is shown in
figure below.
Based on equation
n1 n2 n2  n1
, we need to calculate the value of p. The
 
p i
r
image is virtual image then i = -5.0mm. The object is facing a concave surface.
Thus, radius of curvature is – 3.0mm. Thus,
1.6 1.0 1.0  1.6


. The actual
p - 5.0
- 3.0
depth of the mosquito head p is 4.0mm.
8.5.3 Thin Lenses
A lens is a transparent object with two refracting surfaces whose central axes
coincide. A lens that causes parallel light to converge to a common point is a
converging lens or convex lens. A lens that causes parallel light to diverge is a
diverging lens or concave lens.
We shall only discuss thin lens, which is the lens that has its thickest part
thin as compared with the object distance p, image distance i, and radii
curvature r1 and r2 of the two surfaces of the lens. We also consider only light
rays that make small angle with central axis.
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08 Wave and Geometric Optics
Equation (8.37) used for spherical mirror is also true for thin lens
calculation. However, the equation for calculating the focal length f of thin lens
of refractive index n surrounded by air is given by
1 1
1
 n  1  
f
 r1 r2 
(8.39)
Equation (8.39) is also called the lens maker’s equation. Here r1 is the radius of
curvature of the lens surface nearer to the object and r2 is that of other surface.
The sign of radius follows the rule mention earlier Section 8.5.2. If the
surrounding medium is not air then the refractive index cannot be 1. Equation
(8.39) needs to be modified to include the refractive index nmedium of the medium.
Thus, equation (8.39) will become
 1 1 
1  n
 
 1  
f  nmedium  r1 r2 
(8.40)
Light rays parallel to central axis passing through convex lens would converge
to a real focal point F2 as shown in Fig. 8.20(a). Light rays parallel to central
axis of concave lens would diverge away. The extension of diverged light rays
pass through a virtual focal point F2 as shown in Fig. 8.20(b).
(a)
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(b)
Figure 8.20: (a) Convergence of light by convex lens and (b) divergence of light by concave
lens
Image formed by thin lens can be real or virtual types. Real image formed at the
opposite side of the object, while virtual image formed at the same side where
the object is situated. Figure 8.21 illustrates the image formed by convex and
concave lenses.
(a)
(b)
(c)
Figure 8.21: Image formed by convex lens and concave lens
In Fig 8.21(a), the object is placed at the point beyond the focal point F 1 of the
convex lens. The image formed at the opposite side of the lens is real, enlarged,
and inverted. Figure 8.21(b) shows that an enlarged virtual upright image is
formed when the objective is placed at the point less than the focal length F1 of
convex lens. Figure 8.21(c) shows that an upright diminished virtual image is
formed at the same side as the object.
8.5.3.1 Two Lens Systems
When an object O is placed in front of a system of two lenses whose central
axes coincide, one can locate the final image by working in step. Let lens 1 be
nearer to the object and lens 2 located further away from object.
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08 Wave and Geometric Optics
Step 1: Let p1 be the distance of object from lens 1. One can then final the i1
value from equation
1 1 1
  or drawing the ray diagram.
f
p i
Step 2: Ignore the presence of lens 1 and treat the image formed by lens 1 as the
object for lens 2. If this object is located beyond lens 2, the object
distance p2 is taken as negative. Otherwise, p2 is taken as positive. To
find the location of the final image, one can use equation
1 1 1
  or
f
p i
drawing the ray diagram.
Step 1 and step 2 procedure can be used for multiple lens systems. The overall
lateral magnification M is the product of m1 and m2 produced by the two lenses.
i.e. M = m1m2.
8.5.4 Optical Instruments
Human eye is a remarked effective organ. Its range can be extended in many
ways by optical instruments such as using a microscope to view micro-object,
seeing distance star using telescope etc. Satellite-borne infrared camera and xray microscope are another two examples that extended human vision.
Mirror and thin lens equations derived earlier will still be used in our
studies of optical instruments, although we know well that some of these
instruments do use thick lens. It is for simplicity of the study that we adopt this
approach.
We will study three optical instruments here. They are simple magnifying
lens, compound microscope, and refracting telescope.
8.5.4.1 Simple Magnifying Lens
Human eye can focus sharp image of an object on the retina if the objective is
located from infinity to a point called near point Pn. If one moves the object to
the eye closer than the near point, the retina perceived a fussy image. The
location of near point Pn normally depends on the age of the person. An old age
person would have a near point further away from the eye than a young person.
For the purpose of study, we take near point of 25cm for human eye.
An object O of height h is placed at the near point Pn of human eye will
occupy an angle  in the eye’s view as shown in Fig. 8.22.
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08 Wave and Geometric Optics
Figure 8.22: Object placed on near point of human eye
For small angle , tan  , thus  is approximately equal to

h
25cm
(8.41)
As illustrated in Fig. 8.23, the object is placed at location Pn is less than 25cm.
The image on the retina would be fussy because the human eye cannot bring the
image to be focus on retina.
Figure 8.23: Object placed closer to the human eye, a distance shorted than near point
To correct the fussy image problem, convex lens is placed in front of eye and
placed the object just inside the focal length of the lens. The image produced
would be an enlarged virtual type and occupied a large angle ’ situated at
infinity than what is done by the object in the case shown in Fig. 8.22. The
illustration is shown in Fig. 8.24. The object distance from the central point is
equal to focal length F1 of the convex lens.
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08 Wave and Geometric Optics
Figure 8.24: Correction for fussy image formed by object placed shorter than Pn closer to
eye and closed to the focal point
From Fig. 8.24, one can obtain equation (8.42), which is
' 
h
f
(8.42)
The angular magnification m of what is seen is
m 
'

(8.43)
Substituting equation (8.41) and (8.42) into equation (8.43), it yields equation
(8.44).
m 
25cm
f
(8.44)
For object placed inside the focal point as illustrated in Fig. 8.25 then the
objective distance p is less than the focal length and the image distance i is
equal to 25cm. Thus, according to lens equation, the objective distance p from
central axis is equal to
p
25 f
f  25cm
(8.45)
The lateral magnification is equal to
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08 Wave and Geometric Optics
m
 25 
i
25 f
 25 /
 1   cm
p
f  25cm 
f 
(8.46)
Figure 8.25: Correction for fussy image formed by object placed shorter than Pn closer to
eye and is inside the focal point
8.5.4.2 Compound Microscope
A compound microscope is shown in Fig. 8.26. It consists of an objective of
focal length fob and an eyepiece of focal length fey. The tube length s can be
approximated as the distance i between the objective and the image I.
Figure 8.26: The structure of a compound microscope
The lateral magnification m is defined earlier as and is
m
i
s

p
f ob
(8.47)
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08 Wave and Geometric Optics
Since the image I is located just inside focal point F1' of the eyepiece, the
eyepiece acts as a simple magnifying lens and the observer see a final virtual
and inverted image I’ through it. The overall magnification M of the instrument
is the product of lateral magnification of the instrument m produced by the
objective given by equation (8.44) and the angular magnification m produced
by the eyepiece shown in equation (8.47). Thus,
M  mm   
s 25cm

f ob f ey
(8.48)
8.5.4.3 Refracting Telescope
Telescope comes in variety of forms. The one described here is the refracting
type that consists of an objective and an eyepiece, which is shown in Fig. 8.27.
Unlike the microscope, telescope is designed to view large object such as
galaxy, star, planet etc at large distance. This makes the difference between a
microscope and a telescope. The second focal point F2 of the objective is made
to coincide with the focal point F1' of eyepiece, whereas in microscope design,
the points are separated by a distance s.
Figure 8.27: Structure of a refracting telescope
The rays from distant object are parallel rays strike the objective at an angle ob
with the axis of telescope and forming a real and inverted image at the common
focal point F2 and F1' . This image acts as object for eyepiece, though which an
observer sees a distant and inverted virtual image I’. The rays defining the
image make an angle ey with telescope axis.
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08 Wave and Geometric Optics
The angular magnification m of the telescope is ey/ob. From Fig. 8.28,
the rays close to the central axis such that ob = h’/fob and ey h’/fey. Thus, the
angular magnification m is given by
m  
f ob
f ey
(8.49)
where negative sign indicates that I’ is inverted.
Figure 8.28: The height of image and angle made by the parallel rays and observer
Tutorials
8.1.
Two identical sinusoidal waves moving in same direction along a
stretched string interfere with each other. The amplitude ym of each string
is 9.8mm and the phase difference  between them is 1000. What is the
amplitude of the resultant wave and what is the type of interference
occurs?
8.2.
A stretch string has linear density  = 525g/m and is under tension  =
450N. A sinusoidal wave of frequency f = 120Hz and amplitude ym =
8.5mm is sent at one end of the string. What is the average wave transport
energy?
8.3.
Two sinusoidal waves y1(x, t) and y2(x, t) have same wavelength and
travel together in same direction. Their amplitudes are ym1 = 4.0mm and
ym2 = 3.0mm, and their phase constants are zero and /3 rad. What are the
amplitude and phase constant of the resultant wave? Write the wave
equation of the resultant wave.
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08 Wave and Geometric Optics
8.4.
The maximum pressure amplitude pm that a human ear can tolerate in
loud sound is about 28Pa. What is the displacement amplitude s m for such
a sound in air density  = 1.21kg/m3, at frequency of 1,000Hz and speed
of 343ms-1?
8.5.
In the figure below, it shows two sound source S1 and S2, which are in
phase and separated by distance D = 1.5 emit identical sound waves of
wavelength .
(a). What is the path length difference of the waves from S1 and S2 at
point P1, which lies on the perpendicular bisector of distance D at a
distance greater than D from the sources? Name the type of
interference.
(b). What is the path length difference and the type of interference at
point P2?
8.6.
Human brain used to determine the direction of a source of sound is the
time delay t between the arrival of the sound at the right ear closer to the
source and arrival at the further left ear. Assume that the source is distinct
so that a wave front from it is approximately planar when reaching you,
and let D represent the separation between your ears. Given that the speed
of sound in air and water are respectively equal to 343m/s and 1,472m/s
respectively.
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08 Wave and Geometric Optics
(a). Find the time delay in terms of D. the speed of sound , and angle 
between the direction of the source and the forward direction.
(b). Suppose that you are submerged in water at 20 0C when the
wavefront arrives directly to your right ear. Based on time delay, at
what angle  from the forward direction does the source seem to be?
8.7.
An electric spark jumps along a straight line of l = 10m, emitting a pulse
of sound that travels radially outward. The power of the emission is P s =
1.6x104W. Note that the spark is said to be a line source of sound.
(a). What is the intensity of the sound when it reaches a distance r = 12m?
(b). What is power Pd sound energy intercepted by an acoustic detector
of area Ad = 2.0cm2, aimed at the spark and located at distance r =
12m from the spark?
8.8.
The sound level 46m in front of the speaker was β2 = 120dB. What is the
ratio of the intensity I2 of the band at that spot to the intensity I1 of a
jackhammer operating at sound level β1 = 92dB?
8.9.
A trooper is chasing a speeder along a straight stretch of road at 160km/h.
Both have same speed. Thus, the trooper sounds a siren of frequency
500Hz. What is the frequency heard by the speeder? You may take the
speed of sound to be 343m/s.
8.10. The 16,000Hz whine of turbine engine in the jet plane moving with speed
200m/s. What is the frequency heard by the pilot of a second plane trying
to overtake it at a speed of 250m/s?
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08 Wave and Geometric Optics
8.11. A bullet is fired with a speed of 685m/s. Find the angle made by the
shock cone with the line of motion of the bullet. You may use 343m/s for
the speed of sound.
8.12. A jet plane passes over you at height of 5,000m and a speed of Mach 1.5.
You may use 343m/s for the speed of sound.
(a). Find the Mach cone angle.
(b). How long after the jet passes directly overhead does the shock wave
reach you. You may use 331m/s for the speed of sound.
8.13. A praying mantis preys along the central axis of a thin symmetric lens
20cm from the lens. The lateral magnification of the mantis provided by
the lens is m = -0.25 and the refractive index of the lens material is 1.65.
Determine
(a).
(b).
(c).
(d).
(e).
(f).
The type of image produced by the lens.
The type of lens
Whether the mantis is inside or outside the focal point.
Which side of the lens the imaged appears.
Whether the image is upright or inverted.
What is the radii of the curvature of the lens?
8.14. The jalapeno seed O is placed in front of two thin symmetrical coaxial
lens 1 and lens 2, in which the focal lengths are f1 = +24cm and f2 =
+9.0cm respectively with the lens separation of L = 10cm. The seed is
6.0cm from lens 1, where is the final image located? And state its type.
8.15. A photographer has 8X magnifier for examining the negative. What is the
focal length of the magnifier?
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