Chapter 08 Wave and Geometric Optics 171

Contents

Page

- i -

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

List of Figures

Page

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Figure 8.1: Illustration of transverse wave ....................................................................... 172 Figure 8.2: Illustration of longitudinal wave .................................................................... 172 Figure 8.3: A moving beetle creates both transverse and longitudinal waves ................. 173

Figure 8.4: Waveform of the wave function )tkxsin(y)t,x(y m ........................... 174

Figure 8.5: Illustrations of the interference displacement of two waves of different phases

........................................................................................................................ 177 Figure 8.6: Phasor representation of two waves of small frequency and the displacement

result of interference ...................................................................................... 179 Figure 8.7: Illustration of a sound wave travels from a point source S ............................ 180 Figure 8.8: Plots of (a) displacement and (b) air pressure at t = 0 ................................... 181

Figure 8.9: Two point sources S1 and S2 emit spherical sound waves in phase ............... 182 Figure 8.10: Illustration of Doppler Effect ......................................................................... 185

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 ................................................................... 187 Figure 8.12: Illustration of a source moving at speed v faster than speed vs of sound

showing Mach cone of wave.......................................................................... 187

Figure 8.13: An extended object O and its virtual image l in a plane mirror ..................... 189 Figure 8.14: An illustration of image formed by a concave mirror.................................... 190 Figure 8.15: An illustration of image formed by a convex mirror ..................................... 190

Figure 8.16: Real image formed by concave mirror........................................................... 191 Figure 8.17: Focal point F and length of a concave mirror ................................................ 192

Figure 8.18: Focal point F and focal length of a convex mirror ......................................... 192 Figure 8.19: Images formed by spherical refracting surfaces ............................................ 194 Figure 8.20: (a) Convergence of light by convex lens and (b) divergence of light by

concave lens ................................................................................................... 197 Figure 8.21: Image formed by convex lens and concave lens ............................................ 197

Figure 8.22: Object placed on near point of human eye..................................................... 199 Figure 8.23: Object placed closer to the human eye, a distance shorted than near point ... 199

Figure 8.24: Correction for fussy image formed by object placed shorter than Pn closer to

eye and closed to the focal point .................................................................... 200


eye and is inside the focal point ..................................................................... 201 Figure 8.26: The structure of a compound microscope ...................................................... 201

Figure 8.27: Structure of a refracting telescope ................................................................. 202 Figure 8.28: 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.

08 Wave and Geometric Optics

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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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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 LT v

d

v

dt .

Thus, the distance d is equal to

LT v

1

v

1/td =

150

1

50

1/10x0.5 3 = 37.5cm.

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).

)tkxsin(y)t,x(y m (8.1)


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where ym is the amplitude, )tkxsin( 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 )tkxsin(y)t,x(y m

The phase of the wave is the argument )tkx( . 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 )kxsin(y)0,x(y m .

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 )kxsin(y 1m = 1m xksiny . This implies

that k = 2. Thus, the angular wave number k is equal to

2k (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 )tsin(y)t,0(y m . It is also equal to )tsin(y)t,0(y m .

At time t = t1 and t = t+T, the displacements are the same as shown in Fig.

8.4(b). Thus, function )tsin(y)t,0(y 1m1 = )Tt(siny 1m . This implies that

T = 2. This shall mean that the angular frequency of the wave is equal to

T

2 (8.3)


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The frequency of the oscillation f is equal to 1/T. Thus, the angular frequency

is also equal to 2f.

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,

0dt

dxk (8.4)

This implies that the wave speed is

kdt

dx (8.5)

Example 8.2

A wave traveling along a string is described by )t72.2x1.72sin(00327.0)t,x(y .

(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.

The velocity of this wave is k

1.72

72.2 =3.77cm/s.

The displacement is )t72.2x1.72sin(00327.0)t,x(y

= )9.18x72.210x5.22x1.72sin(00327.0 2 = )rad18.35sin(00327.0

= )Rad18.35sin(00327.0 = 0.00327x0.583 = 1.91mm.


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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

l

m (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 )tkxsin(y)t,x(y m1 and the second one is given by

)tkxsin(y)t,x(y m2 , where is the phase constant. When these two waves

interference, the resultant displacement is given by

)tkxsin(y)tkxsin(y)t,x(y mm

' (8.8)

which is equal to

2

1tkxsin

2

1cosy2)t,x(y m

' . Note that

2

1cos

2

1sin2sinsin . From this equation, the amplitude at

interference is

2

1cosy2 m and oscillation term is

2

1tkxsin . If is equal

to zero then at interference, the magnitude is 2ym. This is a case of fully


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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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The kinetic energy dK associated with a string element of mass dm is given

by

dmu2

1dK 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

dt

tkxsin(yd

dt

)t,x(ydu m

= )tkxcos(ym (8.10)

Equation (8.9) becomes equation (8.11) after substituting equation (8.10) and

dm = dx from equation (8.6).

dxtkxcosy2

dK 22

m

2

(8.11)

Diving equation (8.11) with dt, it yields the rate of energy transmission equation,

which is shown in equation (8.12).

dt

dxtkxcosy

2dt

dK 22

m

2

(8.12)

Note that dt

dxis equal to , the traveling speed of the wave. The average rate at

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,

2

m

2

avg

y4

1

dt

dK

(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.

2

m

2

avg y2

1P (8.14)


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8.1.6 Phasors

Supposing two sinusoidal waves of functions )tkxsin(y)t,x(y 1m1 and

)tkxsin(y)t,x(y 2m2 , when they interfere each other, the displacement of

interference is equation (8.15).

)tkxsin(y)t,x(y 1m

' + )tkxsin(y 2m (8.15)

The displacement of the interference is also equal to )tkxsin(y)t,x(y '

m

' .

The amplitude '

my and phase constant can be calculated from phasor

representation of the waves, which are illustrated in Fig. 8.6. The amplitude '

my can be calculated using equation (8.16).

22m

2

1m2m

'

m sinyycosyy (8.16)

The phase constant can be calculated using equation (8.17).

1m2m

2m1

ycosy

sinytan (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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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

V/V

pB

(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-1

s-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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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

)tkxcos(s)t,x(s m (8.20)

or it can be )tkxsin(s)t,x(s m . As the wave moves, the air pressure at

position x varies sinusoidally. Thus, the description of air pressure variation is

)tkxcos(p)t,x(p m (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).

mm s)(p (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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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 S2 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).

L2 (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.

2m , where m = 0, 1, 2, 3, ……. (8.24)

This is also meant that

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.

1m2 , where m = 0, 1, 2, 3, ……. (8.25)


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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,

A

Pl (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).

2

m

2s2

1l (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 4r2. 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

2

S

r4

P

l (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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Instead of deal with intensity of sound, it is better and convenient to use

sound level , which is defined as

0

logdB10l

l (8.29)

Here dB is denoted as decibel, which is unit for sound. l0 is the standard

reference intensity that has value 10-12

W/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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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).

S

D'

ff (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).

D' ff (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

S

'

ff (8.32)


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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 S

D'

ff and apply it to this scenario.

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

S

D'

ff

Since vD is equal to zero, the f’ measured by the detector mounted on the

pole is 242343

343x250,1

S

'

ff = 4,250Hz.

vS is equal to zero, the frequency f” measured by the detector mounted on

the rocket is 343

242343x4250" D

ff = 7,240Hz.

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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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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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).

sst

tsin

v

v

v

v (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

h

'hm (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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p

im (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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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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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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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

rf2

1 (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).

ipf

111 (8.37)


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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 ipf

111 .i.e.

0.2p

111

pf.

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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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).

r

nn

i

n

p

n 121 2 (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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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 r

nn

i

n

p

n 121 2 , we need to calculate the value of p. The

image is virtual image then i = -5.0mm. The object is facing a concave surface.

Thus, radius of curvature is – 3.0mm. Thus, 3.0-

1.61.0

5.0-

1.01.6

p. The actual

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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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

21

111

1

rrn

f (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

21medium

111

1

rrn

n

f (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 F1 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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Step 1: Let p1 be the distance of object from lens 1. One can then final the i1

value from equation ipf

111 or drawing the ray diagram.

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 ipf

111 or

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 x-

ray 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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Figure 8.22: Object placed on near point of human eye

For small angle , tan , thus is approximately equal to

cm25

h (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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eye and closed to the focal point

From Fig. 8.24, one can obtain equation (8.42), which is

f

h' (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).

f

cm25m (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

cm25

25

f

fp (8.45)

The lateral magnification is equal to


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ff

f

p

i 251

cm25

25/25m cm (8.46)


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

obf

s

p

im (8.47)


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Since the image I is located just inside focal point '

1F 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,

ey

cm25mmM

ff

s

ob

(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 '

1F 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 '

1F . 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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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

eyf

fobm (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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8.4. The maximum pressure amplitude pm that a human ear can tolerate in

loud sound is about 28Pa. What is the displacement amplitude sm 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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(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 200C 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 Ps =

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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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). The type of image produced by the lens.

(b). The type of lens

(c). Whether the mantis is inside or outside the focal point.

(d). Which side of the lens the imaged appears.

(e). Whether the image is upright or inverted.

(f). 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?

Documents

Chapter 08 Wave and Geometric Optics 171