sem2 optik nota

1

1. OSCILLATIONS

- Oscillatory motion is a repetitive motion back and forth

about an equilibrium position – periodic

- simple harmonic motion (SHM) – basic form of

oscillatory motion.

-

Describing Oscillations

displacement (x or y) – distance from equilibrium

restoring force (F) – the force to bring back to

equilibrium position.

amplitude (A) – maximum displacement from

equilibrium.

period (T) 9 – the time for one cycle

frequency (f) – the number of cycles per unit time.

Unit (Hertz).

angular frequency ( ) -

- instantaneous velocity is zero at the point where x =

(turning point of motions)

2

- maximum speed is reached as the object passes

through the equilibrium position (x = 0)

- velocity is positive as the object moves to the right of

the equilibrium and negative when it is to the left

-

important relations

- how is max speed related A?

- how is f or T related to spring mass m, spring const k

and A?

- is oscillation a consequence of Newton’s laws?

Simple Harmonic Motion (SHM)

- Restoring force is proportional to the displacement

from equilibrium

Hooke’s Law

– spring constant – unit (N/m)

- the acceleration is

but Newton’s Law

- harmonic oscillator – a body undergoing SHM

3

Equation of SHM

- simple harmonic motion is the projection of circular

motion onto a diameter.

- displacement

- acc of circular motion is constant

- acc

√

√

A

Q

P

O

4

√

Displacement, Velocity and Acceleration in SHM

phase constant

If at , the phasor OQ makes and angle (phi) with

the positive x-axis then at a later time t, the angle would

be ( )

( )

( )

( )

To find

(

)

√

5

Graphical Representative of SHM

****************************************

( )

( )

( )

6

Ex: An air track glider is attached to a spring, pulled 20.0 cm

to the right, and released at t = 0 s. It makes 15 oscillations

in 10.0 s. (assume SHM)

a. what is the period of oscillations?

b. what is the object maximum speed?

c. what is the position and velocity at t = 0.800 s?

- the oscillation freq is

The oscillation amplitude A = 0.200 m. Thus

( )

The object starts at x = +A at t = 0. The position at t = 0.800

s is

(

) ( ) (

( )

)

( ) ( )

The velocity at this time is

7

(

) ( ) (

(

)

( ) ( )

SHM & CIRCULAR MOTION

(to describe oscillations with other initial conditions)

- imagine a turntable with a small ball glued to the edge.

When light is projected, the shadow of the ball move

oscillates back and forth as the turntable rotates.

(uniform circular motion projected onto one dimension

is SHM)

Thus the x-component of a particle in circular motion is SHM

A

( )

angular velocity

8

The Phase Constant

If the initial condition be , then the angle at a

later time will be be

particle projection will be

( ) ( )

velocity will be

( ) ( ) ( )

( ) is the phase of oscillation

is the phase constant

Ex: An object on a spring oscillates with a period of

0.80 s and amplitude of 10 cm. At t = 0 s, it is 5.0 cm to the

left of equilibrium and moving to the left. What are its

position and direction of motion at t = 2.0 s?

Soln: Find the phase constant of the motion from initial

condition:

9

(

) (

)

moving to the left at t = 0, thus it is in the upper half of the

circular motion diagram and must have a phase between 0

and rad. Thus

Thus the object’s position at 2.0 s is

( ) ( )

( ) (

( )

)

( ) ( )

The object is 5.0 cm to the right of equilibrium.

which way is it moving?

Calc the velocity:

( ) ( )

velocity is positive, so the motion is to the right.

10

Energy in SHM

A particle undergoing SHM, has purely potential energy at

and purely kinetic energy at (equilibrium)

( )

√

Earlier using kinematics we obtain;

Potential energy:

Kinetic energy:

Total mechanical energy:

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For spring system with spring const k and object mass m

√

√

√

(period and freq do not depend on amplitude)

Since energy is conserved:

The total mechanical energy E is constant

( )

thus we can also relate:

√

( ) √

√

√

( )

The Pendulum

- two forces acting: string tension T and gravity FG

12

a. Simple Pendulum

net force =

angular freq √

Period T = √

13

b. Physical Pendulum

Torque in the gravitational arm

From Newton’s second law for rotational motion.

acceleration :

√

14

√

the above equation can be made the basis for

experimental determination of moment of inertia of

body of complicated shape.

Moment of inertia about a pivot point:

Moment of inertia

Thin rod about a perpendicular

axis through its midpoint

Pivot about one end

Damped Oscillation

- an oscillation the runs down and stops – friction –

mechanical energy transforms into thermal energy.

- there must be a damping force (drag force)

15

using

and

, and rearranging

The solution to the above equation is

( ) ( )

The angular frequency is

√

√

for no damping

16

- ( ) is a slowly varying amplitude

time constant is the characteristic time during which

the energy of oscillation is dissipated – lifetime of

oscillation.

has unit kg/s

( )

the mechanical energy

( )

( )

( )

(

)

- the oscillator’s mechanical energy decreases

exponentially with time constant.

Ex: A 500 g mass swings on a 60 cm string as

pendulum. The amplitude is observe to decay to half

its initial value after 35.0 s

a. what is the time constant for the oscillator?

b. At what time will the energy have decayed to half it

its initial value?

Soln: a. The initial amplitude at t = 0 s is A

at t = 35 s, amplitude is (1/2)A

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The amplitude of oscillations is

( )

in this case

Solve for (

)

can calculate damping const

- b. The energy at time t is given by

( )

the time at which the energy has dissipated to half its

initial value is called the half life,

( )

(

)

The half energy is gone at

**************************************

Resonance:

natural frequency

18

driving frequency

resonance: when

Ex: singers can break crystal goblet

army marching in step on a bridge cause it to collapse

pushing a child on a swing

Pushing a person in a swing in time with the natural

interval of the swing (its resonant frequency) will make the

swing go higher and higher (maximum amplitude), while

attempts to push the swing at a faster or slower tempo will

result in smaller arcs. This is because the energy the swing

absorbs is maximized when the pushes are 'in phase' with

the swing's oscillations, while some of the swing's energy is

actually extracted by the opposing force of the pushes when

they are not.

1

CHARACTERISTICS OF WAVES

- MECHANICAL WAVES

Learning outcomes:

1. the different variety of mechanical waves

2. how to use the relationship among speed,

frequency, wavelength for a periodic wave

3. how to interpret and use the mathematical

expressions for a sinusoidal periodic wave

4. how to calculate energy transportation rate

5. what happens when waves overlap

6. the properties of standing waves and their analysis

Definition:

- A disturbance that travels through some material or

medium. As it travels, the particles of the medium

undergo displacement of various kinds depending on

the nature of the wave......

Examples of m. waves:

- ripples on ponds, musical sounds, seismic tremors

Types of Mechanical Waves

- transverse: displacement direction of motion

- longitudinal: displacement direction of motion

Water waves?

Waves created by soccer fans at the stadium?

2

Periodic Transverse Waves

-

- When a sinusoidal wave passes through a medium,

every particle in the medium undergoes SHM.

- speed of periodic wave,

4

Periodic Longitudinal Wave

6

Mathematical description of a wave

when but moves to the right with time

, motion of x at time t can be described as

motion at x = 0 at an earlier time of

wave number

periodic wave

General equation:

(moving in +ve x-direction)

- should be able to differentiate movement of particles and movt of waves.

7

Particle velocity and acceleration

Speed of Transverse Wave Wave speed on string

8

Impulse-momentum relation

- impulse is equal to change in total transverse component of momentum of the moving part of the string.

Transverse impulse =

The mass of the moving portion of the string =

= mass per unit length

9

Transverse momentum =

Momentum increases with time because more mass is brought into motion. Impulse is the total change of momentum

Speed of waves on strings

Examples: 1. Transverse waves on the string have wave speed 8.00

m/s, amplitude 0.0700 m and wavelength 0.320 m. The waves travel in the –x direction, and at t = 0 the x = 0 end of the string has a maximum upward displacement. (a) Find the frequency, period and wave number of

these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of particle at

x = 0.360 m at time t = 0.150 s. (d) How much time must elapse from the instant in

part (c) until the particle at x = 0.360 m next have maximum upward displacement?

10

Soln:

Given: v = 8.00 m/s, A = 0.0700 m, λ = 0.320 m

(a) v = f λ

(b)

(c)

Plug in the value: x = 0.360 m and t = 0.150 s

4.95 cm

(d) The general equation for this wave can be taken to be;

or

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for n = 4, t = 0.1150 s (max just before condition (c))

for n = 5, t = 0.1550 s (next max )

time lapse is = 0.1550 - 0.1500 = 0.0050 s

2. A 1.50 m string of weight 1.25 N is tied to the ceiling at its upper end and the lower end supports a weight of W. When you pluck the string slightly the waves travelling up the string the equation

(a) How much time does it take a pulse to travel the full length of the string?

(b) What is the weight W? (c) How many wavelengths are there on the string? (d) What is the equation for the wave travelling down

the line?

Soln:

From the equation:- A = 8.50 mm, k = 172 m-1,

(a)

12

(b)

(c) . This is the length of

one wavelength. For the whole length of the string;

(no of wavelengths)

(d)

Energy in Wave Motion

How is energy transferred from one portion of the string to another?

- When a point in a string moves in the y-direction, the force F does work on this point and therefore transfers energy into the part to the right of it.

- The corresponding power at that point

13

but and

maximum instantaneous value

average power

Wave Intensity (sound, seismic)

SUPERPOSITION OF WAVES

Interference – overlapping of waves

15

The Principle of Superposition

Example:

A wave pulse on the string has a dimension shown in the figure at t = 0. The wave speed is 40 cms-1.

(a) If point O is a fixed end, draw a total wave on the string at t = 15 ms, 20 ms, 25 ms, 30 ms, 40 ms, 45 ms.

16

(b) Repeat part (a) for the case in which point O is a free end.

Soln:

Standing Waves on a String

- the principle of superposition applies to the incident and reflected waves.

- the resultant is an interference pattern - constructive or destructive interference

17

For a wave from reflected from a fixed end

incident wave travelling to

the left.

reflected wave travelling to

the right.

18

- position of nodes: when . This occurs when

using

Normal Modes on a String

Consider a string of length L fixed at both ends

- when plucked, standing wave will result

possible wavelengths

possible standing wave frequencies

this is called the fundamental

these are called the harmonics

A normal mode – motion in which all the particles move sinusoidally with the same frequency.

19

Example:

fundamental frequency f1

second harmonics, f2

third harmonics f3

fourth harmonics f4

1

SOUND

Objectives:

1. to describe a sound wave in term of particle

displacement of pressure fluctuation.

2. to calculate the speed of sound, intensity and

frequency wave in matter

3. to understand and resonance & interference and able

to solve problems related to them.

4. to understand Doppler shift from a moving source.

‘Sound is the physical sensation created by longitudinal

waves that stimulates or ears.’

‘When you put your ear flat against the ground, you can

hear an approaching train or truck’

‘sound cannot travel through vacuum’

‘speed of sound is different for different medium’

Material Speed (ms-1)

Air 343

water 1440

Sea water 1560

Iron 5000

2

- longitudinal wave in a medium

- audible frequency range 20 to 20,000 Hz

ultrasonic – above audible

infrasonic – below audible

- loudness – relate to intensity

- pitch – relate to frequency

- musical sound – harmonic content.

- noise – combination of all frequencies

- one-dimensional sound wave equation

( ) ( )

pressure wave

- fluctuations of the pressure wave above and below

atm pressure in a sinusoidal variation.

- pressure fluctuations depends on the difference

between the displacement at neighbouring points in

the medium.

- the fractional volume change can be related to

pressure fluctuation by its bulk modulus B

( )

( )

- the volume change

( )

3

( ) ( )

( ) ( )

pressure amplitude.

- pressure amplitude is proportional to displacement

amplitude.

- waves of shorter wavelengths have greater pressure

variation for a given amplitude.

4

Perception of sound waves

- for a given frequency, the greater the pressure

amplitude the louder is the sound.

- a sound at one frequency may seem louder than the

one or equal amplitude at another.

Same loudness

Frequency Min press amplitude

1000 Hz 3 x 10-5 Pa

200 Hz or 15,000 Hz 3 x 10-4 Pa

5

- pitch – quality that we identify as low or high –

frequency related.

- musical sound consists of waves more complex than

sine wave – fundamental + many harmonics at the

same time.

- tones (quality, timbre) from different instrument

might have same fundamental frequency but sound

different because of different harmonic content –

similar to human voice.

- noise: combination of all frequencies not just the

multiples of the fundamentals.

Speed of Sound Waves

- Speed of mechanical wave

√

inertia is related to mass which can be described as density

or mass/volume.

in fluid

- the quantity of fluid set in motion in time

- longitudinal momentum ( )

- bulk modulus

the net force on moving fluid is

6

longitudinal impulse

( )

√

**************************************

Example:

In a sinusoidal sound wave of moderate loudness the maximum

pressure variation are of the order 3.0 x 10-2 Pa above and below

atmospheric pressure (1.013 x 105 Pa). Find the corresponding

maximum displacement if the frequency is 1000 Hz. In air at

normal atmospheric pressure and density, the speed of sound is

344 m/s and the bulk modulus is 1.42 x 105 Pa.

Soln:

7

The maximum displacement

and

( )

( )( )

- What is the wavelength of these waves?

- For 1000 Hz waves in air, what displacement amplitude

would be needed for the pressure amplitude to be at the

pain threshold of 30 Pa?

(

) ( ) (

)

- For what wavelength and frequency will waves with a

displacement amplitude of produce a

pressure of 1.5 x 10-3 Pa?

so

(

) (

)

8

thus

********************

Speed of sound in solid

- in solid rod or bar, slight expansion will occur when it is

compressed longitudinally.

- the reasoning is the same as in fluid

√

Y = Young’s modulus

***********************

Example:

An 80 m brass rod is struck at one end. A person at the other end

hears 2 sounds as a result of two longitudinal waves, one travelling

in air and the other in metal rod. What is the time interval between

the two sounds? The speed of sound in air is 344 m/s and the,

density of brass is 8.6 x 103 kg/m3 and Young’s modulus for brass is

9.0 x 1010 Pa.

Soln:

in air

in brass: √

√

9

*****************************

Speed of Sound in an ideal Gas

- √

Sound Intensity

- travelling sound wave transfer energy from one region of

space to another.

- intensity is the energy being transported per unit area.

- the particle velocity:

( ) ( )

( )

( ) ( ) [ ( )][ ( )]

- velocity of wave as a whole is not the same as particle

velocity.

- the intensity is the time average of ( ) ( ), thus

with and

10

√

the equation shows why in a stereo system, a low-

frequency woofer has to vibrate with much larger

amplitude than a high frequency tweeter to produce the

same sound intensity.

- intensity and pressure amplitude

√

- sinusoidal sound wave of the same intensity but different

frequency have different displacement amplitude A but

the same pressure amplitude

- if the sound distributes itself in all directions equally, the

intensity decreases according to inverse square law.

- an architect’s job when designing an auditorium is to tailor

sound reflections so that the intensity is uniform all

around.

The decibel scale

- the sound intensity level is defined as

( )

it is expressed in decibel, dB

11

Example:

Two identical machines are positioned the same distance

from a worker. The intensity of sound delivered by each operating

machine at the worker’s location is 2.0 x 10-7 W/m2.

a. Find the sound level heard by the worker if one machine is

operating.

b. Find the sound level heard when both the machines are

operating.

Soln:

a.

(

)

b.

(

)

12

when the intensity is doubled, the sound level increases by 3 dB

only. The 3 dB increase is independent of the original sound level.

Loudness is a psychological response to a sound. It depends on

both the intensity and the frequency of sound.

- there is no simple relation between physical measurement

and psychological measurement.

-

Standing Sound Waves and Normal Modes

- when longitudinal waves propagate in a fluid in a pipe with

finite length, the waves are reflected back just as in

transverse waves – standing waves

- sound wave in fluid is described by the displacement of the

fluid or the pressure variation of the fluid.

13

- adjacent nodes are separated by λ/2

- a particle at a displacement node does not move whilst at a

displacement antinode oscillates with maximum

amplitude.

14

- particles on opposite side of the nodes vibrates in opposite

phase gas undergoes the maximum amount of

compression and expansion

- particles on oppt side of the antinode vibrates in phase

no variation in pressure or density.

- pressure node describes the standing wave, pressure

antinode describers greatest pressure or density variation.

A pressure node is always a displacement antinode and a pressure

antinode is always a displacement node.

Organs Pipes and Wind Instruments

1. Open pipes

fn = nf1 (n = 1,2,3,4.....)

15

2. Stopped Pipes

(n = 1, 3, 5.......)

Example:

On a day when the speed of sound is 345 m/s, the fundamental

frequency of a stopped organ pipe is 220 Hz

a. How long is this stopped pipe?

b. The second overtone of this pipe has the same wavelength as

the third harmonic of an open pipe. How long is the open

pipe?

Soln:

a. For a stopped pipe,

16

b. The frequency of the first overtone of stopped pipe is

, and the frequency of the second overtone is

if the wavelength and frequency are the same, the frequency

of the third harmonic of the open pipe is also 1100 Hz.

The third harmonic of an open pipe is at ( )

(

)

Note:

- the open pipe is longer than the stopped pipe.

- its third harmonic is thus 1100 Hz, making its fundamental

is 1100/3 = 367 Hz. This is higher than the fundamental of

the stopped pipe.

Resonance and Sound - same discussion as with transverse wave

- a system can be made to oscillate with a frequency equal to

the frequency of the applied force (driving freq)

- applying an external force periodically along the natural

freq of the system will increase the amplitude.

- for a system with many normal modes, resonance occurs

when the variably applied force coincides with any

particular modes.

- resonance occurs when the magnitude of the forced

oscillations is maximum.

17

Interference of Waves

- when two or more waves overlap in the same region of

space interference.

- standing waves – combination of waves from opt direction

The figure below shows what it meant by combination of waves

- not quite the standing waves we had earlier in transverse

waves – there is still a flow of energy from the speakers to

the surrounding.

- constructive interference – the dist travelled by the two

waves differ by a whole number of wavelength (0, λ,2λ..)

- destructive interference – λ/2, 3λ/2, 5λ/2...

18

- interference effects are used to control noise from very

loud sound sources – use another source, place at the right

distance – create destructive interference.

Example:

Soln:

the dist from speaker A to P is [( )]

the dist from speaker B to P is [( )]

the path difference d = 4.47 m - 4.12 m = 0.35 m

Given v = 345 m/s

a. For what frequencies does

constructive interference occur at P?

b. For what frequencies does destructive

interference occur at P

19

a. constructive interference

( )

, 3000 Hz....

b. destructive interference

( )

as the frequency, the sound at P alternates between large and

small amplitude.

Beats

- fluctuations in amplitude produced by two sound waves of

slightly different frequencies.

- this amplitude variation causes variation of loudness -

beats

- the frequency different is the beat frequency.

20

- alternative method- adding up the two waves

( ) ( )

( ) ( ) [

( )( ) ]

( )( )

- beats can still be heard up to beat frequency 6 Hz

- application of beats – tuning musical instruments.

***********************

Example:

Two guitarists attempt to play the same note of wavelength 6.50

cm at the same time, but one of the instruments was slightly out of

tune and plays a note of wavelength 6.52 cm instead. What is the

frequency of the beat these musicians hear when they play

together?

Soln: and

(

) (

) (

)

there are 16 beats per second.

************************

The Doppler Effect

- when source of sound and observer are in relative motion,

the sound heard is not the same as the source frequency.

21

moving listener

- Speed of propagation relative to the listener,

or

(

) ( )

Moving Source and Moving Listener

- wave crest emitted by a moving source are crowded

together in front of the source and stretched out behind it.

- in the region to the right of source (in front of source)

22

the region behind

the frequency heard by the listener

( )

- the Doppler effect explains why the siren on a fire or

ambulance has a high pitch ( ) when it is

approaching us and low pitch when it is moving away.

Example:

A police siren emits a sinusoidal wave with frequency

. The speed of sound is 340 m/s.

a. Find the wavelength of the wave if the siren is at rest in

air.

b. If the the siren is moving at 30 m/s, find the wavelength

of the waves in front and behind the source.

Soln: a) at rest

b)

23

- If the listener is at rest and the siren is moving away at 30

m/s, what frequency does the listener hear?

( )

- if the siren is at rest and the listener is moving away from

the siren at 30 m/s, what frequency does the listener hear?

( )

( ) 274 Hz

- if the siren is moving from the listener with a speed of 45

m/s relative of air and the listener is moving towards the

siren with a speed of 15 m/s relative to the air, what

frequency does the listener hear?

24

( )

************

Shock Waves

- motion of airplane through the air produces sound; if

as , λ approaches zero, wave crest pile up

occurs large increase in the aerodynamic drag

- supersonic

- after time t the crest emitted by appoint S1 has spread to a

circle of radius and the plane has moved to a greater

distance to position S2. The circular crests interfere

constructively at points along the blue line.

25

- this causes a very large amplitude crest – shock wave.

is called the Mach number

- when an airplane travels at supersonic speeds, the noise it

makes and its disturbance of the air forms into a shock

wave containing a tremendous amount of sound energy.

- when an aircraft approaches the speed of sound, it

encounters a barrier of sound waves in front of it. To

exceed the speed of sound, the aircraft needs extra thrust

to pass through the sound barrier.

- this causes a very loud noise - the sonic boom.

- application of shock wave -

breaking of kidney stones

Applications

Sonar

- the reflection of sound to determine distance.

- also called ‘pulse echo’

- used to locate underwater objects

-

*********************************

Example:

An airplane is flying at Mach 1.75 at an altitude of 8000 m where

the speed of sound is 320 ms-1. How long after te plane directly

overhead will you hear the sonic boom.

Soln:

26

the speed of the plane , ( )( )

From the figure

( )( )

- you here the boom 20.5 s after the airplane passes

overhead and at that time it has travelled 11.5 km.

************************************

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 36

Diffraction


Goals for Chapter 36

• To define and explain Fresnel and Fraunhofer

diffraction

• To consider single-slit diffraction

• To summarize and then apply an understanding of

diffraction gratings

• To consider the atomic example of x-ray diffraction

• To study circular apertures and resolving power

• To introduce holography


Introduction

• It’s intuitive that sound can diffract (and travel around corners). Light doesn’t “show its poker hand” so easily.

• If you shine light from a point source to a ruler and look at the shadow, you’ll see the edges are … well … not sharp. A close inspection of the indistinct edge will reveal fringes.

• This phenomenon may not sound useful yet but stay with us until the end of Chapter 36. This line of thinking has shown the way for advances in DVD technology and applications in holography.


Fresnel and Fraunhofer diffraction

• According to geometric optics, a light source shining on an

object in front of a screen will cast a sharp shadow. Surprisingly,

this does not occur.


Diffraction and Huygen’s Principle

• Diffraction patterns can be analyzed as we did in Section 33.7 using

Huygen’s Principle. Recall, every source of a wave front can be

considered to be the source of secondary waves. Superposition of these

waves results in diffraction.

• If the source and the screen are close to the edge causing the

diffraction, the effect is called “near-field” or Fresnel diffraction. If

these objects are far apart, so as to allow parallel-ray modeling, the

diffraction is called “far-field diffraction” or Fraunhofer diffraction.


Diffraction from a single slit

• The result is not what you might expect. Refer to Figure 36.3.


Dark fringes in single-slit diffraction

• Consider Figure 36.4 below.

• The figure illustrates Fresnel and Fraunhofer outcomes.


Fresnel or Fraunhofer?

• The previous slide outlined two possible outcomes but didn’t set conditions to make a choice.

• Figure 36.5 (below) outlines a procedure for differentiation.


Fraunhofer diffraction and an example of analysis

• Figure 36.6 (at bottom left) is a photograph of a Fraunhofer pattern from a single slit.

• Follow Example 36.1, illustrated by Figure 36.7 (at bottom right).


Intensity in a single-slit pattern

• Following the method we used in Section 35.5, we can derive an expression for the intensity distribution.


Intensity maxima in a single-slit pattern

• The expression for peak maxima is iterated for the strongest peak.

• Consider Figure 36.9 that shows the intensity as a function of angle.


Interference from multiple slits

• The approximation of sin θ = θ is very good considering the size of the slit and the wavelength of the light.

• Consider Figure 36.10 at the bottom of the slide.

• Follow Example 36.2.



Multiple slit interference

• The analysis of intensity to find

the maximum is done in similar

fashion as it was for a single slit.

• Consider Figure 36.12 at right.



Several slits

• Consider Figure 36.14

at right.

• Consider 36.15 below.


A range of parallel slits, the diffraction grating

• Two slits change the intensity profile

of interference; many slits arranged

in parallel fashion are now termed as

rulings.

• Consider Figures 36.16 and 36.17 at

right.




The grating spectrograph

• A grating can be used like a prism, to disperse the wavelengths of a light source. If the source is white light, this process is unremarkable, but if the source is built of discrete wavelengths, our adventure is now called spectroscopy. Chemical systems and astronomical entities have discrete absorption or emission spectra that contain clues to their identity and reactivity. See Figure 36.19 for a spectral example from a distant star.


The grating spectrograph II—instrumental detail

• Spectroscopy (the study of light with a device such as the spectrograph shown below) pervades the physical sciences.


X-ray diffraction

• X-rays have a wavelength commensurate with atomic structure. Rontgen had only discovered this high-energy EM wave a few decades earlier when Friederich, Knipping, and von Laue used it to elucidate crystal structures between adjacent ions in salt crystals. The experiment is shown below in Figure 36.21.


Ionic configurations from x-ray scattering

• Arrangements of cations and anions in salt crystals (like Na+ and Cl– in Figure 36.22 … not shown) can be discerned from the scattering pattern they produced when irradiated by x-rays.


X-ray scattering set Watson and Crick to work

• An x-ray scattering pattern recorded by their colleague Dr. Franklin led Watson and Crick to brainstorm the staircase arrangement that eventually led to the Nobel Prize.

• Follow Example 36.5, illustrated by Figure 36.25 below.


Circular apertures and resolving power

• In order to have an undistorted Airy disk (for whatever purpose), wavelength of the radiation cannot approach the diameter of the aperture through which it passes. Figures 36.26 and 36.27 illustrate this point.


Using multiple modes to observe the same event

• Multiple views of the same event can “nail down” the truth in the observation.

• Follow Example 36.6, illustrated by Figure 36.29.


Holography—experimental

• By using a beam splitter, coherent laser radiation can illuminate an object from different perspective. Interference effects provide the depth that makes a three-dimensional image from two-dimensional views. Figure 36.30 illustrates this process.


Holography—theoretical

• The wavefront interference creating the hologram is diagrammed in Figure 36.31 below.


Holography—an example

• Figure 36.32 shows a holographic image of a pile of coins. You can view a hologram from nearly any perspective you choose and the “reality” of the image is astonishing.

1

Polarization of EM Waves

- the EM waves emitted by a TV station all have the

same polarization.

- the EM waves emitted by any common source of

light eg sunlight or bulb are polarized randomly or

unpolarized.

- the direction of polarization is the direction in

which the electric field is vibrating.

- linearly polarized – when the electric field vibrates

in the same direction at all times at a particular point.

Methods to get polarized light

i. Selective Absorption

- using Polaroid sheet- fabricated material of long chain

hydrocarbon.

- the direction perpendicular to the molecular chain is the

transmission axis

The plane of oscillation

of a polarized EM wave.

2

- in an ideal polarizer all light with E parallel to the transmission

axis is transmitted and all light with E perpendicular to the

transmission axis is absorbed.

ii. Reflection

- when light is incident on a reflecting surface of the polarizing

angle the reflected light is linearly polarized.

3

- at the incident angle known as polarizing angle, the light for

which E lies in the plane of incidence is not reflected at all but

is completely refracted.

- at polarizing angle, the reflected and refracted arays are

perpendicular to each other.

iii. Scattering

- scattering of light by the earth’s atmosphere depends on

wavelength.

- sky is blue – blue and violet are scattered more than red and

orange.

- near sunset - sunlight has to travel a long distance through the

earth’s atmosphere, a substantial fraction of the blue light is

Brewster’s Law

Clouds are white

because they efficiently

scatter light of all

wavelengths.

4

removed by scattering. White light minus blue looks yellow or

red.

- Scattering: removing of energy from an incident wave by a

scattering medium and the reemission of some of the portion of

that energy in many directions.

- Rayleigh scattering: small dimension particles (compared to

wavelengths) act as scattering centers eg scattering of

sunlight from oxygen and nitrogen molecules in the

atmosphere.

- Scattering from larger particles such as clouds, fogs, powdered

materials appears as white light in contrast to Rayleigh

scattering.

- The scattered radiation is wavelength independent

- Scattered radiations may also be polarized

- Refer: Stimulated Raman, Rayleigh, and Brilloiun scattering-

modern optics

5

iv. Double Refraction /Birefringence

Material: able to cause double refraction due to the existence of

two different indices of refraction. eg Calcite, quartz

Manifestation of birefringence in materials

- The two beam emerging are linearly polarized in orthogonal

orientation

- The ordinary ray is polarized perpendicular to OA and so

propagates with the refractive index

- The extraordinary ray emerged polarized in a direction

perpendicular to the polarization of the ordinary ray

The perpendicular component propagates with speed

- The other component propagates with refractive index

- some materials such as glass and plastics become birefringent

when stressed.

6

Intensity of Transmitted Polarized light

(selective absorption)

- if the original wave is unpolarized, the sum of the

horizontal and vertical components are equal. Thus

when one component is absorbed, the other

component is transmitted which a reduced intensity

than the incident light.

- supposed the light reaching the polarizing sheet is already

polarized

7

Example:

In the above figure, when unpolarized light falls onto two

crossed polaroids (axes at 90°) no light passes through. What

happens if a third Polaroid with axis at 45° to each of the

other two is placed between them?

Soln:

When the third Polaroid is placed in between the two

polaroids, we have to consider the light path from the

unpolarized stage to the final light emerging at the end.

Out of the first polarizer;

out through the middle polarizer.

out through the third polarizer

8

Thus 1/8 of the original intensity get transmitted.

Application of polarized light:

1. Engineers often used the optical stress analysis, in

designing structures ranging from bridges to small tools -

birefringence

2. LCD- liquid crystal display

- made up of many tiny rectangles called pixels (picture

elements)

- they are actually organic materials that at room temp exists in

a phase that is neither fully solid nor fully liquid

- colour TV and computer LCD – a colour pixel consists of three

subpixels each covered with a red, green or blue filter.

- varying brightness of these three primary colours can yield

almost any natural colour.

9

- a good quality consists of a million or more pixels.

- behind the array of pixels is a light source.

- the light passes through the pixels or not depending on the

voltage applied to each subpixel.


PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 33

The Nature and Propagation of Light


Goals for Chapter 33

• To overview light and its properties

• To study reflection and refraction

• To examine the unique phenomenon of total

internal reflection

• To consider the polarization of light

• To see polarization by reflection

• To study Huygens’s Principle


Introduction

• A coating of oil on water or a

delicate glass prism can create a

rainbow. A rainstorm among open

patches of daylight can cast a

conventional rainbow. Both

effects are beautiful and arise from

the wavelength dependence of

refraction angles.

• Eyeglasses or contact lenses both

use refraction to correct

imperfections in the eyeball’s

focus on the retina and allow

vision correction.


“Light is a wave,” “Light is a particle”

• The wave–particle duality of light was not well understood until

Albert Einstein won his Nobel Prize in the early 20th century. It

was a tenacious understanding of light that led to quantum

mechanics and modern physics.

• Consider Figures 33.1 and 33.2 below. Black-body and laser

radiation look so different but are brother and sister.


Wave fronts and rays

• Light is actually a nearly uncountable number of electromagnetic

wave fronts, but analysis of refraction or reflection are made

possible by treating light as a ray.


Reflection and refraction

• Figure 33.5 illustrates both reflection and refraction at once. The

storefront window both shows the passersby their reflections and allows

them to see inside.


We will consider specular reflections

• A real surface will scatter and reflect light. Diffuse reflection is the rule, not the exception. We will use specular reflection as we used the ray approximation, to make a very difficult problem manageable.

• Consider Figure 33.6.


Laws of reflection and refraction

• Angle of incidence = angle of reflection.

• Snell’s Law of Refraction considers the slowing of light in a medium other than vacuum … the index of refraction.

• Consider Figures 33.7 and 33.8.


Why should the ruler appear to be bent?

• The difference in index of refraction for air and water causes your eye to be deceived. Your brain follows rays back to the origin they would have had if not bent.



Why should sunsets be orange and red?

• The light path at sunset is much longer than at noon when the sun is directly overhead.



Tabulated indexes of refraction


Total internal reflection I

• As the angle of incidence becomes more and more acute, the light ceases to be transmitted, only reflected.



Total internal reflection II

• Using clever arrangements of glass or plastic, the applications are mind boggling.


Total internal reflection III

• Diamonds sparkle as they do because their index of refraction is one of the highest a transparent material can have. Nearly all light that enters a surface ends up making many passes around the inside of the stone. The effect is only amplified by cutting the surfaces at sharp angles. See Figure 33.17.

• Follow Conceptual Example 33.4.


Dispersion

• From the discussion of the prism seen in earlier slides, we recall that light refraction is wavelength dependent. This effect is made more pronounced if the index of refraction is higher. “Making a rainbow” is actually more than just appreciation of beauty; applied to chemical systems, the dispersion of spectral lines can be a powerful identification tool.

• Refer to Figures 33.18 and 33.19 (not shown).


Special case—dispersion and atmospheric rainbows

• As a person looks into the sky and sees a rainbow, he or she is actually “receiving light signals” from a physical spread of water droplets over many meters (or hundreds of meters) of altitude in the atmosphere. The reds come from the higher droplets and the blues from the lower (as we have seen in the wavelength dependence of light refraction).


Selecting one orientation of the EM wave—the Polaroid

• A Polaroid filter is a polymer array that can be thought of like teeth in a comb. Hold the comb at arm’s length with the teeth pointing down. Continue the mental cartoon and imagine waves oscillating straight up and down passing without resistance. Any “side-to-side” component and they would be blocked.


Polarization I

• Read Problem-Solving Strategy 33.2.


• By reflection from a surface? Read pages 1139 and 1140 and then refer to Figure 33.27 below.


Polarization II


• Follow Example 33.6, illustrated by Figure 33.29.


Scattering of light

• The observed colors in the sky depend on the scattering phenomenon. Deep blue sky comes from preferential scattering of photons of shorter wavelength in the visible spectrum.

• Clouds are white because they scatter all wavelengths efficiently.

• Refer to Figures 33.32 and 33.33 (not shown).


Huygens’s Principle

• From the work of Christian Huygens in 1678, the geometrical analysis reveals that every point of a wave front can be considered to be a source of secondary wavelets that spread with a speed equal to the speed of propagation of the wave.

• Consider Figures 33.34 and 33.35.


Huygens’s Principle II

• Huygens’s work can form an explanation of reflection and refraction.

• Refer to Figures 33.36 and 33.37.

1

THE ELECTROMAGNETIC WAVES

- the dual nature of light

- transverse wave

- time varying electric and magnetic fields that can propagate

through space from one region to another.

Maxwell’s equations

(Gauss’s Law)

(Gauss’s Law for magnetism)

(Ampere’s Law)

(Faraday’s Law)

- these equations predict the existence of EM waves that

propagate in vacuum at the speed of light.

permittivity of free space =

James Clerk Maxwell (1831- 1879)

- the first person who truly

understand the

fundamental nature of light

2

permeability of free space = 8.85418 x 10-12 C2/N.m2

Key properties of Electromagnetic wave

- Wave is transverse. E and B are perpendicular to one

another and to the direction of propagation (E x B).

- ratio between the magnitudes E = cB

- require no medium for propagation

- constant speed in vacuum.

- speed in medium (dielectric) less than in vacuum;

K = dielectric function

Km = relative permeability of dielectric

3

The Electromagnetic Spectrum

Wavelength of visible light

400 to 440 nm violet

440 to 480 nm blue

480 – 560 nm green

560 – 590 nm yellow

590 – 630 nm orange

630 – 700 nm red

- this is what we usually term as light

4

- sinusoidal electromagnetic waves (+ve x-direction)

- wave function:

individually:

Units: E (volts per meter), B (Tesla)

5

Wave travelling in the negative x- direction

Energy density:

Energy Flow:

For area perpendicular to the direction of wave motion:-

- energy transferred per unit time per unit cross-sectional

area or power per unit area.

6

- the energy flow per unit time per unit area, S

- as vector quantity – Poynting Vector in vacuum

- the intensity can be taken as the average value of the

Poynting vector

Example:

A radio station on the surface of the earth radiates a sinusoidal

wave with an average total power of 50kW. Assuming that the

transmitter radiates equally in all directions above the ground, find

the amplitudes Emax and Bmax detected by the satellite at a distance

of 100 km from the antenna.

Soln:

The surface area of a hemisphere of radius r = 100 km

all the radiated power passes through this surface, so the

average power per unit area;

but so

7

Note that the value of Emax is comparable to fields

commonly seen in the lab, but Bmax is extremely small. For

this reason, most detectors of EM radiation respond to the

effect of the electric field and not the magnetic field.

EM momentum and Radiation Pressure

- EM waves transport energy. They also carry momentum, p

which is the property of the wave.

- As the momentum is absorbed by the some surface, pressure is

exerted on the surface – radiation pressure

totally absorbed

totally reflected.

- Radiation pressure of stars can have dramatic effect on the

material surrounding them.

8

Standing Wave

- nodal planes and antinodal planes

nodal planes of E:

nodal planes of B:

-

-

- A typical microwave oven sets up a standing EM wave with

a wavelength that is strongly absorbed by

the water in the food. Because the wave has nodes spaced

at apart, the food must be rotated while cooking.

Otherwise the portion that lies at the node will remain

cold.