Vibrations and Waves Simple Harmonic Motion Wave Interactions Mechanical Waves (eg, Sound)...

Vibrations and Waves

Simple Harmonic MotionWave Interactions

Mechanical Waves (eg, Sound)Electromagnetic Waves (eg, Light)

Properties of Waves• Amplitude – Maximum displacement from equilibrium

• Period – Time to complete one cycle (wavelength) of motion. Represented by T; Units of sec.

• Frequency – Number of cycles (wavelengths) per unit time. Represented by f; Units of Hz or sec-1 kHz (AM radio station), MHz (FM radio station), GHz

(radar, microwaves), etc..

• Wavelength – Distance between two adjacent corresponding points on a wave (e.g., crests, troughs, etc.). Represented by lambda, Units of length (m, ft, etc.)

Note: Period = 1 / Frequency and Frequency = 1 / Period

Equilibrium Position

WavelengthAmplitude

Waves and Wave Motion

• What is a Wave??– The motion of a disturbance!

– Example: • One person on each end of a long spring (or rope) • A pulse is produced in the spring….• Wave pulse moves from one end of the spring to the other, • BUT no part of the spring is being carried from one person

to the other.

Direction of WaveDirection of Medium (Spring)

Periodic Motion

• Definition: Back and forth motion over the same path

• Examples: – Mass - Spring System

• Bungee Jumping• Shock Absorbers on Vehicles

– Pendulums:• Child on a swing; • Trapeze Artists• Pendulum of a grandfather clock• Wrecking Ball

Simple Harmonic Motion

• Definition:– Vibration about an equilibrium position in which a

restoring force is proportional to the displacement from equilibrium

– Sine waves describe particles vibrating with SHM

• Examples: – Mass – Spring System (Hooke’s Law)– Pendulum (small angles, <15 degrees)

• Examples: Visible light, radio waves, microwaves, x-rays, etc.

Question---

Is there a direct relationship between the displacement of a mass on a spring and the elastic force of that spring?

Is Felastic proportional to x?... i.e, Felastic Constant x?

At equilibrium, • Net force is zero• So,

Fg + Felastic = 0

Fg = - Felastic

Fg = - Constant * x

x = displacement in meters

Fg

Fel

Mass-Spring SystemWorksheet

x = displacement in meters

Fg

Fel

Mass (kgs) x Fg

0.000

0.250

0.500

0.750

1.000

1.250

1.500

1.750

NOTE:

• Fg = force due to gravity (Fg = m*g)

•Fel = Elastic Force of the springUse 250g, 500g, 1000g masses. Eight

combinations, including 0 mass.

Mass-Spring SystemPlot Fg vs. x

Fg,

(i

n N

ewto

ns)

Displacement, x (in meters)

Hooke’s LawFor a Spring-Mass System,

Robert Hooke established the relationship between Force and Displacement:

Felastic = - kx

where, k is known as the “Spring

Constant”, measuring the “stiffness” of the spring. Units for k is N/m.

Hooke’s Law (con’t)

Example 1: If a mass of 0.55kg attached to a vertical spring stretches the spring 2 cm from its equilibrium position, what is the spring constant?

Given: m = 0.55 kgx = -0.02 mg = -9.8 m/s2

Solution:

Fnet = 0 = Felastic + Fg

0 = - kx + mg or, kx = mg k = mg/x = (0.55 g)(-9.8 m/s2)/(-0.02 m) = 270

N/m

x = -0.02 m

Fg

Fel

Fg

Fel

Period of a Mass-Spring System:

T = 2√

Where, k is the spring constant and

m is the mass

NOTE: Changing the amplitude of the vibration (x) does NOT affect the period or frequency of vibration.

Mass-Spring System

m

k

Example: Mass-Spring System

A body of a 1275 kg car is supported on a frame by four springs, each of which has a spring constant of 2.0 x 104 N/m. Two people riding in the car have a combined mass of 153 kgs. Find the period of vibration of the car when it is driven over a pothole in the road.

Solution: k = 2 x 104 N/m

m = 1275 kg + 153 kg = 1428 kg

But the mass is evenly distributed over 4 springs, so meff = 1428/4 = 357 kgs

T = 2 * *(357 kgs/2 x 104 N/m)1/2

= 2 * * ( 0.01785 s2)1/2

= 0.84 s

Period of a Pendulum System:

T = 2√Where, L is the length of the pendulum

arm

g is the acceleration due to gravity

NOTE: Changing the amplitude of the vibration () does NOT affect the period or frequency of vibration.

Pendulum System

L

g

Pendulums and Spring-Mass Systems

The period and frequency of motion for each of these systems is INDEPENDENT of:

Pendulum:Amplitude ()Mass on swinging arm

Mass-Spring System:Amplitude (x)

Example: Pendulum

You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling almost touches the floor and that its period is 12 s. How tall is the tower?

Given: T = 12 s g = 9.8 m/s2

Solution: Use the equation for the period of the pendulum and solve for L.

T = 2 * * (L / g )1/2

T2 = 4 * 2 * (L / g)

(T2 * g) / (4 * 2) = L

((12 s2)2 * 9.8 m/s2) / (4 * 2) = 35.7 m tall

Review and Revisit…

You are sightseeing in Europe…and curious about the architectural structures….

What would be other ways to determine the height of a tower given minimal pieces of data??

Data Solution Strategy

1. Period of pendulum (suspended from ceiling to floor)

Solve for L in equation for period, T.

2. Time for an object to fall from tower

Kinematics: y = ½ gt2

3. Angle and distance, x Trig. Functions (xTan)

4.

Wave Speed, v

Speed of wave (v) depends upon:– Medium– Frequency, f– Wavelength,

Wave Speed = wavelength x frequency

or

Equation: v = f

Think About It….

• Given the equation for the speed of waves:

v = f

Does this mean, for example, that high frequency sounds (high pitches), travel faster than low frequency sounds????

NO!!! Wavelength and frequency vary inversely to produce the same speed of all sounds

Doppler EffectSTATIONARY

SOUND-GENERATING OBJECT

MOVING

SOUND-GENERATING OBJECT

Waves are created at point source and radiate outward creating a wave front with the same frequency as that of the source.

Velocity, v

A B

Although the frequency of the sound generating object remains constant, wave fronts reach the observer at Point B more frequently than Point A.

Doppler Effect

Doppler Effect:The frequency shift that is the result of relative motion between the source of waves and an observer.

Higher frequency: Object approachingLower frequency: Object receding

Some Applications:Echolocation (e.g., Submarines, Dolphins, Bats, etc.)Police RadarWeather Tracking

Resonance

Every object (all matter!!) vibrates at a characteristic frequency – resonant (“natural”) frequency.

Resonance: A condition that exists when the frequency of a force applied to a system matches the natural frequency of the system.

Examples:– Pushing a swing– Tuning a radio station– Voice-shattered glass.– Tacoma Narrows Bridge Collapse in 1940.—High winds set

up standing waves in the bridge, causing the bridge to oscillate at one of its natural frequencies.

Wave Interactions

• Unlike Matter, more than one wave or vibration can exist – at the same time and – in the same space.

• This is known as SUPERPOSITION.

• Superposition Principle: – The method of summing the displacements

(amplitudes) of 2 or more waves to produce a resultant wave.

– Applies to all waves types – mechanical and electromagnetic.

Interference Patterns

The individual waves can overlap and produce interference patterns.

The resultant wave is the sum of the displacements from equilibrium (ie the amplitude) at each point for the individual waves.

+ =

+ =

Constructive Interference = Reinforcement

Destructive Interference = Cancellation

Superposition Principle

Position Displacement Conveyed Resultant wave displacement

Red Dots Up 2 units Up 4 units

Green Dots Up 1 unit Up 2 units

Yellow Dots Down 2 units Down 4 units

Purple Dots 0 Displacement 0 Displacement

Black Dots Up 2 units/Down 2 Units 0 Displacement

+ =

+ =

Constructive and Destructive Interference Definitions

• Constructive Interference:– Interference in which individual displacements on the

SAME SIDE of the equilibrium position are added together to form the resultant wave.

• Destructive Interference: – Interference in which individual displacements on

OPPOSITE sides of the equilibrium position are added together to form the resultant wave.

Wave Superposition-- Demo

Demo1:

(1) Using a long coiled spring, generate a transverse pulse wave(s).

a. First, from one end while other end fixed.

b. Then from both ends of the spring simultaneously and in the same direction.

Observe that the amplitudes of traveling waves add as the waves pass one another.

Wave Superposition– Activities

Using a long coiled spring, generate transverse pulse wave(s) from each end of the spring simultaneously. Observe the pulse that reaches your hand after the pulses have passed through one another.

Experiment with the following variables:a. Displacements in opposite directions; same directionsb. Pulses of different amplitudesc. Combinations of a. and b.

What did you observe? Which examples were constructive and which were destructive? What can you conclude?

Observe that the pulses that pass through from one side to the other are unaffected by the presence of the other pulse!!

f1

f2

Beat Frequency equals:

fbeat = f1 – f2

Standing Waves

Standing Waves:

Resultant wave created by the interference of two waves traveling at the same frequency, amplitude and wavelength in opposite directions.

Standing Waves have Nodes and Antinodes

Nodes and Antinodes

• Nodes: – Points in the standing wave where the two

waves cancel – complete destructive interference– creating a stationary point!

• Antinodes– Point in the standing wave, halfway between

the nodes, at which the largest amplitude occurs.

Standing Waves

Only certain frequencies of vibration produce standing waves for a given string length!!! ….More later when we get to SOUND…

The wavelength of each of the standing waves depends on the string length, L

n = 2L/n

Wavelength,

1 = 2L

2 = L

3 = 2L/3

4 = 2L/4 or ½ L

Standing Waves on a Vibrating String

n = 2L/n

Wavelength,

1 = 2L

2 = L

3 = 2L/3

4 = 2L/4 or ½ L

Frequency, f

f1 = v / 1

f2 = 2 f1

f3 = 3 f1

f4 = 4f1

Fundamental Frequency or 1st Harmonic

2nd Harmonic

3rd Harmonic

4th Harmonic

fn = n v/2L n = 1, 2, 3, …

A

A A

A A A

A A A A

N

NN

N

N

N

N

N N

N

N

NNN

• Fundamental Frequency:– The lowest frequency of vibration of a standing wave:

f1 = v / 1 = v / 2L

Where,

v is the speed of waves on the vibrating string (NOT the speed of the resultant waves in air!!!!)

L is the portion of the string that is vibrating

Standing Waves on a Vibrating String

Harmonic Series of Standing Waves on a Vibrating String

A series of frequencies that includes the fundamental frequency and integral multiples of the fundamental frequency.

fn = n v / 2L, n = 1,2,3,….

Frequency = harmonic number x (speed of wave on the string) / (2 x length of the vibrating string)

Standing Waves in an Air ColumnOPEN at BOTH ENDS

n = 2L/n

Wavelength,

1 = 2L

2 = L

3 = 2L/3

Frequency, f

f1 = v / 1

f2 = 2 f1

f3 = 3 f1


2nd Harmonic

3rd Harmonic

fn = n v/2L n = 1, 2, 3, …

A ANNN

A A AN

N N

N

A A A AN

NNNN

Example: Organ Pipes; Flute

Standing Waves in an Air ColumnCLOSED at ONE END

n = 4L/n

Frequency, f

f1 = v / 1

f3 = 3 f1

f5 = 5 f1


3rd Harmonic

5th Harmonic

fn = n v/4L n = 1, 3, 5,…

NA A

N

A A A

N NN

N

A A A A

NNN

Example*: Clarinet, Saxophone, Trumpet

Wavelength,

1 = 4L

3 = 4L/ 3

5 = 4L/5

Waves Types• Pulse Waves – A Single non-periodic

disturbance• Periodic Waves -- A wave whose source

is a form of periodic motion

• Transverse Waves– A wave whose particles vibrate perpendicular to the direction of wave motion.

• Longitudinal Waves – A wave whose particles vibrate parallel to the direction of the wave motion

Wave Motion

– Mechanical Waves– Propagation requires a medium

• Examples: Sound waves; ripples in water, etc

– Electromagnetic Waves – Propagation does NOT require a medium; can travel in a vacuum

LIGHT

Characteristics of “Light”– Electromagnetic Wave:

• A TRANSVERSE wave • Consisting of alternating electric and magnetic fields

at right angles to each other,• Travels through a vacuum • At the speed of light, c (3 x 108 m/s)

– Wave-Particle Duality (more later!)– Light can also be described as a “Particle”

See Holt T63

Visible Light

• Visible Light:– Small Part of EM Spectrum

– Wavelengths:

700 nm (red) > > 400 nm (violet)

– Frequencies:

4.3 x 1014 Hz < f < 7.5 x 1014 Hz

Electromagnetic Spectrum

• All EM radiation travel at the speed of light in a vacuum…– but their wavelengths and frequencies will

vary!

• Wave Speed Equation:

c = * f

Speed of light = wavelength x frequency

Speed of Light

Interactions of EM Radiation with Matter

• Radiation interacts with matter in 3 principal ways:

– Scattered ….from the material’s surface– Absorbed ….by the material– Transmitted ….through the material, often

changing direction in the process.

Polarization of Light

• Unpolarized light: Randomly oscillating charges (electric and magnetic fields)

• Linear Polarization:The alignment of the electromagnetic waves in such a way that the vibrations of the electric fields in each of the waves are parallel to each other.

For example, certain processes can separate waves with electric field oscillations in the vertical direction from those in the horizontal direction.

• Light can be linearly polarized through:– Transmission, and/or– Reflection and Scattering

See Holt T 70, 71

Polarization of Light via Transmission

Direction of Wave

The transmission axis of the substance is parallel to the plane of polarization of the light–

Light passes through freely and “brightly”!

Transmission Axis

Polarization of Light via Transmission

Direction of Wave

The transmission axis of the substance is perpendicular to the plane of polarization of the light–

NO Light passes through

As the angle between the plane of polarization for the light and the transmission axis of the substance increases from 0 to 90 degrees, amount of light passing through decreases from 100% to 0%,

Transmission Axis

Polarization of Light via Reflection

• When light is reflected a certain angle from a surface, the reflected light is completely polarized parallel to the reflecting surface.

• For example, if the reflecting surface is parallel to the ground, then the light is polarized horizontally.– Eg, roadways, car hoods, bodies of water– Sunglasses application…

• filter out horizontally polarized “glare” with a “vertical” polarizer.

Polarization of Light by Scattering

• Scattering of light (the absorption and re-radiation of light) by particles in the atmosphere can also cause polarization.

• Example: SUNLIGHT:– When unpolarized beam of sunlight strikes air

molecules in the atmosphere, the electrons in the molecules begin to vibrate in the same plane as the electric field of the incoming wave.

– The re-radiated light is polarized in the direction of the electron oscillations.

Physics of Color• All kinds of interactions of light with matter

(scattering, absorption, and transmission) depend on the wavelength of the EM radiation.

• Rules governing the scattering of EM Waves:

1. If the object causing the scattering is much smaller than the wavelength of radiation, then shorter wavelengths are scattered much more strongly than longer ones.

2. If the object causing the scattering is much larger than the wavelength of incoming radiation, then all wavelengths are scattered equally.

Why is the Sky Blue??

• Sunlight is scattered by air molecules in the atmosphere.

• Since the size of molecules (tenths of nanometers) is much less than the wavelength of visible light (hundreds of nms), we expect short wavelengths (blue light) to be much more scattered than longer ones (red light).

• The sky appears blue!

Why are Clouds White?

• When sunlight strikes the clouds, it scatters from droplets of water. These water droplets vary in size, but are typically much larger than the wavelength of visible light.

• All colors are scattered equally, and the clouds appear white.

Why are Sunsets Red?

• Light from the Sun is white…containing all the colors of the visible spectrum.

• As the light travels through the atmosphere, blue light is scattered…making the sky appear blue.

• In the evening, as the sun sets over the horizon, it has to travel a longer distance through the atmosphere, so more scattering occurs. Once the blue light has been removed, yellow and green follow, leaving red.

• This gradual filtering explains the appearance of the sun at Sunset.

Waves as “Rays”

• Simplified representation of the light wave:– Visualize the direction the wave is moving.– The line that traces the motion of the wave is

called the light “RAY”.

“LIGHT RAY” = Direction of Wave

Magnetic Field Oscillations

Electric Field Oscillations

Scattering – Diffuse Reflection

• Diffuse Scattering (or Diffuse Reflection):– Light is reflected from a “rough”, textured surface in all

directions.

– “Rough” must be defined relative to the incidence EM wavelength -- Short wavelengths (eg visible light) require smoother surfaces than long wavelengths (eg radio waves)

• Examples/Applications:– Reading the pages of a book

• Can be read from any angle– Heating in a microwave oven

• Even heating throughout oven

Scattering – (Specular) Reflection

• Specular Reflection: – A beam parallel light rays encountering a smooth, mirrored

surface are scattered from that surface in one direction only, leaving the surface as parallel light rays.

The angle of incidence = the angle of reflection = ’

– Applies to other forms of EM radiation, but the “smoothness” of the surface is dependent upon wavelength of radiation.

– Radio waves: Smooth = wire mesh surface– Visible Light: Smooth = mirrored surface

’Incident Ray

Reflected Ray

Reflection with Flat Mirrors

• Flat Mirrors:– A Virtual image is formed by rays that appear to

intersect at the image point behind the mirror.– The virtual image appears as the same height, h’, as

the real image, h. – The distance from the virtual image to the mirror (q) is

the same as that of the real image to the mirror (p), except that it appears behind the mirror.

p

h h’

OBJECTVIRTUAL IMAGEq

Flat Mirrors – Image Location using Rays

• Pick a point on the object and draw two incident rays to the mirror surface and their reflected rays:– 1st ray perpendicular to mirror surface – 2nd ray at an angle from the perpendicular to the mirror surface.

• Trace both rays back to the point from which they appear to have originated behind the mirror (apparent rays designated with dotted lines)

• The point at which the apparent rays intersect is the location of the virtual image point.

p

h h’

OBJECT VIRTUAL IMAGE

q

Curved Mirrors

• Convex Mirrors:– Bowed outward– Images appear smaller than the

object– Ex. Used in Stores for

surveillance, etc.

• Concave Mirrors:– “Caved” inward– Images appear larger than the

object– Ex. Used in telescopes, satellite

dishes, etc.

Concave Parabolic Mirrors• A Concave parabolic mirror focuses incoming parallel

rays at a focal point.

• The distance between the focal point and the mirror is the focal length.

Focusing the reflected rays at the focal point concentrates the signal, making a weak signal much

stronger to detect!!

Focal Point

Mirror Equation

Focal Length: -Distance from focal point to mirror; For Concave Mirrors: Positive Value (focal point in front of mirror)For Convex Mirrors: Negative Value (focal point behind mirror)

Object distance, do: Distance from mirror to object (Pos. number)

Image distance, di: -Distance from mirror to image“Real”: -Able to be projected on a screen (in front of

mirror); inverted“Virtual”: -Right-side up behind mirror

1 11

Focal Length Object Distance

Image Distance

+=

1 11

f do di

+=

Example 1: Reflection Plane Mirror

Example 1: Plane Mirror

Sitting in her parlor one night, Gerty sees the reflection of her cat Whiskers, in the living room window. If the image of Whiskers makes an angle of 40 deg with the normal, at what angle does Gerty see him reflected?

SOLUTION:

Angle of incidence = angle of reflection

So Gerty must see Whiskers reflected at a 40 deg angle

EXAMPLE: Convex (Diverging) Mirror

Wendy the Witch is polishing her crystal ball. It is so shiny that she can see her reflection when she gazes into the ball from a distance of 15 cm.

a. What is the focal length of Wendy’s crystal ball if she can see her reflection 4.0 cm behind the surface?

b. Is the image real or virtual?

Example 2: Reflection, Convex Mirror

EXAMPLE: Convex (Diverging) Mirror

Solution: Find f

Given: do = 15 cmdi = -4 cm (behind the mirror)

Equation: 1/f = 1/ do + 1/di

1/f = 1/15 + 1/(-4)1/f = 4/60 - 15/60 = -11/60

f = -60/11 = -5.5 cm

Minus sign indicates that this is a focal length of a convex (diverging) mirror.

It is a virtual image behind the mirror.

Example 2: Reflection

Example 3: Concave Converging Mirror

With his face 6.0 cm from his empty water bowl, Spot sees his refection 12 cm behind the bowl and jumps back.

a. What is the focal length of the bowl?

b. What was surprising about Spot’s reflection that may have caused him to jump?

Solution: Find fGiven: do = 6.0 cm

di = -12 cm (behind the mirror!)

Equation: 1/f = 1/d0 + 1/di

1/f = 1/6 + 1/(-12)1/f = 2/12 - 1/121/f = 1/12

f = 12 cm

Since the object’s distance is closer to the mirror than 1 focal length, the image is enlarged!!!...frightening Spot!

Example 3: Concave Converging Mirror

Concave Mirrors and Focal Length

A. do < f:Object EnlargedObject UprightVirtual Image (Behind

mirror)

B. do > fObject ReducedObject InvertedReal Image

(In front of mirror)

Scenario A.

do < f

Scenario B.

do > f

What happens when the object is placed ON the focal point??

Refraction

• Refraction:– When light is transmitted through a material

substance, its path and speed may change significantly, causing…

The bending of light waves as it passes at an angle from one medium to another.

– Refraction occurs when the velocity of the light changes.

Sidewalk

Grass

When the barrel rolls onto grass from sidewalk, the grass slows it down, causing the barrel to turn.

Air

H2O

Index of Refraction• Index of Refraction, n:

– The ratio of the speed of light in a vacuum to the speed of light in a particular substance.

n = c / v

Index of refraction = (speed of light) / (speed of light in a particular medium)

– The index of refraction for light in air is nearly that in a vacuum, so we approximate it as n = 1.00.

– As n increases, more bending from the “normal” occurs.

See Holt reference pg 564

Refraction: Snell’s Law

The angle to which the light will bend on going from one medium to the next depends on:

– The index of refraction for each medium (n), and– The light’s angle of incidence

n1 sin 1 = n2 sin 2

Where, 1 is the angle of incidence and

2 is the angle of refraction

n1 and n2 are the indices of refraction for medium 1 and medium 2 , respectively

Snell’s Law: Special Case-- Critical Angle!

Special Case for Refraction:

1 is a critical angle whereby 2 (the refracted beam) is 90 degrees (from the normal)

1

2

Glass

Air

c

2

Glass

Air

What happens when 1 exceeds c ???

NORMAL REFRACTION SPECIAL CASE

Total Internal Reflection

• When the angle of incidence exceeds the critical angle, Total Internal Reflection occurs…100%!!!...

• Applications: Optical Fibers• Machinists, physicians-- to view hard-to-reach areas!• Communications – replacing electric circuits and microwave links

– More information can be carried in high frequencies of visible light than lower frequencies of electric current.

• Critical Angles:– Glass ~ 43 degrees– Diamond = 24.6 degrees … smallest of all known substances.

1 > c 2 = 1 Glass

Air

Refraction-- Example 1

While fishing out on a lake one summer afternoon, Amy spots a large trout just below the surface of the water at an angle of 60.0 deg to the vertical, and she tries to scoop it out of the water with her net.

a. Draw the fish where Amy sees it.

b. At what angle should Amy aim for the fish nwater = 1.33

Solution: Solve for 1

Given: n1 = 1.33 (water)

n2 = 1.00 (air)

2 = 60 deg

The fish will appear to be straight ahead, but since light travel slower in water than air, it is actually closer than she thinks.

Sin 1 = (n2sin2)/n1 = (1.00)(sin 60)/1.33 = 0.651

1 = Sin-1(0.651) = 40.6

Refraction-- Example 1

2

1

Example 2: Refraction

Binoculars contain prisms inside that reflect light entering at an angle larger than the critical angle. If the index of refraction of a glass prism is 1.58, what is the critical angle for light entering the prism?

Given: n1 = 1.58 (glass)

n2 = 1.00 (air)

2 = 90 deg

Equation: n1 sin c = n2 sin 90

Solution: sin c = (1.00)(sin 90)/1.58 = 0.633

c = sin-1 0.633 = 39.3 deg

Refraction: Lenses

Lens: • Transparent (translucent) object that refracts light rays,

causing them to converge or diverge to create and image.

• Images can be real or virtual..

Applications:• Optical instruments

– Cameras, – Telescopes– Microscopes– Magnifying glass– Binoculars

• Human Eyeball – Lens converges light on retina

Types of Lenses

Converging Diverging

Principal AxisFF

ff

F

ff

Object at infinite distance appears as parallel lines entering the lens.

These parallel lines (also parallel to the principal axis) will pass through the focal point --- thereby identifying the focal length!

**Focal Length is the image distance for an object at infinite distance.**

Ray Diagrams

Rules for Drawing Reference Rays

Ray From Object to Lens From Converging Lens to Image

Parallel Ray Parallel to Principal Axis Passes thru the focal point, F

Central Ray To the center of the lens From the center of the lens

Focal Ray Passes thru the focal point, F Parallel to Principal Axis

Ray From Object to Lens From Diverging Lens to Image

Parallel Ray Parallel to Principal Axis Directed away from the focal point, F

Central Ray To the center of the lens From the center of the lens

Focal Ray Proceeding toward back focal point , F

Parallel to Principal Axis

Characteristics of Lenses• Can produce real or virtual images• See Handout Holt, pg 571 (or Hewitt Ch 30, pg 469)

Some Conditions for Converging Lenses

Object Position Image Properties

Select Examples: Technology/Application

Infinity Point at F Burning a hole / magnifying glass

Object beyond 2F Real, Smaller Lens of a camera, human eye

Object at 2F Real, same size Inverting lens - Field telescope

Object between F and 2F

Real Magnified Slide Projector, Compound microscope (objective lens)

Object a F At Infinity Lighthouse, search lights

Object inside F Magnified, Virtual (same side as object)

Magnifying glass, eyepiece lens of binoculars, telescope, and microscope

Example: Magnifying Glass

Conditions:– Converging Lens– Object closer to lens than focal point, F

FF

ff

Focal Ray

Parallel Ray

Central Ray

RESULT: Image is magnified and remains on the same side of the object (VIRTUAL Image!!!)

Magnification of a Lens

Magnification = image height / object height = (distance from image to lens) (distance from object to lens)

M = hi / ho = - di / do

+ -dO (or p) Object in front of lens Object in back of lens

di (or q) Image in back of lens (real)

Image in front of lens (virtual)

f Converging lens Diverging lens

Sign Conventions

-M = Real and inverted +M = Upright and virtual

Thin Lens Equation

Identical to the Mirror Equation!!!!

Thin Lens Equation:

1/(focal length) = 1/(object distance) + 1/(image distance)

1/f = 1/do + 1/di

Assumption: • Lens is “very thin”…thickness is much less than focal

length. • This allows one to measure the focal length from the

center of the lens or the surface of the lens.

Example: Human Eyeball

Conditions: – Converging Lens– Object outside 2F– Generates real, smaller, inverted image

FF

ff

Focal Ray

Parallel Ray

Central Ray2F

Example: Human Eyeball

Conditions: – Converging Lens– Object outside 2F

RESULT: Generates a Real, smaller, inverted image between F and 2F

FF

ff

Focal Ray

Parallel Ray

Central Ray

2F2F

Human Eyeball Lens

• Focusing difference between Camera and Eyeball:– Camera: Alters distance between lens and film – Eyeball Lens Changes Shape and Thickness to

focus light on retina through the action of the ciliary muscles.

– Process is called Accommodation

Normal Distance Vision Normal Close Vision

Vision DefectsFarsightedness

• Can see clearly at a distance• Image is formed behind retina

—eyeball to short• Remedy: Converging Lens

Nearsightness• Can see clearly close up• Image is formed in front of

retina – eyeball too long• Remedy: Diverging Lens

Dispersion

• The average speed of light is less than that of c in a transparent medium.

• The magnitude of this speed is dictated by the medium and the frequency of the incoming wave.

• Frequencies closer to the natural frequency of the electron oscillators in the medium travel more slowly through the medium due to more interactions with the medium

• The natural (resonant) frequency of most transparent materials is in the UV part of the EM spectrum….Thus visible light of higher frequencies travel slower than those of lower frequencies….causing….

Visible Light to split into the Colors of the Rainbow!!

Diffraction and Interference

• Diffraction is another process whereby light is bent…– distinct from refraction and reflection.

• Diffraction involves the bending of light as it:– Passes through a small slit/opening– Passes around an object– Passes by sharp edges

Terminology

• Monochromatic – – Light waves composed of a single wavelength– Waves do not have to maintain a constant phase

relationship

• Coherent—– Light waves of a identical wavelengths– Light waves maintain a constant phase relationship

• Interference– – superposition of light waves constructively or

destructively

Interference Examples

• Soap Bubble• Thin layer of Gasoline (or oil, transmission

fluid, etc) on paved surface• CD’s

In all cases above, light waves interfere to form bands of color.

This interference pattern depends on the difference in distance traveled between the interfering waves (ie, path difference)

Path Difference Conditions for Interference of Light Waves

• See Holt Transparencies #80-85

d

l

sin = opposite / hypotenuse

sin =l / d

d sin = l

d

l

The path length

difference, l, between the two waves equals a

whole-number multiple of the two waves’ wavelengths.

Constructive Interference

• Equation for Constructive Interference:

d*sin = m where, m= 0, +/-1, +/-2, …

The path length difference between the two waves equals a whole-number multiple of the

two waves’ wavelengths.


• Equation for Destructive Interference:

d*sin = (m + ½) where, m= 0, +/-1, +/-2,

…

The path length difference between the two waves equals an odd number of half

wavelengths


Examples: Diffraction Grating

• Monochromatic light shines at the surface of a diffraction grating with 5.0 x 103 lines/cm. The first order maximum is observed at a 15 deg angle. Find the wavelength?

Given: d = 1/(5 x 103 lines/cm) = 1/(5 x 105 line/m) = 15 degm = 1

Solution: d sin = m1/(5 x 105 line/m)) * sin (15) =

520 nm = 5.2 x 10-7 m =

Monochromatic light from a He-Ne laser ( = 632.8 nm) shines at a right angle to the surface of a diffraction grating that contains 150,500 line/m. Find the angles at which one would observe the 1st and 2nd order maxima.

Given: = 632.8 nm = 6.328 x10-7

d = 1/(150,500 line/m)1, 2 ???

Solution: d sin = m

sin = 1(150500)(6.328 x10-7 ) sin= 9.524 x 10-2

1 = 5.465 degrees

So, 2 = 10.98 degrees

Examples: Diffraction Grating

Application…X-Ray Crystallography

• Diffraction of x-rays by the “crystal structure”

of a compound…– The atoms of the

molecules in the crystalline lattice act as a diffraction grating.

– Result: Identification of molecular structures!!!

Application/Devices: Spectrometers

• Spectrometers: Separate light from a source into its monochromatic components– Light passes through a grating, – Diffracted beams are collected at various

angles– Wavelengths of light calculated– Chemical Composition of Light Source

Identified!!!! …………………….HOW??

From Classical Physics

to

Atomic Physics or Quantum Mechanics

Some Encouraging Quotes…

The very attempt to conjure up a picture of elementary particles and think of them in visual terms is wholly to misinterpret them... Atoms are no things. The electrons, which form an atom’s shell, are no longer things in the sense of classical physics, things which could be unambiguously described by concepts like location, velocity, size. When we get down to the atomic level, the objective world in space and time no longer exists. Werner Heisenberg, Physics and beyond (1971).


• All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics invented more than fifty years ago. It has survived all tests. We suppose that it is exactly correct. Nobody understands it but we all know to use it and to apply it to all problems: thus we have learned to live with the fact nobody can understand it. Murray Gell-Mann.


• Physics is not about the real world, it is about 'abstractions' from the real world, and this is what makes them so scientific. Anthony Standen, Science is a Sacred Cow (1958).


• Nothing is more curious than the self-satisfied dogmatism with which mankind at each period of its history cherishes the delusion of the finality of its existing modes of knowledge… At this moment scientists and skeptics are the leading dogmatists. Advance in detail is admitted; fundamental novelty is barred. This dogmatic common sense is the death of philosophic adventure. A.N. Whitehead, Dialogues, recorded by L. Price (1956).

ENERGY is QUANTIZED!!!

Max Planck – Classical Physics cannot adequately explain…

Electromagnetic Radiation and Thermodynamics

Blackbody Radiation: electromagnetic radiation emitted by a blackbody, which absorbs all incoming radiation and then emits radiation based only on temperature.

• At low temperatures, radiation is in IR region.• As temperature increases, radiation shifts the visible

region (higher energy). • However, Planck discovered that energy is QUANTIZED!!!

Energy is absorbed or emitted in discrete packets of light energy called quanta (photons) by jumping from one

energy level (quantum state) to another adjacent level.

• Energy of a Photon:

E = n*h*f or E = n*h*c /

Where, h = Planck’s constant: 6.626 x 10-34 J secf = frequencyc = speed of light = wavelengthn = 1, 2, 3, ….

(n = 1 for a single quantum of light)Note: 1eV = 1.60 x 10-19 J

Planck’s Quantum Theory

WavelengthIn

ten

sity

E-Mag Radiation at ~4000K

Classical

Planck’s Theory

Quantum Energy --Exercises

Example 1: How much energy in Joules is carried by 1000 photons of light of the following frequencies:(a) 3.0 x 1014 Hz (infrared)(b) 5.0 x 1014 Hz (orange light)(c) 6.0 x 1014 Hz (UV)

Solution: E = nhf

(a) = (1000)(6.63 x 10-34 Js)(3.0 x 1014 Hz) = 2.0 x 10-16J(b) = (1000)(6.63 x 10-34 Js)(5.0 x 1014 Hz) = 3.3 x 10-16J(c) = (1000)(6.63 x 10-34 Js)(6.0 x 1014 Hz) = 4.0 x 10-16J

• Example 2: A photon has 2 eV of energy. What are its frequency and wavelength?

• Given: E = hf or hc/c = 3 x 108 m/sh = 6.63 x 10-34 Js1eV = 1.60 x 10-19 J; 2eV = 3.2 x 10-19 J

• Solution:(a) 3.2 x 10-19 J = (6.63 x 10-34 Js) * f

(3.2 x 10-19 J) / (6.63 x 10-34 Js) = f = 4.8 x 1014 Hz

(b) = c/ f= (3.0 x 108 m/s)/4.8 x 1014 Hz = 6.2 x 10-7 m = 620 nm


Example 3. A quantum of a certain color of visible light is found to have an energy of 5 x 10-19 J. What is the wavelength and color of this light?

Given: E = hc / E = 5 x 10-19 Jc = 3 x 108 m/sh = 6.63 x 10-34Js

Solution: = hc/E = (6.63 x 10-34Js)(3 x 108 m/s) / (5 x 10-19 J)= 3.98 x 10-7 m = 398 nm ….VIOLET!


Energy Absorbed --Electron Jumps to

next Energy Level

EXCITED STATEPositively

Charged Protons; Neutrons

K Shell

L Shell

M Shell

XX

XX XX

XX

X X

X

N Shell Energy In

X

X

Energy Out

Energy Released --Electron

Drops down to

lower Energy Level

GROUND STATE

Bohr Model of the Hydrogen Atom

• Electron Moves in circular orbits around nucleus

• Electric force between the positively charged protons in the nucleus and the negatively charged electrons holds the electron in orbit.

• Only certain orbits are stable…– Electrons “never” found in between these orbits– Electrons jump between orbits

• Energy radiated out when jumping from an outer orbit to an inner one

• The frequency of the radiation is related to the change in the atoms energy

E = Efinal – Einitial = hf

• Transitions between stable orbits (r1, r2, etc) with different energy levels (E1, E2, etc.) account for discrete spectral lines.

• Transitions between any two levels are allowed-- resulting in emission or absorption spectra.

– Example: If an atom has 4 possible energy levels, how many different spectral lines could be emitted?

E4-E1, E3-E1, E2-E1, E4-E3, E4-E2, E3-E2

• Transitions that result in emissions in the visible range are called the Lyman Series.

Bohr Model of the Hydrogen Atom

4

3

2

1

Positively Charged Protons; Neutrons

K Shell

L Shell

M Shell

N Shell

X

Photons of Energy Absorbed

Photons of Energy Emitted

N

M

L

K

Absorption and Emission Spectra

Bohr Model of Hydrogen Atom

Note: Transitions to E1 (K) are in

the UV region Sample Emission spectral lines

Emission of electrons from a surface that occurs when light of certain frequencies shines on that surface.

Emitted Electrons: PhotoelectronsResponsive Surfaces: Photosensitive Surfaces

The maximum kinetic energy that the emitted electron can have equals the incoming photon energy minus the energy required to remove the electron from the metal (overcoming the force that binds it to the metal, “work function”).

KEmax = hf - hf t or KEmax = hf - W

f = frequency of incoming photon

ft = threshold frequency specific to the metalh = Planck’s constant

Photoelectric Effect

Example 1: Photoelectric Effect

Example 1: A sodium surface is illuminated with light with a frequency of 1 x 1015 Hz. The work function of sodium is 2.28 eV. Find the maximum kinetic energy of the photoelectrons in electron volts.

Given: f = 1 x 1015 Hz hft = 2.28 eV = Wh = 6.63 x 10-34 Js (1eV = 1.6 x 10-19 J)Find KEmax ???

Solution: KEmax = hf – W

KEmax = [(6.63 x 10-34 Js)(1 x 1015 Hz) / (1.6 x 10-19 J/eV)] – 2.28

eV= 4.14 eV – 2.28 eV

KEmax = 1.86 eV


Example 2: Which of the following metals will exhibit the photoelectric effect when light of 7.0 x 1014 Hz frequency is shined on it?

(a) Lithium, W = 2.3 eV(b) Silver, W = 4.7 eV(c) Cesium, W = 2.14 eV

Given: h = 6.63 x 10-34J*s (or 4.14 x 10-15 eV*s)f = 7.0 x 10 14 Hz (428 nm, Violet!)

KEmax = hf – W

Example 2: Photoelectric Effect…Solution

KEmax = (4.14 x 10-15 eVs)(7 x 1014Hz) – 2.3 eV= 2.89 eV – 2.3 eV = 0.60 eV for Li YES

KEmax = (4.14 x 10-15 eVs)(7 x 1014Hz) – 4.7 eV= 2.89 – 4.7 eV = -1.8 eV for Ag NO

KEmax = (4.14 x 10-15 eVs)(7 x 1014Hz) – 4.7 eV= 2.89 – 2.14 eV = 0.75 eV for Cs YES

Li and Cs will exhibit the photoelectric effect with this frequency of light, but Ag will not…The input energy is insufficient to overcome silver’s binding energy of its

electrons.


Example 3: Light of wavelength 350 nm (UV) falls on a potassium surface, and the photoelectrons have a maximum KE of 1.3 eV. What is the work function and the threshold frequency for potassium?

Given: h = 6.63 x 10-34J*s (or 4.14 x 10-15 eV*s) = 3.5 x 10-7 mKEmax = 1.3 eV

KEmax = hf – W, where W = hft

Example 3: Photoelectric Effect…Solution

Solution: 1.3 eV = [(4.14 x 10-15 eV*s)(3.0 x 108 m/s)/(3.5 x 10-7 m)] -

W1.3 eV = 3.54 – WW = 3.54 – 1.3 eV = 2.24 eV Work function

W = hft

2.24 eV = (4.14 x 10-15 eV*s)*ft

2.24 ev / (4.14 x 10-15 eV*s) = ft

5.41 x 1014 Hz = ft , the threshold frequency(554 nm = Green light!)

Arthur H. Compton (1892-1962)… “The Compton Shift”

Further Evidence for Quantization of Light– (i.e., Light behaves as Particles (Photons))

Theorized–

If light behaved like a particle, then a collision between an electron and a photon should be similar to the collision between two billiard balls!

• Photons should have momentum and energy• Momentum and Energy are both conserved in collisions

The Compton Shift

Ex: If a photon collides with an electron at rest, then the photon should transfer some of its energy to the electron, leaving the scattered photon with lower energy and lower frequency, yet longer wavelength.

Compton Shift: The change in wavelength between incoming and scattered electromagnetic waves.

= (h / mec )(1-cos)

Compton Wavelength = h / mec where me = 9.109 x 10-31kg

Note: The wavelength change is very small and difficult to detect in visible region; observed with shorter wavelengths (x-rays).

e- e-

Stationary electron

p = 8 x 10-22 kg m/s p=0 p = ?? p = 2.7 x 10-22

kg m/s


Cannot be explained by Classical Physics!!!

Classical Predictions

Experimental Evidence

Whether electrons are emitted depends upon…

The intensity of the light

The frequency of the light

The kinetic energy of ejected electrons depends upon…

The intensity of the light

The frequency of the light

At low intensity, electron ejection…

Takes time Occurs almost instantaneously above a certain frequency

Connections…

• Glowing Objects:– As a hot object glows, the color of its glow

depends on the object’s temperature. As the temperature increases, the color turns from red to orange to yellow to blue to white.

Classical physics cannot explain. What explanation would be given by quantum mechanics??

Connections…

• Photoelectric effect:– Even though bright red light delivers more

total energy per second than dim violet light, the red light cannot eject electrons from a certain metal surface, while the dimmer violet light can.

How does Einstein’s photon theory explain this observation?

Wave-Particle Duality

Electromagnetic Radiation

WAVE

EVIDENCE

Interference

Refraction

Diffraction

EVIDENCE


Luminescence: Fluorescence, Phosphorescence

Compton Effect

Best Used for L-o-n g Wavelengths

(eg, Radiowaves)

Best Used for Short Wavelengths

(eg, Gamma Rays)

Wave Particle Duality

Consider L-O-N-G wavelengths...Can one Observe their Particle properties???

Ex: Radiowaves at 2.5 Mhz (2.5 x 106 Hz)

E = hf =(6.63 x 10-34 Js)(2.5 x 106 Hz)

= 1.7 x 10-27 JThis energy is too small to be detected as a single

photon!!!

• A sensitive radio receiver might receive 1010 photons per second to produce a detectable signal!!!

• With such a large number of photons reaching the detector per sec, one would not be able to detect a single photon

• …it would appear as a continuous wave.

Particles as Waves!!!

Louis DeBroglie:

“If waves can have particle properties, cannot particles have wave properties??....”

…question posed as a student…earned him a PhD in Physics…with

hesitation!!! (Einstein intervened!) …1st experimental confirmation – electrons show

interference patterns – electron diffraction …won the Nobel Prize in physics 5 yrs after PhD

“DeBroglie Wavelength”: The wavelength of a particle…!!!

All particles of matter – electrons, protons, atoms, marbles, and even YOU and I – have a wavelength that is related to the momentum of the particles by:

Wavelength = h / momentum = h / p = h / mv,

where h is Planck’s constantp = momentum, m = mass, v =velocity

As the particle’s velocity (or mass) increases, its wavelength decreases!!

Matter Waves…DeBroglie Wavelength

“DeBroglie Frequency”: The frequency of matter waves…!!!

DeBroglie Frequency = Energy / Planck’s Constant

f = E / h

Dual nature of matter represented:– Particle concepts (E and mv), and – Wave concepts ( and f).

Matter Waves…DeBroglie Frequency

Particles as Waves

Wavelength vs. Size (mass) of the Particle

1. Macroscopic World:For “larger mass” objects traveling at ordinary speeds:

Ex.: mass of ball bearing = 0.02 kgspeed = 330 m/s

Wavelength = (6.63 x 10-34Js)/((0.02 kg)(330 m/s)) = 10-34 m …

1024 times smaller than the diameter of an H atom!!!!!Wavelengths are below detection limits (too SMALL!)

Wave properties not “observable”!!

Particles as Waves

Wavelength vs. Size (mass) of the Particle

2. Microscopic World: e.g. ELECTONS

For “tiny particles” traveling at ordinary speeds:

Ex.: mass of an electron = 9.109 x 10-31 kgspeed = 0.02*c

Wavelength = (6.63 x 10-34Js) / ((9.109 x 10-31 kg)(0.02 * 3 x 108

m/s)) = 1.2 x 10-10 m …

Roughly equal to the diameter of an H atom!!!!!Wavelengths and wave properties are DECTECTABLE!!!

Wave - Particle Duality

Matter and Electromagnetic Radiation***Looking at Extremes for Particle and Wave Properties***

Waves Observable

Long Wavelengths (Radio waves)

Low Frequency

------------------

Low Energy

Small Mass

Particles Observable

Short Wavelengths (Gamma Rays)

High Frequency

------------------

High Energy

Large Mass

Wave Concepts

Particle Concepts

Application of DeBroglie Waves

• Electron Diffraction:3 years after his theory was proposed, interference patterns were experimentally confirmed using an electron beam!!!....

• (Scanning) Electron Microscopes 1000x shorter wavelengths than visible light, yielding much more detail….i.e., better resolution

• Refined the Bohr Planetary Model of the Atom – Early Wave Model….“Electron Clouds”

De Broglie’s Early Wave Model of the Atom…

Bohr’s Planetary Model– explained...– Atomic spectra

• Elements only emit/absorb certain frequencies of light • corresponds to electron transitions between discrete energy

levels

Question: Why must the electrons reside in set orbits or energy levels at discrete distances

(radii) from the nucleus?

DeBroglie’s Response…the electrons travel in orbit as waves rather than particles…

De Broglie’s Early Wave Model of the Atom…

OK!...6 whole waves yield Standing Wave!!

NO GOOD!....4.5 waves will not yield a Standing Wave (Out of Phase)

Wrap the wave around the circumference (length = 2r) until the ends meet… Standing Waves exhibit Constructive Interference for an

integral number of wavelengths at each energy level

Multiple Energy Levels

Review for TestWave-Particle Duality (of Electromagnetic Radiation)• Diffraction, Interference (Evidence for Light as “Waves”)

– Spectrometers– dsin = m(for constructive interference)– dsin = (m + ½)(for destructive interference)

• Quantization of Light (photons) – Light as “Particles”– E = n*hf; E = n*hc/(Recall, f = c/)– Bohr Model of Atom;

• Atomic Spectra (Fluorescence; Phosphorescence); • e.g., E2 – E1 = hf

– Photoelectric Effect• KEmax = hf – hft or hf – W

– Compton Shift = (h/mec)(1-cos)

Matter as Waves – – DeBroglie Wavelength and Frequency

= h/p or = h/mv- f = E/h

- DeBroglie’s Wave model of atom

Vibrations and Waves Simple Harmonic Motion Wave Interactions Mechanical Waves (eg, Sound)...

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