Chapter 22 Wave Optics

© 2013 Pearson Education, Inc.

Chapter 22 Wave Optics

Chapter Goal: To understand and apply the wave model of light.

Slide 22-2


Models of Light

The wave model: Under many circumstances, light exhibits the same behavior as sound or water waves. The study of light as a wave is called wave optics.

The ray model: The properties of prisms, mirrors, and lenses are best understood in terms of light rays. The ray model is the basis of ray optics. Chapter 23

The photon model: In the quantum world, light behaves like neither a wave nor a particle. Instead, light consists of photons that have both wave-like and particle-like properties. This is the quantum theory of light. Physics 43

Slide 22-29

Diffraction depends on SLIT WIDTH: the smaller the width,

relative to wavelength, the more bending and diffraction.

Diffraction depends on SLIT WIDTH: the smaller the width,

relative to wavelength, the more bending and diffraction.

Ray Optics assumes that λ<<d , where d is the diameter of the opening.

This approximation is good for the study of mirrors, lenses, prisms, etc.

Wave Optics assumes that λ~d , where d is the diameter of the opening.

d>1mm: This approximation is good for the study of interference.

Ray Optics: Ignores Diffraction

and Interference of waves!

Geometric RAY Optics (Ch 23)

i r 1 1 2 2sin sinn n

James Clerk Maxwell

1860s

Light is wave.

8

0

13.0 10 /

o

c x m s

Speed of Light in a vacuum:

186,000 miles per second

300,000 kilometers per second

3 x 10^8 m/s

Intensity of Light Waves

E = Emax cos (kx – ωt)

B = Bmax cos (kx – ωt)

max

max

E ω Ec

B k B

2 2

max max max maxav

2 2 2I

o o o

E B E c BS

μ μ c μ

I 2

m axE

The Electromagnetic Spectrum

Visible Light

• Different wavelengths

correspond to

different colors

• The range is from red

(λ ~ 7 x 10-7 m) to

violet (λ ~4 x 10-7 m)

Limits of Vision

112.4 10e x m

Electron

Waves

Interference of Light

Wave Optics

Double Slit is VERY IMPORTANT because it is evidence

of waves. Only waves interfere like this.

If light were made of

hard bullets, there

would be no

interference pattern.

In reality, light does show

an interference pattern.

Light acts like a

wave going through

the slits but arrive at

the detector like a

particle.

Photons

Particle Wave Duality

Photons & Electrons!

Interference pattern builds one

electron at a time.

Electrons act like

waves going through

the slits but arrive at

the detector like a

particle.

112.4 10e x m

Double Slit for Electrons

shows Wave Interference!

Key to Quantum Theory!

Electron Diffraction with Crystals

Electron Microscope

Electron microscope picture of a fly.

The resolving power of an optical lens depends on the wavelength of

the light used. An electron-microscope exploits the wave-like

properties of particles to reveal details that would be impossible to see

with visible light.

Intereference of 2-D

Coherent Light Waves

Loud Max

Quiet Min

Quiet Min

Loud Max

RECALL: Intereference of 2-D

Coherent Sound Waves

0

2Phase Difference at P: , 0r

Intereference of 2-D

Coherent Light Waves

Young’s Double Slit

• To observe interference in light waves, the following two conditions must be met:

1) The sources must be coherent

• They must maintain a constant phase with respect to each other

2) The sources should be monochromatic

• Monochromatic means they have a single wavelength

Fig. 37-3, p. 1086

2Phase Difference at P: r

Constructive : 2 , , 0,1,2,3...

1Destructive : (2 1) , ( ), 0,1,2,3...

2

m r m m

m r m m

Fig. 37-5, p. 1087


Constructive : 2 , , 0,1,2,3...

1Destructive : (2 1) , ( ), 0,1,2,3...

2

m r m m

m r m m

Fringe Equations

• Maxima: bright fringes

• Minima: dark fringes

sin

, ,

bright ( 0 1 2 )

d θ mλ

λLy m m

d

sin

, ,

dark

1

2

1( 0 1 2 )

2

d θ m λ

λLy m m

d

“m” is the fringe order.

Red light (=664nm) is used in Young’s double slit as

shown. Find the distance y on the screen between the

central bright fringe and the third order bright fringe.

Problem

bright ( 0 1 2 ), ,λL

y m md

Measuring the wavelength of light

A double-slit interference

pattern is observed on a

screen 1.0 m behind two

slits spaced 0.30 mm apart.

9 bright fringes span a

distance of 1.7cm. What is

the wavelength of light?

y L

dFringe Spacing.

bright ( 0 1 2 ), ,λL

y m md

Double Slit

The image shows the light intensity on a

screen behind a double slit. The slit

spacing is 0.20 mm and the wavelength of

light is 600 nm. What is the distance from

the slits to the screen?

bright ( 0 1 2 ), ,λL

y m md

Double Slit Interference

Dependence on Slit Separation

bright ( 0 1 2 ), ,λL

y m md

A laboratory experiment produces a double-slit interference pattern

on a screen. If green light is used, with everything else the same,

the bright fringes will be

QuickCheck 22.5

A. Closer together

B. In the same positions.

C. Farther apart.

D. There will be no fringes because the conditions for

interference won’t be satisfied.

Slide 22-44

bright ( 0 1 2 ), ,λL

y m md


on a screen. If green light is used, with everything else the same,

the bright fringes will be

QuickCheck 22.5

dy

Land green light has a shorter wavelength.

A. Closer together.


C. Farther apart.



Slide 22-45

bright ( 0 1 2 ), ,λL

y m md

A laboratory experiment produces a double-slit interference

pattern on a screen. If the slits are moved closer together, the

bright fringes will be

QuickCheck 22.6

A. Closer together.


C. Farther apart.



Slide 22-46

bright ( 0 1 2 ), ,λL

y m md

A laboratory experiment produces a double-slit interference

pattern on a screen. If the slits are moved closer together, the

bright fringes will be

QuickCheck 22.6

A. Closer together.


C. Farther apart.



y L

dand d is smaller.

Slide 22-47

bright ( 0 1 2 ), ,λL

y m md


on a screen. What is the path difference the light travels from the two

slits to the point on the screen marked with a dot?

A. 1.0 .

B. 1.5 .

C. 2.0 .

D. 2.5 .

E. 3.0 .

QuickCheck 22.3

Slide 22-35

bright ( 0 1 2 ), ,λL

y m md


on a screen. What is the path difference the light travels from the two

slits to the point on the screen marked with

a dot?

A. 1.0 .

B. 1.5 .

C. 2.0 .

D. 2.5 .

E. 3.0 .

QuickCheck 22.3

Slide 22-36

bright ( 0 1 2 ), ,λL

y m md

Start Using Exam 2 Equation

Sheet!

Fringe Equations

• Maxima: bright fringes

• Minima: dark fringes

sin

, ,

bright ( 0 1 2 )

d θ mλ

λLy m m

d

sin

, ,

dark

1

2

1( 0 1 2 )

2

d θ m λ

λLy m m

d

“m” is the fringe order.

Single Slit Interference Is called

Diffraction

Light interferes with itself

Single Slit Diffraction

Dark fringe: sin 1, 2, 3,...m ma

Single Slit Problem

A narrow slit is illuminated with sodium yellow light

of wavelength 589 nm. If the central maximum

extends to ±30, how wide is the slit?

Dark fringe: sin 1, 2, 3,...m ma

If the incident light has =690nm (red) and a = 4 x 10-6m,

find the width of the central bright fringe when the screen is 0.4m away.

Dark fringe: sin 0,1,2,3,...m ma

2w y

tanm my L

Width of the Central Maximum

Dark fringe: sin ma

Compare Fringe Equations for

Single and Double Slits

: sin , ,

sin , ,

maxima ( 0 1 2 )

1minima: ( 0 1 2 )

2

d θ mλ m

d θ m λ m

minima: sin , 1, 2, 3,...darka m m

Light Intensity: Ignoring Single Slit

Diffraction

• The interference

pattern consists of

equally spaced

fringes of equal

intensity – it is an

idealization

• This result is valid

only if L >> d and for

small values of θ and

is an approximation of

the inside fringe

pattern.

Reality: Combined EffectsInterference maximum coincides with the first diffraction minimum.

Light Intensity: Ignoring Diffraction


pattern consists of

equally spaced

fringes of equal

intensity

2

max cos ( / 2)I I


sin

, ,

bright ( 0 1 2 )

d θ mλ

λLy m m

d

The power transmitted through a closed surface by a wave is

proportional to the amplitude of the wave.

Recall: Sound Power

2

~Intensity Amplitude

RECALL: Superposition of

Sinusoidal Waves• Assume two waves are traveling in the same direction,

with the same frequency, wavelength and amplitude

• The waves differ in phase

• y1 = A sin (kx - wt)

• y2 = A sin (kx - wt + )

• y = y1+y2

= 2A cos (/2) sin (kx - wt + /2)

Resultant Amplitude Depends on phase:

Spatial Interference Term

Intensity Distribution Resultant Field

• The magnitude of the resultant electric field

comes from the superposition principle

– EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]

• This can also be expressed as

– EP has the same frequency as the light at the slits

– The amplitude at P is given by 2Eo cos (φ / 2)

• Intensity is proportional to the square of the

amplitude:

• The maximum intensity at P

is 4 times one source!!!!!!!!!

2 cos sin2 2

P o

φ φE E ωt

cos

2 2 242

P P o

φI E E

Amplitude is twice as big

But intensity is proportional to

the amplitude SQUARED so

the Intensity is four times as

big as the source. This is

energy being conserved! The

light energy is redistributed

on the screen.

2 cos sin2 2

P o

φ φE E ωt

cos

2 2 242

P P o

φI E E

Light Intensity: Ignoring Diffraction


pattern consists of

equally spaced

fringes of equal

intensity

2

max cos ( / 2)I I


sin

, ,

bright ( 0 1 2 )

d θ mλ

λLy m m

d

Intensity

In a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the screen is 150 cm. What wavelength (in nm) is needed to have the intensity at a point 1 mm from the central maximum on the screen be 80% of the maximum intensity?

WARNING! We are going to ignore the intensity

drop due to distance in inverse square law. We

assume that the amplitude remains constant over

the short distances considered. We will only

considering the intensity change due to

interference! This is not a bad approximation

using lasers as sources.

2 2

W

4 m

PI

r

Combined EffectsInterference maximum coincides with the first diffraction minimum.

Two-Slit Diffraction Patterns,

Maxima and Minima• To find which interference

maximum coincides with the first diffraction minimum

– The conditions for the first interference maximum

• d sin θ = mλ

– The conditions for the first diffraction minimum

• a sin θ = λ

sin

sin

d θ mλ dm

a θ λ a

Intensity of Two-Slit Diffraction

The Complete Equation

Section 38.2

2

2

max

sin sin sin cos

sin

/I I

/

πa θ λπd θ

λ πa θ λ

Hyperphysics

Multiple Slits: Diffraction Gratings

Section 37.3

For N slits,

the intensity

of the

primary

maxima is

N2 times

greater than

that due to

a single slit.

The Diffraction Grating

The figure shows a diffraction

grating in which N slits are

equally spaced a distance d

apart.

This is a top view of the

grating, as we look down on

the experiment, and the slits

extend above and below the

page.

Only 10 slits are shown here,

but a practical grating will

have hundreds or even

thousands of slits.Slide 22-53

Multiple Slits: Diffraction Gratings

Reflection Gratings

In practice, most diffraction

gratings are manufactured

as reflection gratings.

The interference pattern

is exactly the same as

the interference pattern

of light transmitted

through N parallel slits.

Slide 22-62


Bright fringes will

occur at angles m,

such that

d sinm = m

where m = 0, 1, 2, 3, …

The y-positions of these fringes are:

Slide 22-54

Circular Aperature

Dark fringe: sin 0,1,2,3,...m mD

Circular Diffraction Patterns

Fig 38-3, p.1207

Penny: Central Bright Spot: Poisson Spot

At the beginning of the 19th

century, the idea that light

does not simply propagate

along straight lines gained

traction. Thomas Young

published his double-slit

experiment in 1807. The

original Arago spot

experiment was carried out

a decade later and was the

deciding experiment on the

question of whether light is

a particle or a wave. It is

thus an example of an

experimentum crucis.

The images show simulated Arago spots in the shadow of a

disc of varying diameter (4 mm, 2 mm, 1 mm – left to

right) at a distance of 1 m from the disc. The point source

has a wavelength of 633 nm (e.g. He-Ne Laser) and is

located 1 m from the disc. The image width corresponds to

16 mm.

Airy DiskDue to diffraction, the

smallest point to which a lens

or mirror can focus a beam of

light is the size of the Airy

disk. Even if one were able

to make a perfect lens, there

is still a limit to the

resolution of an image

created by such a lens. An

optical system in which the

resolution is no longer

limited by imperfections in

the lenses but only by

diffraction is said to be

diffraction limited.

Fig 38-13, p.1215

Diffraction Resolution

Resolution

• The ability of optical systems to distinguish between closely spaced objects is limited because of the wave nature of light

• If two sources are far enough apart to keep their central maxima from overlapping, their images can be distinguished

– The images are said to be resolved

• If the two sources are close together, the two central maxima overlap and the images are not resolved

Resolving Power

min

1.22

Rayleigh Criterion

D

The diffraction limit: two images are just resolvable when the center of the diffraction

pattern of one is directly over the first minimum of the diffraction pattern of the other.

D: Aperture Diameter

Space Shuttle Resolution

What, approximately is the distance between objects on Earth that the astronauts can resolve by eye at 200 km height from the space shuttle? Assume l = 500 nm light and a pupil diameter D = 0.50 cm. Ignore the index of refraction from vitreous eye fluid.

min

1.22

Rayleigh Criterion

D

http://www.smithsonianmag.com/science-

nature/prehistoric-pointillism-long-seurat-ancient-

artists-chiseled-mammoths-out-dots-180962306/

http://www.smithsonianmag.com/science-nature/prehistoric-pointillism-long-seurat-ancient-artists-chiseled-mammoths-out-dots-180962306/

http://www.smithsonianmag.com/science-nature/prehistoric-pointillism-long-seurat-ancient-artists-chiseled-mammoths-out-dots-180962306/

Standing back from a Georges Seurat painting, you can cannot resolve the dots but a camera,

at the same distance can. Assume that light enters your eyes through pupils that have

diameters of 2.5 mm and enters the camera through an aperture with diameter of 25 mm.

Assume the dots in the painting are separated by 1.5 mm and that the wavelength of the light

is 550 nm in vacuum. Find the distance at which the dots can just be resolved by a) the

camera b) the eye.

Georges Seurat’s 1884 pointillist

masterpiece

A Sunday on La Grand Jatte

Thin Film

Interference

http://hometown.aol.com/_ht_a/surfacant/spzoom3.html


RGB Color Theory

What is color? What is light?

“Light” is what we see.

An optical Tuning Fork

I am Watching YOU!!

Human Retina

Sharp Spot: Fovea

Blind Spot: Optic Nerve

Optical Antennae: Rods & ConesRods: Intensity Cones: Color

Visible Light

• Different wavelengths

correspond to

different colors

• The range is from red

(λ ~ 7 x 10-7 m) to

violet (λ ~4 x 10-7 m)

Limits of Vision

112.4 10e x m

Electron

Waves

How is Light Made?

Atoms are EM Tuning Forks

They are ‘tuned’ to particular

frequencies of light energy.

Visible light ( and UV and IR) is

emitted when an electron in an atom

jumps between energy levels either by

excitation or collisions.

Light Emission

Atomic Emission of Light

Each chemical element produces its own

unique set of spectral lines when it burns

Dispersion with a Diffraction

Grating

Dispersion via Diffraction

: sin , 0,1,2,3constructive d m m

Hydrogen Spectra

http://www.assumption.edu/users/bniece/CHE131/LineSpectra/HiResolution/H2.pdf

http://www.assumption.edu/users/bniece/CHE131/LineSpectra/HiResolution/H2.pdf

Incandescent Light Bulb

Full Spectrum of Light

All frequencies excited!

If you pass white light through a prism,

it separates into its component colors.

R.O.Y. G. B.I.Vlong wavelengths short wavelengths

Dispersion via Diffraction


Diffraction Grating Problem

White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (l = 640 nm) appear in first order?



Diffraction gratings are

used for measuring the

wavelengths of light.

If the incident light

consists of two slightly

different wavelengths,

each wavelength will be

diffracted at a slightly

different angle.

Slide 22-56

Resolving Power of a Diffraction Grating

Resolving Power of a Diffraction Grating• For two nearly equal wavelengths, λ1 and λ2, between

which a diffraction grating can just barely distinguish, the resolving power, R, of the grating is defined as

• Therefore, a grating with a high resolution can distinguish between small differences in wavelength

• The resolving power in the mth-order diffraction is

• R = Nm

– N is the number of slits

– m is the order number

• Resolving power increases with increasing order

number and with increasing number of illuminated

slits

2 1

ave aveλ λR Nm

λ λ λ

Grating Resolution

Determine the number of grating lines necessary to resolve the 589.59 nm and 589.00 nm sodium lines in second order.

2 1

ave aveλ λR Nm

λ λ λ

Intensity of a Diffraction Grating

Because intensity

depends on the

square

of the amplitude, the

intensities of the bright

fringes are:

Slide 22-55

QuickCheck 22.8

In a laboratory experiment, a diffraction grating produces an interference pattern

on a screen. If the number of slits in the grating is increased, with everything else

(including the slit spacing) the same, then

A. The fringes stay the same brightness and get closer together.

B. The fringes stay the same brightness and get farther apart.

C. The fringes stay in the same positions but get brighter and narrower.

D. The fringes stay in the same positions but get dimmer and wider.

E. The fringes get brighter, narrower, and closer together.

Slide 22-57

QuickCheck 22.8

In a laboratory experiment, a diffraction grating produces an interference pattern

on a screen. If the number of slits in the grating is increased, with everything else

(including the slit spacing) the same, then

A. The fringes stay the same brightness and get closer together.

B. The fringes stay the same brightness and get farther apart.

C. The fringes stay in the same positions but get brighter and

narrower.

D. The fringes stay in the same positions but get dimmer and wider.

E. The fringes get brighter, narrower, and closer together.

Slide 22-58

Reflection Gratings in Nature

Naturally occurring

reflection gratings are

responsible for some

forms of color in nature.

A peacock feather

consists of nearly

parallel rods of

melanin, which act

as a reflection grating.

Slide 22-63

Diffraction & Interference

Iridescence

http://graphics.stanford.edu/courses/cs348b-competition/cs348b-01/butterfly_big.jpg

http://graphics.stanford.edu/courses/cs348b-competition/cs348b-01/butterfly_big.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/rheten6.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/rheten6.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/ole5.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/ole5.jpg

Thin Film

Interference



Radiation of Visible Sunlight

Additive Primary Colors

Red, Green, Blue

RGB Color Theory

Additive Complementary Colors

Yellow, Cyan, Magenta

The color you have to add to get white light.

Red + Green = Yellow

Blue + Green = Cyan

Red + Blue = Magenta

Red + Blue + Green = White

White light – yellow light = ??

White light – red light = ??

FYI: Mixing Colored Pigments

Subtractive ColorsPigments subtract colors from white light.

Yellow + Cyan = Green

Cyan + Magenta = Purple

Yellow + Magenta = Red

Yellow + Cyan + Magenta = Black

Why are some materials colored?

Why is a Rose Red?

Colored materials absorb certain colors that

resonate with their electron energy levels and

reject & reflect those that do not.

Shine cyan light on a red rose and

what color do you see?

Why is the Ocean Cyan?

White light minus cyan is red. Ocean water absorbs red.

Thin Film

Interference



The Index of Refraction• Refraction: Light Bends in

Transmission

• The speed of light in any

material is less than its speed

in vacuum

• The index of refraction, n,

of a medium can be defined

as

speed of light in a vacuum

nspeed of light in a medium n

c λ

v λ

in vacuum

in a mediumn

λ λn

λ λ

• For a vacuum, n = 1

– We assume n = 1 for air

also

• For other media, n > 1

• n is a dimensionless number

greater than unity, not

necessarily an integer

Some Indices of Refraction

Variation of Index of Refraction with Wavelength

• This dependence of n on λ

is called dispersion

• The index of refraction for

a material generally

decreases with increasing

wavelength

• Violet light bends more

than red light when passing

into a refracting material

speed of light in a vacuum

nspeed of light in a medium n

c λ

v λ

Interference in Thin Films• Light will be cancelled when

reflecting off a thin film due to phase changes.

• Find the total path difference due to wave 2 travelling 2t and any phase changes of waves.

• The wavelength of ray 1 in the film is /n

• Apply interference conditions




Thin transparent films,

placed on glass surfaces,

such as lenses, can control

reflections from the glass.

Antireflection coatings on

the lenses in cameras,

microscopes, and other

optical equipment are

examples of thin-film

coatings.

Application: Thin-Film Optical Coatings

Slide 21-96

Interference in Thin Films

Depends on Medium and

index of refraction!When reflecting off a medium of greater refractive index, a

light wave undergoes a phase shift of ½ a wavelength.

When reflection off a medium of lesser refractive index, a

light wave undergoes NO phase shift.

From Low to

High, a phase

change of pi!

From High to

Low, a phase

change? NO!

Reflections Phase Shift Chant

Thin Film Interference

The light reflected from a soap bubble (n = 1.40) appears red ( = 640 nm). What is the minimum thickness (in nm)?

Constructive or destructive?

constructive interference

2t = (m + ½) /n (m = 0, 1, 2 …)

Interference in Thin Films with the thin film on air

• The wavelength of ray 1 in the film is /n

• For constructive interference

2t = (m + ½) /n (m = 0, 1, 2 …)

This takes into account both the difference in optical path length for the two rays and the 180°phase change

• For destructive interference

2t = m/n (m = 0, 1, 2 …)


Example 21.9 Designing an Antireflection

Coating

Slide 21-98


Example 21.9 Designing an Antireflection

Coating

Slide 21-100


The phase difference between the two reflected waves is:

where n is the index of refraction

of the coating, d is the thickness,

and is the wavelength of the

light in vacuum or air.

For a particular thin-film, constructive or destructive

interference depends on the wavelength of the light:

Application: Thin-Film Optical Coatings

Slide 21-97

The equations in the text is for ONLY the first case

where the coating lies on the glass with a higher

index of refraction. BUT…

What if the film floats on water? Or in air?

Just don’t use those equations!! DERIVE THEM!!!!!!

MORE Applications of Interference:

Interference: Holography

The diffracted reference beam reconstructs the original scattered wave.

As you look at this diffracted wave, from the far side of the hologram, you “see” the object exactly as if it were there.

The view is three dimensional because, by moving your head with respect to the hologram, you can see different portions of the wave front.

Slide 22-104

https://www.youtube.com/watch?v=0ics3RVSn9w



Michelson Interferometer

• A ray of light is split into two rays by the mirror Mo

– The mirror is at 45o to the incident beam

– The mirror is called a beam splitter

• It transmits half the light and reflects the rest

• After reflecting from M1 and M2, the rays eventually recombine at Moand form an interference pattern

• The fringe pattern shifts by one-half fringe each time M1 is moved a distance λ/4

• The wavelength of the light is then measured by counting the number of fringe shifts for a given displacement of M1

Michelson Interferometer

The interference changes from a max to a min and back to a max every time L increases by half a wavelength. The number of maxima that appear as the Mirror shifts a length

https://www.youtube.com/watch?v=j-u3IEgcTiQ


Interferometer Problem

Monochromatic light is beamed into a

Michelson interferometer. The movable

mirror is displaced 0.382 mm,

causing the interferometer pattern to

produce 1 700 fringes. Determine the

wavelength of the light. What color

is it?

James Clerk Maxwell 1860sLight is wave. The medium is the Ether.

8

0

13.0 10 /

o

c x m s

Rotate arms to produce interference fringes and find

different speeds of light caused by the Ether Wind, due to

Galilean Relativity: light should travel slower against the

Ether Wind. From that you can find the speed of the wind.

https://www.youtube.com/watch?v=7qJoRNseyLQ

https://www.youtube.com/watch?v=7qJoRNseyLQ

http://www.youtube.com/watch?v=4KFMeKJySwA&feature=related

http://www.youtube.com/watch?v=XavC4w_Y9b8&feature=related

http://www.youtube.com/watch?v=ETLG5SLFMZo

http://www.youtube.com/watch?v=Z8K3gcHQiqk&feature=related





http://www.youtube.com/watch?v=ETLG5SLFMZo

http://www.youtube.com/watch?v=Z8K3gcHQiqk&feature=related

Clocks slow down

and rulers shrink

in order to keep the

speed of light the

same for all

observers!

Time is Relative!

Space is Relative!

Only the SPEED

OF LIGHT is

Absolute!

On the Electrodynamics of Moving Bodies

1905

LIGO: Gravitational Interferometry

http://www.youtube.com/watch?v=RzZgFKoIfQI&feature=related

https://www.youtube.com/watch?v=HwC5IYw5uAE



https://www.youtube.com/watch?v=HwC5IYw5uAE

LISA

The Laser Interferometer Space

Antenna

http://www.youtube.com/watch?v=DrWwWcA_Hgw&feature=related




http://www.youtube.com/watch?v=tUpiohbBv6o



Physics Diffraction Fun on an

Airplane

Always sit on the side opposite the

sun when traveling north-south!!

Interference of Light

Wave Optics

Chapter 22 Wave Optics

Documents

Transcript of Chapter 22 Wave Optics