Light as a wave

We've spent some time developing a powerful set of tools for analyzing the physics of waves. So far, we've treated cases like waves on a string, or sound waves as they propagate through air. But, there are many more phenomena in the natural world that can be explained and analyzed using wave physics. A major one that persists throughout the development of science and human history has been light. From ancient texts to modern communication technology, the quest to understand light has been a driving force in science. We'll take a look now at how light can be understand as a wave. In the early years of scientific thought, the study of light was very problematics. Several of the conveniences we have now didn't exists, and this made light a rather difficult subject to study. For example, we know now the speed of light to be 186,000 miles per second. (Or 300,000,000 m/s) So, imagine trying to measure something that moved this fast using only pendulum clocks or sundials. It would be hard. Also, when we study light now, we use lenses and other tools to shape it and make it behave. Such things were not available in the beginning of natural science. So, most of the work on light was done using reasoning and deduction, rather then measurements. Of course, some experiments were done, but they were often hard to interpret correctly. A major argument that persisted in the scientific community was weather light was a particle, or a wave. At times, it seems to act like a waves from a pebble on the surface of a pond, other times, like rubber balls moving in straight lines. We'll study both approaches and see how they offer different advantages for understanding the natural world.

Huygens' Princinple 1. Each point on a wave front is the source of a spherical wavelet that spreads out at the speed of the wave. 2. At a later time, the shape of the wave is the curve that is tangent to all the wavelets. Christian Huygens [16291695] One of the first scientists to make a good case for 'light as a wave' was the Dutch physicist Christian Huygens. Around 1678, he wrote a treatise about light in which he described a wave based understanding about the propagation of light. Although the details of the mechanism he proposed were not exactly correct, the basic conceptual understanding provided by the model remains useful. At this point in time, people weren't sure if light moved at infinite speed (instaneously) or if it had a finite velocity. Huygens firmly believed had a finite speed and was convinced of this by astronomical observations provided by Olaus Roemer. He also could explain certain aspects by treating light as a wave. Essentially, Huygen's principle states that a wave front can be created by many closely spaced, coherent point sources that are all spreading out in a spherical manner. (Coherent means they have the same initial phase) plane wave

1. Each point on a wave front is the source of a spherical wavelet that spreads out at the speed of the wave. 2. At a later time, the shape of the wave is the curve that is tangent to all the wavelets.

t=0

t>0

1. Each point on a wave front is the source of a spherical wavelet that spreads out at the speed of the wave. 2. At a later time, the shape of the wave is the curve that is tangent to all the wavelets.

Index of Refraction The index of refraction (or refractive index) of a material is a dimensionless parameter, n , used to describe how light moves in a particular medium. n=

speed of light in vacuum c = v speed of light in material

The index of refraction of vacuum is n = 1.000 and is exactly 1. Medium

Index

Vacuum

1 (exactly)

Air (0ºC, 1 atm) 1.00029 Water

1.33

Glass

1.52

Saphire

1.77

Diamond

2.42

[*Note for labs: The index of refraction can change based on the wavelength of the light]

Index of Refraction The velocity of the wave in a medium is given by: v =

c n.

From the previous visualization, we can see that the wavelength will also change. Thus we can write λn =

λ n

What about the frequency?

Refraction

n1

Refraction involves the bending of light at an interface between two medium with different indices of refraction. We'll see that the following relation holds:

n2

sin θ1 n2 = n1 sin θ2

Refraction can be understood as a direct consequence of the change in velocity of a wave as it enters a new medium at an angle.

Law of Refraction

Derive the law of refraction

Big time implications n1

light vacuum (n = 1)

n2

n1

light vacuum (n = 1)

n2

Two light waves enter two different materials.

Upon exiting the material: The waves are no longer in phase! Yes, but by how much? out of phase!

n1

light vacuum (n = 1)

n2

Question: A single light beam is split into two equal beams, denoted A and B. Beam A travels through a medium with a higher index of refraction than the medium that beam B travels through. After both beams exit their media are back into the air, how do their wavelengths compare? a) Beam A has a longer wavelength b) Beam B has a longer wavelength c) Both beams have the same wavelength

Based on the length of the materials, the indices of refraction, and the wavelength of the light in vacuum, we can determine the phase difference (actually the difference in number of wavelengths) between the two light waves. N2 − N1 =

L (n 2 − n 1 ) λ

We can get this relation in the following way. The wavelength in the medium will be different by wavelength in vacuum:

λn1 =

λ n1

Also, if the medium has a length L, then there will be N1 wavelengths present in the medium:

N1 =

Ln1 L = λn1 λ

N2 =

Ln2 L = λn2 λ

Similary, for the waves in medium 2:

Subtracting N2 - N1 gives us the desired relation.

Color

frequency of wave [Hz] 1018

10-10

1014

10-7 10-6

1010

10-2

radio/tv

400 Terahertz

x rays

780 Terahertz

microwave

The color of light that we perceive is based on the frequency (or wavelength) of the light.

108

1

wavelength of wave [m] ~400 nm

~500 nm

~700 nm

The human eye has three types of cells that are sensitive to different frequencies of light. The triggering of these cells by the different frequencies results in color perceptions. They are responsive to only a very narrow set of frequencies however. The complete spectrum of light contain frequencies and wavelengths that do not trigger the cells our eyes. We'll see this spectrum again, after we go through electricity and magnetism, since those topics will contain the material to really understand what light is.

The light from the sun

The physical processes happening in the Sun create light of many different frequencies. This chart shows the intensities emmited frequencies as a function of wavelength. It is another example of a spectrum. We saw a spectrum graph when we covered sound waves. There we plotted the intensity of the sound waves as a function of the frequency (pitch) of the sound. This graph contains similar information. From it, we can see the most of the light that the sun produces is around 500 nm wavelength. Fortunately, this is around the frequency that our eyes are very sensitive too.

index of refraction

Chromatic Dispersion n = 1.55

n = 1.45

200 nm

400 nm

600 nm

800 nm

1000 nm

1200 nm 1400 nm

wavelength

Here is a qualitative plot of the index of refraction for a hypothetical material as a function of the wavelength of the incident light. When light enters a medium, the index of refraction it encounters will depend slightly on the wavelength of the light. This effect is called Chromatic Dispersion.

Dispersion n>1

The implications are that the angle of refraction is now dependent on the wavelength of the light because n carries this dependence: n → n(λ) . sin θ1 n 1 (λ) = sin θ2 n 2 (λ)

Diffraction - Intro The spreading out of the waves after passing through the opening is called diffraction. Diffraction is a general properties of all wave phenomenon. Water waves are particularly easy to see diffraction effects because the wavelengths are about as big as the obstacles which cause the diffraction. Check out the Panama canal entrance: Link to Map

Wave interference crest point 1

Two speakers?

trough

L2

L1

L2

L1 point 2

The physics of light borrows much from the treatment of sound waves. We figured out how to calculate where load and quite spots should be in the case of two interfering sound waves. A similar approach will give us information about the interference of light.

Thomas Young There were two competing theories about what exactly light was. Newton's experiments led him to postulate that light was a particle (he called them 'corpuscles'.) While Huygens' and others had done work which seemed to show that light acted as a wave. Thomas Young was an English Scientist. Around 1801 he did some experiments with light with the goal of demonstrating its wave-like nature.

Young's double hole experiment double slit (hole)

screen

max min max min max min max min max min max min

The basic set up is as follows. Light (monochromatics and coherent) passes through two small slits. From each slight emerges a diffracted beam. These beams undergo wave interference and the resulting light that appears on a screen away from the slits shows the interference pattern.

+y

y3 P

y2 y1

r1

y0

r2

m=3

y’2

m=2

y’1

m=1

y’0

m=0 m=1

L

slits

m=2 m=3 screen

P

y = L tan

ΔL = d sin θ

r2 slits

y=0 r1

L screen

When constructive: d sin θ = mλ for m = 0, 1, 2

th r 2

path leng

d

path leng

th r 1

and for destructive: d sin θ = (m + 1/2)λ for m = 0, 1, 2

∆r = path length difference

Question: A beam of monochromatic light with a wavelength of 660 nm is directed at a double slit. Consider the five locations labeled in the drawing, (the central maximum is labeled “B.”) Which one of these fringes is produced when the path difference is 1320 nm?

Example Problem:

Light from a Laser has a wavelength of 633 nm. This is passed through two small slits spaced .4 mm apart. A viewing screen is 2.0 m behind the slits. What are the distances between the two m = 2 bright fringes and between the two m = 2 dark fringes, on the screen?

Question: A laboratory experiment produces a double-slit interference pattern on a screen. If the screen is moved farther away from the slits, the fringes marked B and C will be:

a) Closer together. b) In the same positions. c) Farther apart. d) Fuzzy and out of focus.

Example Problem:

A double slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. Ten bright fringes span a distance of 1.7 cm. What is the wavelength of the light?

Intensity plot

I

Of course, as scientists, we would rather have our data in a more useable form. To this end, we can convert our 'screen' images of dark and light spots to an intensity plot. This lets us put units, like y W/m 2 for intensity as a function of image datadistance, and use functions to desribe the plot.

The same information is present -- it's just easier to work with.

Question: A laboratory experiment produces a double-slit interference pattern on a screen using red light. If green light is used, with everything else the same, the bright fringes will be: a) Closer together. b) In the same positions. c) Farther apart. d) Fuzzy and out of focus.

Waves and Reflections We saw what happened when a wave on a string reflected off a boundary. Play/Pause

Play/Pause

Boundary is not perfect Here we have a wave on a rope traveling towards a partially reflecting boundary. The 'boundary' is really a connection between a low density and high density section of rope. Play/Pause

We can see that most of the wave is reflected, and inverted, at the boundary, however, some of the energy of the wave is transmitted across the boundary and continues into the higher density region. Additionally, since v = ‾μ‾τ , the wave √ speed is different for the reflected and the transmitted waves.

Boundary is not perfect Likewise, if the wave travels from a higher density region to a lower density region, most of the wave's energy will be transmitted, while some will be reflected. Play/Pause

Again, the lower density section of rope will have a higher wave speed than the high density region. This time, there is no amplitude inversion. vid

Phase change due to reflection Since the index of refraction of a material is somewhat analogous to the density of a string, as far as waves go, (they both are parameters that affect the speed of the waves), we would expect a similar phenomena with light as it reflects from boundaries. Indeed.

incident wave

transmitted wave

reflected wave

Depending on if the light is going reflecting off a higher index or a lower index, we will obtain the following phase shifts for the reflected waves. Reflection

incident wave

transmitted wave

reflected wave

Phase shift incident wave

Reflects off a lower index

n1 > n2

n1 < n2

0

Reflects off a higher 0.5 index wavelength

transmitted wave

reflected wave

Phase Shift Sources We now have 3 sources of phase shifts for a light wave: 1. Reflection: depending on the n of the two media, the reflected wave may or may not be phase shifted 2. Path Length: The analysis of the double slit showed that path length difference can also lead to a phase shift. 3. Index of Refraction: Light traveling through media with different n can undergo a phase shift.

Thin Film Interference nair= 1

air

nx= 1.33

nglass= 1.5

thin film

incident wave

glass

transmitted wave

t

reflected wave (from first surface)

reflected wave (from glass surface)

Thin Film Interference n1

n2

n3

r2 3

r1

2 1

incident ray

The original incident ray is both reflected and transmitted at the first boundary. This reflection becomes r1 . The transmitted wave goes until the next boundary, , where it is also transmitted and reflected. The reflected ray proceeds back through material 2, and it then transmitted through the boundary at . This last transmitted ray is r2 .

By figuring out all the phase shifts along the way, we can tell if rays r1 and r2 will be in phase or out of phase. That is, will they constructively interfere or destructively interfere?

Constructive interference n2

air

air

In this situation, we have n 1 < n 2 > n 3 . Since at the interface the ray is reflecting off a large n , the reflected ray r1 will have a phase shift of λ/2 .

r2 3

r1

r2 will have a path length difference compared to r1 of 2L. Thus to get constructive interference between r1 and r2 , we need

2 1

incident ray 2L = (

odd number 1 λ λn2 = (m + ) for m = 0, 1, 2, … ) 2 2 n2

Destructive Interference n2

air

air

If we where looking to have destructive interference between the two reflected rays, we would need:

r2

2L = integer × λ

3

r1

2 1

. Thus, with the same substitutions, we can arrive at:

incident ray 2L = (integer) λn2 = m

λ for m = 0, 1, 2, … n2

Example Problem:

What is the thinnest film of MgF2 (n = 1.38) on glass that produces a strong reflection for light with a wavelength of 600nm.

Example Problem:

Two 15-cm long flat glass plates are separated by a 10 μm thick spacer at one end, leaving a thin wedge of air between the plates. A monochromatic light (λ = 589 nm) from above illuminates the plates. Alternating bright and dark fringes are observed. What is the spacing between two bright fringes.

n1

n2

n3

r2 3

r1

If the film is much smaller than the wavelength of the light passing through, (i.e t < .1λ ), then we can ignore the path length created by the film and only concern our analysis with the reflections of the light at the boundaries.

2 1

incident ray

t < .1 l

soap film incident

transmitted

reflected incident reflected

transmitted

Since the criteria for constructive and destructive interference depends on wavelength, light of different colors will respond differently to a thin film, like a soap bubble or oil slick. Some colors will experience constructive interference, while others will have destructive interference

after the two reflections.

Interferometer We’ve seen that the amount of light visible is very sensitive to the path difference between two rays. This means we can make a device to measure very small distances.

emitters

detector

The smallest length we can measure using this device will be given by the wavelength of the light used.

Interferometer mirror

We know the path length difference is: L1

Δr = 2L2 − 2L1 source

splitter

L2

detector

mirror

When Δr is equal to a whole number of wavelengths, ie: Δr = mλ , then the interference between the two paths will be perfectly constructive. Any change in the amplitude of the final, detected light, implies a change in either the wavelength of the light during one arm of travel, or a change in the length of one arm.

Uses of interferometry: 1. Disproved the Luminiferous Aether 2. Surface topography of samples. 3. Measured Newton’s Gravitational Constant, G. 4. Maybe gravitational waves??? (lookup LIGO - http://www.ligo.caltech.edu/)

Question: Which one of the following statements best explains why the diffraction and interference of sound is more apparent than that of light under most circumstances? a) Sound waves are longitudinal, and light waves are transverse. b) Sound requires a physical medium for propagation. c) Light waves can be represented by rays while sound waves cannot. d) The speed of sound in air is six orders of magnitude smaller than that of light. e) The wavelength of light is considerably smaller than the wavelength of sound.

The diffraction of light

Our use of the ray model relied on the assumption that the light rays didn't interact with objects that had features in the same length scale as the wavelength of the light. Clearly, this limits the range of topics we can discuss, and it limits the depth of phenomena we can understand.

Let’s go back to this: A monochromatic light source passing through a small slit. The wavelength of this light is roughly the same size as the slit through which it is passing. Play/Pause

The set up is the same as the double slit, except there is only one slit. Again, we see a diffraction pattern emerge on the screen. This one is a bit different however.

Huygens-like analysis of a single slit A

Ⓐ The wavelets from each point on the initial wavefront overlap and create an interference pattern

B

Ⓑ These all travel the same distance and create the central maximum

C

Ⓒ These travel different distances and create the fringes.

1 3 5 2 4 6

a/2 a

∆r

The path length difference between wavelet 1 and 2 is: Δr12 =

a sin θ 2

Which is another way of saying that wavelet 2 travels Δr more to get the screen at the angle θ . If Δr12 = λ/2 , then so are all the other ∆r values, and the interference at the screen will be destructive, meaning there will be a dark spot there. The locations of the dark spots are then: a λ sin θ = 2 2 or, a sin θ = λ

The other minimums can be figured out in a similar way. We get the following for the angular locations of the dark regions. a sin θ = mλ

single slit

mλL a

thus: w=

1λL 1λL 2λL + = a a a

y

w

a

The width of the central maximum is easily figured by looking at the distance between the first minimums. y=

screen

central maximum L

m=2 m=1 m=1 m=2

Question: Light of wavelength λ illuminates a single slit so that the width of the central diffraction maxima is l . The slit width is then decreased to a/2. What is the width of the central diffraction maxima? a) l/2 b) 2l c) l/4 d) 4l e) l

Question: This time you keep the same slit width, but use another monochromatic light of wavelength 500 nm. How does the broadness (width) of the central bright fringe change compared to that produced by the 600 nm wavelength? a) Increases b) Decreases c) Stays the same d) Depends on the exact value of slits

Multiple Slits What happens when we increase the number of slits? This system is called a diffraction grating. We can think of it as a multi-slit device.

d d

towards

screen

d plane waves approaching from left

We can increase the number of slits per inch. Here's 5 slits and a sample of what the diffraction pattern would look like. screen

5 slit grating

Using the same analysis as before, we can find the locations on the screen where the path length difference leads to constructive interference.

y2

m=2

y1

m=1

0

m=0

Δr = d sin θ = mλ

y1

m=1

(for m = 1, 2, 3)

y2

m=2

Again, note that this value is dependent on the wavelength.

The quality (sharpness and intensity) of the diffraction pattern is determined by the number of slits in the grating. As the number of slits, n, increases, the width of the fringes decreases, but their intensity increases.

Different wavelengths will have maximums at different positions along the screen.

Awesome Tool: The spectrometer Build your own: DIY spectrometer kit here

Circular Aperture Diffraction circular aperture

screen

incident light

Monochromatic light which passes through a very small circular aperture will create a circular diffraction pattern like this.

D light intensity

L

The location of the first minimum can be found by: θ1 = 1.22

λ D

The lens as an aperture A lens can be considered a type of aperture. There will be diffraction of the light waves as it passes through.

Two distant stars

D

Rayleigh's Criterion Two point sources can only be resolved if their angular separation, α is greater than: θR = sin −1

1.22λ 1.22λ ≈ D D

Example Problem:

Two lightbulbs are 1.0 m apart. From what distance can these lightbulbs be resolved by a small telescope with a 4.0 cm diameter objective lens. Assume that the lens is diffraction limited and λ = 600nm.

Question: A spy satellite is at an altitude 650 km above the Earth’s surface. How large must the satellite’s camera lens be so that its resolution is 25 cm? Assume the average wavelength of light of 480 nm. a) 1.8 m b) 2.7 m c) 0.55 m d) 1.5 m e) 0.85 m

Resolving Power: In general, we can say the resolution for any optical instrument is limited by the wavelengths of the light used.

Guess what color the light from a Blu-ray player laser is? Why?