wave optics

Wave OpticsUnit 11

Light

• In the last unit, we studied several properties of light, including refraction and reflection.

• However, these were geometric properties, and had nothing to do with the fact that light is a wave.

• Now we will no longer be able to ignore this reality.

Interference

The Double-Slit Experiment

Interference

• One of the most conclusive examples of the wave nature of light was the Double Slit Experiment.

• Early physicists, such as Newton, had thought of light as a particle.

• However, Thomas Young was able to show light acts as a wave, and was even able to measure the wavelength of visible light.

Interference

• Young created a setup that allowed light from a single source to fall on two closely spaced slits.

• If the particle picture of light were correct, we would expect to see two bright lines on a screen placed behind the slits.

Interference

• However, instead a series of bright lines were observed on the screen.

• Young explained this effect as a wave-interference phenomenon.

Interference• To see how this effect occurs,

let’s consider Young’s setup.

• Plane waves of light with a single wavelength are striking the slits.

• Light of this type is called monochromatic, meaning “one color.”

Interference• When the waves strike the

slits, they spread out in all directions on the other side.

• This is analogous to what happens when two rocks are thrown into a pond, or when two speakers are placed close together.

Interference• Recall from the speaker problem

that the waves from the two slits will have traveled different distances depending on where they strike the screen.

• As a result, the waves may arrive out of phase.

• This can lead to interference when the waves add back together.

Interference• Let’s consider

monochromatic light with wavelength λ striking slits S1 and S2, which are separated by a distance d.

• If we look at the center of the screen, we see a bright spot.

Interference• This is not surprising since the

light from each source travels the same distance.

• The two waves arrive at the screen in-phase, resulting in constructive interference.

• The brightness of the spot is brighter than the light from either slit.

Interference• Likewise, if we look at

certain points on the screen that are off the axis, we also see a bright spot.

• These are also points where constructive interference is occurring.

Interference• From our experience with

sound waves, we know that this means the path length difference must be some multiple of a wavelength.

• We will quantify this in a moment.

Interference• At other points, no light

can be seen.

• These points indicate destructive interference is occurring.

• The path length difference must be a multiple of a half wavelength.

Interference• To develop a formula for

determining the location of the bright (or dark) fringes, let’s look at the geometry of the system.

• We want to know how far each ray will have to travel to reach a point at an angle θ above the center.

Interference• Notice that ray 2 travels farther

than ray 1.

• This is the path length difference.

• Notice also that if forms a side of a right triangle with the spacing between the slits.

Interference• Based on this, we can use

trigonometry to write the path length difference in terms of d and θ, quantities we can easily measure:

Interference• This gives us the formula

for constructive interference:

Interference• This also gives us the

formula for destructive interference:

Interference• Notice also that there is a

second right triangle that involves the distance from the center on the screen.

• From this we can see that

Interference• This lets us rewrite our

formulas:

Notes

• These formulas are only valid when the distance between the slits (d) is very small compared to the distance to the screen (L).

• As a result, θ must be very small.

• The value of m is called the order of the interference fringe.

Example

Two slits are 0.1 mm apart and are located 1.2 m from a viewing screen. Light from a He-Ne laser beam (λ = 633 nm) is shined on the slits.

a)How far from the center is the second order interference minimum located?b)What angle above the horizontal is this?

You Try

The He-Ne laser from the last example is replaced with a new laser. You measure the location of the first order interference maximum to be 0.6 cm above the axis. What is the wavelength of the light being emitted by this laser? Express your answer in nm.

Homework

• Read 24-3.

• Do problems 1, 2, 3, and 5 on page 692.

Announcements

• Paper final draft due next Monday.

• Presentations May 2 – May 5.

Problem Day

• Do problems 5, 10, and 11 on page 692.

• We will whiteboard at the end of class.

Homework

• Read 24-1 and 24-5.

• Do problem 12 on page 692.

Huygens’s Principle and Diffraction

Huygens’s Principle

• Christian Huygens was a Dutch scientist who lived about the same time as Newton.

• He proposed a wave model for light that is still used today to describe many phenomena.

• Recall that for 2D and 3D waves, we describe the position of the wave using the wave front.


Every point on a wave front can be considered to be a source of tiny wavelets that spread out in the forward direction at the speed of the wave itself.

The new wave front is the envelope of the wavelets.


• Note that this is a model of how a wave propagates.

• However, it is extremely useful for explaining phenomena such as diffraction.

Diffraction

• In 1819, the French scientist Augustin Fresnel used Huygens’s principle to explain how light could “bend” around the edges of solid objects.

• This phenomenon is known as diffraction.

• As we will see in a moment, diffraction produces results similar to interference. The two effects are related, but are not the same.

Diffraction• Fresnel considered a circular disk that was

illuminated by monochromatic light.

• Using Huygens’s principle, we can consider the points around the edge of the disk to be point sources of wavelets.

Diffraction• These wavelets spread out in all directions.

• On the outside of the disk, the wavelets add together with other parts of the beam.

Diffraction• However, the wavelets that spread out toward

the inside of the disk, do not encounter each other until they reach the screen.

• Thus, Fresnel predicted a bright spot at the center of the shadow of a disk.

Diffraction

See also the pictures on p 673.

Question

• What effect do those bright/dark fringes remind you of?

• In fact, these fringes are due to the interference of the diffracted light waves with themselves.

• The image as a whole is called a diffraction pattern.

Single-Slit Diffraction• To see how a diffraction pattern arises, we will

consider the simplest case of monochromatic light passing through a single, narrow slit.

• We will assume the rays striking the slit are parallel and the viewing screen is very far away.

Single-Slit Diffraction• From Huygens’s Principle, we can model the

slit as being made up of a bunch of point sources.

• These sources send out wavelets in all directions.

Single-Slit Diffraction• For the wavelets that continue on straight

ahead, all the waves travel the same distance to the screen.

• This means they arrive in phase and interfere constructively, producing a bright spot.

Single-Slit Diffraction• However, if we move off axis by an angle θ,

then the rays from the top of the slit will have to travel a longer distance than the rays from the bottom.

• This leads to interference at the screen.

Single-Slit Diffraction• Consider first an angle where the top ray

travels one λ farther than the bottom ray.

• This means the middle ray will travel λ/2 farther than the bottom ray.

Single-Slit Diffraction• This means the middle ray will interfere

destructively with the bottom ray.

• The ray just above the bottom will interfere destructively with the ray just above the middle.

Single-Slit Diffraction• Thus all the rays interfere with each other at

this angle.

• The result is a dark spot on the screen.

Single-Slit Diffraction• If we increase the angle so that now the top

ray travels 3λ/2 farther than the bottom, we still get some destructive interference.

• The bottom 2/3 of the light from the slit cancel out as before.

Single-Slit Diffraction• However, now there are no rays to cancel the

light coming from the top third of the slit.

• This light reaches the screen, resulting in a bright spot.

Single-Slit Diffraction• If we increase the angle further so the path

difference is 2λ, we get total destructive interference again.

• The bottom fourth cancels with the second fourth and the third fourth cancels with the top fourth.

Single-Slit Diffraction• The result is a diffraction pattern with a

central bright spot and several dimmer spots seen as you go off the axis in either direction.

Single-Slit Diffraction• To locate the diffraction minima, we need to

look at the geometry again.

• Notice that, as with the double-slit, the path difference is equal to Dsinθ, where D is the width of the slit.

Single-Slit Diffraction• Notice that, unlike with the double-slit, the

diffraction minima occur when the path length difference is equal to a whole number multiple of the wavelength (i.e. 1λ, 2λ, etc.).

Single-Slit Diffraction• This allows us to write a formula for locating

the diffraction minima:

Things to Notice

• This formula looks extremely similar to the formula for finding interference maxima.

• However, there are important differences.

• First, this formula gives the diffraction minima.

Things to Notice

• Second, the minima only occur for m ≥ 1. There is a bright spot for m = 0.

• Third, here D is the width of a slit rather than the distance between two infinitely narrow slits.

• Lastly, this formula cannot be used to find diffraction maxima, which are more complicated.

Example: Single-Slit

Light of wavelength 750 nm masses through a slit that is 1.0 x 10-3 mm wide.

a) What is the angular distance to the firstminimum?b) If the pattern is viewed on a screen 20

cm away, how far from the central axis is the minimum?

In-class Work

• Do problems 17, 18, 20, 22, and 23 on page 693.

• Your homework is to finish these problems.

Problem Day

• Do problems 25 and 62 on pages 693-694.

• We will whiteboard these problems and the homework problems in ~25 mins.

Homework

• Study your notes on interference and diffraction.

• Do problems 24 and 67 on pages 693-695.

Dispersion

The Visible Spectrum: A Review

• There are two aspects of light that can be explained by the wave theory.

• The brightness of the light corresponds to the intensity of the electromagnetic wave.

• The color of the light corresponds to the frequency (or wavelength) of the light.


• The human eye is sensitive to EM waves with frequencies from 4 x 1014 Hz to 7.5 x 1014 Hz.

• In air, this corresponds to wavelengths of 400 nm to 700 nm.


• This region of the EM spectrum is referred to as the visible spectrum.

• Light with wavelengths shorter than 400 nm is called ultraviolet (UV).

• Light with wavelengths longer than 750 nm is called infrared (IR).

Prism Demo

Prisms• When light passes from one

medium to another, it is refracted according to Snell’s Law.

• However, the prism seemed to separate the white light into many different colors.

• Why did this happen?

Prisms• It turns out that the index

of refraction of a material depends on the wavelength of light that shines on it.

• The index is greatest for light that has a shorter wavelength (blue-purple).

Prisms• As a result (see Snell’s Law), the

blue-purple light gets refracted at a larger angle than the red light.

• Thus, if we send white light (which contains all colors) through a prism, each color is refracted at a different angle.

• This spreading of light is called dispersion.

Rainbow• We see dispersion on a regular basis (well, not

in Phoenix) through rainbows.

• A rainbow appears when you look at falling water droplets with the sun behind you.

Rainbow• The sunlight contains all wavelengths of light.

• When light strikes the water droplets, it is refracted due to the change in medium.

• The angle of refraction depends on the wavelength of the light.

Rainbow• The light is reflected off of the back of the

droplet and travels toward your eye.

• The red light is bent the least, and so you see red light from droplets higher in the sky.

Rainbow• The purple light is bent the most, so you see

purple light from droplets that are lower in the sky.

• The other colors are seen from droplets in between, resulting in a rainbow.

Homework

• Read 24-4, 24-6 and 24-7.

• Do problems 14 and 15 on page 693.

Multiple Slit Interference and Diffraction

Where We Are…

• We have already looked at interference effects for a two-slit setup and diffraction effects for a single slit.

• We also observed that a real-world two-slit system has both interference and diffraction effects present.

• Now we will look at what happens when light shines on more than 1 slit.

Multiple Slit Interference

• When monochromatic light is shined on a series of 3 slits, we can still see an interference pattern.

• However, the maxima are narrower than for the two slit case.

• There are also smaller maxima that can be seen in between the larger one.


• As the number of slits is increased, the large maximums become even narrower.

• There are more little maximums, but these become dimmer and dimmer as more slits are added.


• Why does this happen?

• For the two slit case, if we move a little distance away from a maximum, the waves from the two slits will be only slightly out of phase.

• As a result, the interference will be almost constructive.


• This means the interference fringes will be fairly wide for two slits.

• However, as more slits are added, the fringes sharpen.

• This is because there are now more waves interfering at each point.


• The waves from two adjacent slits might be only slightly out of phase.

• But they will be canceled out by waves from other slits.

• As a result, bright fringes only occur when all the waves arrive in phase.


• The smaller maximums are where waves from 2 or 3 slits arrive in phase.

• As you increase the number of slits, the brightness of these smaller maximums decreases to zero.

Multiple Slit Interference – Real World

• When we looked at the double-slit experiment in the lab, we noticed that some of the interference fringes were brighter than others.

• This is because the slits in the lab are not infinitely narrow. They have a finite width.


• As a result, both interference and diffraction effects are present.

• The pattern we see is an interference pattern where the brightness is determined by the diffraction effect.

• The spacing of the interference fringes is determined by the spacing of the slits. The brightness is determined by the width of each slit.


• The same thing happens when we view multiple slit interference in the real world.

• We see an interference pattern contained within a diffraction pattern.

• This will help us analyze diffraction gratings.

Diffraction Gratings


• A diffraction grating is a collection of a large number of slits (usually around 104).

• When the number of slits is large, the primary maximums of the interference pattern become very narrow.

• At the same time, the intermediate maximums become so dim they disappear.


• If we shine monochromatic light on a diffraction grating, we will see series of narrow fringes on the screen.

• Because the interference fringes are so narrow, the diffraction grating is a much better instrument for measuring the wavelength of light.

Diffraction Gratings• Since the position of the fringes result from

interference, we can find the location of each fringe using the old formula

• Here, d is the distance between two adjacent slits on the grating.

Example

A diffraction grating contains 10,000 lines/cm. Determine the location of the 2nd order maxima for

a) 400 nm light.b) 700 nm light.

Diffraction Grating – White Light

• A more interesting question is what happens when white light is shined on a diffraction grating?

• As we know, white light is composed of all the wavelengths of the visible spectrum.

• To see what happens, let’s first consider a simple case where the incident light is a mixture of two wavelengths.


• Suppose the incident light had only two wavelengths: 400 nm (purple) and 700 nm (red).

• We already know that the location of the diffraction maximums depends on wavelength.


• Suppose the incident light had only two wavelengths: 400 nm (purple) and 700 nm (red).

• We already know that the location of the diffraction maximums depends on wavelength (except for the central fringe).


• The location central fringe will be the same for both wavelengths (it’s at 0°).

• But the higher order fringes for the red and blue lines will be at different locations since the two colors correspond to different wavelengths.


• What we are seeing is basically two different interference patterns on the same screen.

• Question, what would happen to the pattern if the incident beam also included green light (say 525 nm)?


• If the incident beam contains all wavelengths, then each is diffracted at a slightly different angle.

• As a result, the light is separated into rainbows for each order.

Diffraction Grating – White Light• Question: why is the light for the second order

fainter than the first?

• The answer is the diffraction effect limits the brightness of the higher orders.

• Notice also that it is possible to have the rainbows overlap depending on the spacing of the slits.

Example: Spectra Overlap

White light strikes a diffraction grating containing 4000 slits/cm. Show that the blue (450 nm) end of the 3rd order spectrum overlaps with the red (700 nm) of the 2nd order spectrum.

Homework

• Reread 27-6.

• Do problems 27, 28, and 29 on page 693.

Thin Film Interference

Thin Film Interference• Interference gives rise to many every-day

phenomena.

• These include the colors reflected off a soap bubble or an oil slick.

Thin Film Interference• To see how this effect

occurs, let’s look at a thin lair of oil on the surface of water (or some other surface with higher n).

• For simplicity, let’s assume the light is monochromatic (for now).

Thin Film Interference• When the beam strikes the

oil, part of the beam is reflected off the surface, and part is refracted into the oil.

• The refracted beam continues until it strikes the water. Again, part is reflected and part is refracted.

Thin Film Interference• The reflected beam passes

back out of the oil and into the air, forming a parallel beam with the first reflected ray.

• However, the second ray has traveled a longer path length than the first.

Thin Film Interference• When these two rays arrive

at your eye, there will be a phase difference that depends on the path length difference ABC.

• If ABC is equal to a wavelength (or 2λ, 3λ, etc), the rays arrive in phase and interfere constructively.

Thin Film Interference• If the thickness of the film is

a multiple of a half-wavelength, the rays will be out of phase when they reach your eye and you will see no light.

• However, we have to be careful, since the wavelength of the light changes in the film.

Thin Film Interference• Therefore the path length

difference must be a multiple of the wavelength in the medium for constructive interference to occur.

• In other words,

Thin Film Interference• To find λn, use the index of

refraction

• Then use the relation

Thin Film Interference• Let’s look at the following situation.

• A curved piece of glass is placed in contact with a flat surface.

• The system is illuminated with monochromatic light.

Thin Film Interference• The air gap between the two pieces of glass

acts like a thin film.

• The two reflected rays have a phase difference due to the path length difference.

Thin Film Interference• However, the width of the film changes as you

move away from the center.

• Thus, at some points ABC is a whole multiple of a wavelength.

Thin Film Interference• At other points it is a half multiple of a

wavelength.

• This leads to a series of bright and dark circles resulting from interference.

Newton’s Rings• These fringes are called Newton’s rings.

• Notice that the central fringe is actually a dark fringe.

Newton’s Rings• This is surprising, since the two surfaces are in contact and

the path difference is zero.

• Yet there is a dark spot, indicating the rays are out of phase.

• It turns out that this is an important statement about the behavior of light.

Reflection and Phase Changes

If a beam of light is reflected off of a surface with a greater index of refraction than the medium it is traveling through, it changes phase by 180° (in other words, the wave flips).


• An interference pattern can also be seen when two pieces of glass are separated by a wedge-shaped region of air.

• Again, the air acts like a thin film, leading to interference that depends on the path difference.


• Because the region of air varies in thickness, the path difference changes.

• As a result, constructive interference occurs at some points and destructive at others.

Homework

• Read 24-8.

• Complete your lab sheets.


• If a thin film is illuminated by white light instead of monochromatic light, a series of different colors is seen.

• Why do you think this happens?


• When white light is used, interference is still occurring.

• However, the type of interference that occurs for a given wavelength depends on the thickness of the film.

• So, different colors will interfere constructively at for different thicknesses.

Soap Bubbles and Oil Slicks

• This explains the colors seen on soap bubbles and oils slicks.

• The film is not the same thickness throughout.


• At at one point the thickness leads to constructive interference for red light.

• At another point, its yellow, (then green, blue, etc.).


• Eventually, the thickness leads to constructive interference for red light again, and the pattern repeats.

• As a result, you see a series of different colors at different points.

wave optics

Documents

monochromatic light

properties of light

light acts

plane waves of light

wave nature of light

wavelength of visible

particle picture of

path length difference