IB DP Physics · Theme C · Wave behavior

Wave phenomena.

C.3 · Reflection, refraction, diffraction and interference
Standard level and higher level

SL + HL · 5 hours Additional HL · 6 hours

The journey

Four parts.

Part 01

Waves at a boundary

Wavefronts and rays, Snell’s law and total internal reflection

Part 02

Diffraction and superposition

Spreading round obstacles, and what happens when waves meet

Part 03

Interference

Coherent sources, path difference and Young’s double slit

Part 04 · HL only

Slits and gratings

The single slit, the modulated double slit and the diffraction grating

C.3 WAVE PHENOMENA · SL + HL

Part 01

Waves at a boundary.

What a wave does when the medium changes

C.3 WAVE PHENOMENA · SL + HL

Two ways to draw a wave

Wavefronts and rays.

A wavefront joins points that are in phase: a crest, say
Neighboring wavefronts are one wavelength apart
A ray shows the direction the wave travels
Rays are always at right angles to wavefronts

NEAR A SOURCE, AND FAR FROM ONE

C.3 WAVE PHENOMENA · SL + HL

Two pictures of the same thing

Drop a stone in a pond and the ripples you see are wavefronts: each ring is a line of crests. The direction in which any part of a ring is moving is outwards, along a radius: that is the ray. You can always draw one from the other, because a ray must cross every wavefront at right angles.

Why distant sources give plane waves

A small piece of a very large circle is nearly straight. Light from the Sun, or from a lamp across the room passing through a small slit, arrives with wavefronts that are flat for all practical purposes, and its rays are parallel.

Which to draw

Rays are best for showing direction: reflection, refraction and lenses. Wavefronts are best for showing wavelength and for anything where the wave nature matters: diffraction and interference. Exam questions ask for both.

When you draw wavefronts, keep them evenly spaced in one medium: the spacing is the wavelength, and it does not change unless the speed does.

Where two media meet

Reflection, refraction and transmission.

Part of the wave is reflected: same angle to the normal
Part is transmitted into the new medium
If its speed changes, it changes direction: refraction
All angles are measured from the normal

ONE RAY IN, TWO RAYS OUT

C.3 WAVE PHENOMENA · SL + HL

Both happen at once

Look at a shop window: you see the goods inside by transmitted light and your own reflection by reflected light. The energy of the incoming wave is shared between the two. For light meeting glass head-on, about 4% is reflected and 96% transmitted; at glancing angles much more is reflected.

Which way it bends

Going into a medium where it travels more slowly, the ray bends towards the normal. Going into one where it travels faster, it bends away. A ray along the normal is not bent at all, though its speed still changes.

Why the normal

Angles could be measured from the surface, but for a curved surface, a lens or a raindrop, “the surface” has no single direction. The normal at the point where the ray lands always does.

Measuring angles from the surface. If a question gives the angle between a ray and the surface, subtract it from 90° first.

Wavefronts at a boundary

Why a wave refracts.

The frequency cannot change: it is set by the source
Slower medium: \(\lambda = \dfrac{v}{f}\) is shorter
Each wavefront must stay joined across the boundary, so it has to turn

INTO A SLOWER MEDIUM

C.3 WAVE PHENOMENA · SL + HL

Why the frequency is fixed

Count the crests arriving at the boundary each second, and the crests leaving it. They must be equal: crests cannot pile up at the surface or be created there. So the frequency is the same on both sides, and if the speed drops, the wavelength must drop in the same ratio.

Why the wave turns

Follow one wavefront as it crosses at an angle. The end that reaches the boundary first slows down first, while the rest of the front is still traveling at full speed. That part gets ahead, and the wavefront swings round to face more nearly into the new medium. A line of soldiers marching at an angle from a road into mud turns for the same reason.

The geometry

Along the boundary the wavefronts on the two sides meet point for point. The distance between those meeting points is \(\frac{\lambda_{1}}{\sin \theta_{1}}\) on one side and \(\frac{\lambda_{2}}{\sin \theta_{2}}\) on the other, and the two must be equal. That is Snell’s law.

Asked what changes when light enters glass: speed and wavelength both fall; frequency stays the same. The color, which depends on frequency, does not change.

How much it bends

Snell’s law.

\(\dfrac{n_{1}}{n_{2}} = \dfrac{\sin \theta_{2}}{\sin \theta_{1}} = \dfrac{v_{2}}{v_{1}}\)

\(n\): the refractive index of a medium, \(n = \dfrac{c}{v}\)
Equivalent form: \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\)

Medium (approximate, for light)	n
Vacuum	1 exactly
Air	1.00
Water	1.33
Glass	1.50
Diamond	2.42

C.3 WAVE PHENOMENA · SL + HL

What the refractive index means

It says how many times slower light travels in the medium than in a vacuum. In glass of index 1.50, light travels at \(\frac{3.00 \times 10^{8}}{1.50} = 2.00 \times 10^{8}\) m s⁻¹. A larger index means slower light and more bending. For visible light, every transparent material has an index greater than 1.

Reading the data booklet’s form

The fraction looks upside down at first: \(\frac{n_{1}}{n_{2}}\) equals \(\frac{\sin \theta_{2}}{\sin \theta_{1}}\). The pattern is that in each medium, the product \(n \sin \theta\) is the same. Where \(n\) is bigger, \(\sin \theta\) and the angle are smaller. Many students find \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\) harder to get wrong.

Not only for light

The right-hand part, \(\frac{\sin \theta_{2}}{\sin \theta_{1}} = \frac{v_{2}}{v_{1}}\), holds for any wave: water waves slowing in the shallows, sound passing from warm air to cold, earthquake waves in rock.

Subscript 1 is the medium the wave comes from, subscript 2 the medium it goes into. Label the diagram before substituting.

Worked example

Light entering glass.

Air (\(n_{1} = 1.00\)) into glass (\(n_{2} = 1.50\)) \(\theta_{1} = 40^{\circ}\) \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\) \(\sin \theta_{2} = \dfrac{n_{1} \sin \theta_{1}}{n_{2}} = \dfrac{1.00 \times \sin 40^{\circ}}{1.50}\) \(\sin \theta_{2} = 0.429\) \(\theta_{2} = \class{res}{25^{\circ}}\)

The speed in the glass \(v = \dfrac{c}{n} = \dfrac{3.00 \times 10^{8}}{1.50}\) \(v = \class{res}{2.00 \times 10^{8} \,\mathrm{m} \,\mathrm{s}^{-1}}\) Light of 600 nm in air is 400 nm in the glass

C.3 WAVE PHENOMENA · SL + HL

Is the answer sensible?

Light is going into the slower medium, so it should bend towards the normal: the angle should be smaller than 40°. It is. Make this check every time; it catches an upside-down fraction immediately.

Coming out again

At the far side of a parallel-sided block the ray meets the glass–air surface at 25° and, by the same equation run backwards, leaves at 40°. It emerges parallel to its original direction but shifted sideways.

Measuring a refractive index

Shine a ray into a glass block at several angles and trace the rays. A graph of \(\sin \theta_{1}\) against \(\sin \theta_{2}\) is a straight line through the origin whose gradient is the refractive index of the glass.

Dividing the angles instead of their sines. \(\frac{40}{1.5} = 26.7^{\circ}\) is close here, but the error grows quickly at larger angles.

Going from slow to fast

Total internal reflection.

\(\sin \theta_\mathrm{c} = \dfrac{n_{2}}{n_{1}}\)

Only on the way into a medium of lower refractive index
Only beyond the critical angle \(\theta_\mathrm{c}\)
Then all the light is reflected

LIGHT INSIDE GLASSGLASS TO AIR: 42°

C.3 WAVE PHENOMENA · SL + HL

Where the critical angle comes from

Leaving glass for air, the ray bends away from the normal, so the angle in the air is always the larger. Increase the angle in the glass and the angle in the air reaches 90° first: the refracted ray skims the surface. Put \(\theta_{2} = 90^{\circ}\) in Snell’s law, \(n_{1} \sin \theta_\mathrm{c} = n_{2} \sin 90^{\circ}\), and \(\sin \theta_\mathrm{c} = \frac{n_{2}}{n_{1}}\). Beyond that angle, Snell’s law would need a sine greater than 1, which is impossible: there is no refracted ray.

“Total” means total

Below the critical angle some light is reflected and some escapes. Beyond it, all of the light is reflected. No mirror does as well, which is why prisms, not mirrors, turn the light in binoculars.

Optical fibers

A thin glass fiber is coated with glass of slightly lower refractive index. Light sent in at one end meets the wall at a large angle, beyond the critical angle, and is totally reflected again and again along the whole length. Fibers carry telephone and internet signals under the oceans, and let surgeons see inside the body.

Expecting total internal reflection when light goes from air into glass. It can only happen on the way out of the slower medium.

Worked example

Two critical angles.

Glass (\(n_{1} = 1.50\)) to air (\(n_{2} = 1.00\)) \(\sin \theta_\mathrm{c} = \dfrac{n_{2}}{n_{1}} = \dfrac{1.00}{1.50}\) \(\theta_\mathrm{c} = \class{res}{42^{\circ}}\)

Water (\(n_{1} = 1.33\)) to air \(\sin \theta_\mathrm{c} = \dfrac{1.00}{1.33}\) \(\theta_\mathrm{c} = \class{res}{49^{\circ}}\) A larger index: a smaller critical angle

C.3 WAVE PHENOMENA · SL + HL

What a diver sees

Looking up from under water, a diver sees the whole world above the surface squeezed into a bright circle overhead, a cone of half-angle 49°. Outside that circle the surface acts as a mirror and shows a reflection of the bottom.

Why diamonds sparkle

Diamond has a refractive index of 2.42, so its critical angle is only \(\sin^{-1} \frac{1}{2.42} = 24^{\circ}\). Light that gets in is likely to strike the inside faces beyond that angle and be reflected several times before it finds a way out. The stone is cut to make sure it does.

A 45° prism

Light entering the short face of a right-angled glass prism meets the long face at 45°. That is more than 42°, so it is totally reflected and turned through a right angle.

If the second medium is not air, keep its index in the equation: for glass to water, \(\sin \theta_\mathrm{c} = \frac{1.33}{1.50}\), giving 62°.

Quick check

Waves at a boundary.

01What angle does a ray make with a wavefront?

90°

02Light enters water from air. Which of \(v\), \(f\) and \(\lambda\) stays the same?

The frequency

03Speed of light in water, \(n\) = 1.33?

\(\frac{3.00 \times 10^{8}}{1.33} = 2.26 \times 10^{8} \,\mathrm{m} \,\mathrm{s}^{-1}\)

04Going into a slower medium, which way does a ray bend?

Towards the normal

05Two conditions for total internal reflection?

Into lower \(n\); beyond the critical angle

06Critical angle for a medium of \(n\) = 2.00 in air?

\(\sin^{-1} 0.500 = 30^{\circ}\)

PRESS → TO REVEAL EACH ANSWER

Part 02

Diffraction and superposition.

Two things that only waves do

C.3 WAVE PHENOMENA · SL + HL

Spreading out

Diffraction through an aperture.

THE SAME WAVE, TWO GAPSWAVELENGTH UNCHANGED

Diffraction is noticeable when the gap is about the same size as the wavelength, and greater still when it is narrower.

C.3 WAVE PHENOMENA · SL + HL

Why you can hear round a corner but not see round it

Speech has wavelengths of about a meter, the size of a doorway, so sound spreads right round the frame and fills the next room. The wavelength of light is about one two-millionth of the width of the doorway, so it passes almost straight through and casts a sharp-edged patch on the far wall. Both are waves; the difference is the ratio of wavelength to gap.

What does not change

The medium is the same on both sides of the barrier, so the speed, the frequency and the wavelength are all unchanged. Only the shape of the wavefronts, and the direction in which the energy spreads, is different. In a sketch, keep the wavefront spacing the same before and after the gap.

Less gets through a narrow gap

The wave beyond a small gap spreads widely but is weaker: less of the incoming wavefront passes through, and its energy is spread over a larger arc.

Drawing the wavefronts closer together after the gap. That would mean a shorter wavelength, which only a change of medium can cause.

Into the shadow

Diffraction round a body.

A wave passing an edge bends into the region behind it
Longer wavelength: more bending
A body much smaller than the wavelength hardly casts a shadow at all

PAST THE EDGE OF AN OBSTACLE

C.3 WAVE PHENOMENA · SL + HL

Radio in the hills

Long-wave radio, with wavelengths of a kilometer or more, bends over hills and round buildings and can be received in a valley. FM radio and television use wavelengths of a few meters and less, which diffract far less round a hill, so a valley may need its own relay transmitter. Mobile phone signals, at about 10 cm, are shadowed even by buildings.

Looking at small things

A wave cannot give a sharp image of anything smaller than about its own wavelength: it simply bends round the object and closes up behind it. This sets the limit of an optical microscope at around half a micrometer. To see atoms you need a wave of much shorter wavelength: X-rays, or the electrons of E.2.

Each point on a wavefront acts as a source of small secondary waves, and these add up to give the next wavefront. Near an edge some of them are missing, and the rest spread sideways. This idea is due to Huygens, of the pendulum clock in C.1.

When waves meet

The principle of superposition.

Where two waves overlap, the displacement is the sum of their separate displacements
Displacements have signs: up and down cancel
Afterwards each wave carries on unchanged

TWO PULSES ON A STRING

C.3 WAVE PHENOMENA · SL + HL

Waves pass through each other

Two balls thrown at each other collide and bounce off. Two pulses on a string meet, add together for a moment and then separate, each exactly as it was. This is how you can follow one conversation in a room full of them, and how the beams of two torches cross without deflecting each other.

An upward and a downward pulse

At the instant they overlap completely, the string is flat: the two displacements add to zero. The energy has not vanished. The string is flat but moving fast: at that instant all the energy is kinetic, and a moment later the two pulses reappear.

How to add them

Work point by point along the string. At each position, add the two displacements, taking upwards as positive and downwards as negative. Exam questions give two pulse shapes on a grid and ask for the resultant at a stated time.

Move each pulse the correct distance first (speed × time), then add. Most errors come from adding the pulses where they started.

Superposition of continuous waves

Constructive and destructive.

IN PHASECONSTRUCTIVE

IN ANTIPHASEDESTRUCTIVE

C.3 WAVE PHENOMENA · SL + HL

Crest on crest, crest on trough

If two waves of the same frequency arrive at a point in phase, their crests coincide and so do their troughs: the resultant has twice the amplitude. If they arrive in antiphase, half a cycle apart, each crest of one meets a trough of the other and the resultant is zero. Anything in between gives something in between.

Energy

The energy of a wave goes as the square of its amplitude. Twice the amplitude is four times the energy, not twice. That does not break conservation of energy: where waves from two sources interfere constructively in some places, they interfere destructively in others. The energy is redistributed, not created.

Noise-canceling headphones use the right-hand picture. A microphone picks up the noise, and the earpiece emits the same wave inverted, so that the two cancel at your eardrum.

Complete cancellation needs equal amplitudes. If they differ, the minimum is the difference of the two amplitudes, not zero.

Quick check

Diffraction and superposition.

01When is diffraction through a gap noticeable?

Gap about the wavelength, or narrower

02Does diffraction change the wavelength?

No

03State the principle of superposition.

Resultant displacement = sum of the displacements

04A +3 cm pulse meets a −1 cm pulse. Displacement where they overlap?

+2 cm

05Two equal waves meet in antiphase. Resultant amplitude?

Zero

06Why does sound bend round a doorway but light does not, noticeably?

Its wavelength is comparable with the gap

PRESS → TO REVEAL EACH ANSWER

Part 03

Interference.

A steady pattern from two sources in step

C.3 WAVE PHENOMENA · SL + HL

Double-source interference

The pattern of two sources.

Along some lines the waves always meet in phase
Between them, always in antiphase
The pattern stays put only if the sources are coherent

Coherent: a constant phase difference, which needs the same frequency.

TWO SOURCES IN STEP

C.3 WAVE PHENOMENA · SL + HL

Why coherence matters

At a given point the waves from the two sources arrive with a certain phase difference. If the sources keep a constant phase relationship, that difference never changes, and the point is permanently bright, or permanently dark. If the sources drift in and out of step, each point flickers between the two faster than anything can detect, and all you see is an even average.

Why two lamps give no fringes

A lamp emits light in short bursts from billions of atoms, each with its own random phase. Two separate lamps have no fixed phase relationship for longer than about a hundred-millionth of a second. The classic way round this is to take light from one source and split it in two: whatever the source does, both halves do together. Two loudspeakers wired to the same signal generator are coherent for the same reason.

“Coherent” does not mean “in phase”. Two sources exactly in antiphase are coherent: their phase difference is constant, at \(\pi\). The bright and dark lines simply swap places.

Bright or dark?

Path difference.

Constructive \(\text{path difference} = n \lambda\) Destructive \(\text{path difference} = \left(n + \tfrac{1}{2}\right) \lambda\) \(n = 0, 1, 2, \dots\) · for sources in phase

TWO ROUTES TO ONE POINT

C.3 WAVE PHENOMENA · SL + HL

Why whole wavelengths give brightness

The two waves leave their sources in step. If one has to travel exactly one wavelength further, it arrives one full cycle late, which is the same as arriving on time: crest still meets crest. The same is true for two, three or any whole number of extra wavelengths. If the extra distance is half a wavelength, or one and a half, it arrives half a cycle late: crest meets trough.

A worked case

Two loudspeakers in phase emit a 1000 Hz note, wavelength 0.34 m. A listener is 3.20 m from one and 3.71 m from the other. The path difference is 0.51 m, which is \(\frac{0.51}{0.34} = 1.5\) wavelengths: destructive interference. The sound there is quiet.

The central line

Every point equidistant from the two sources has zero path difference (\(n = 0\)) and is a maximum.

Method: find the path difference, divide by the wavelength, then look at the answer. A whole number: constructive. A whole number and a half: destructive.

The story · London, 1801

Background story · not assessed

Thomas Young against Newton.

Painted portrait of Thomas Young, a man with short dark hair in a dark coat and white cravat — **Thomas Young**, after Thomas Lawrence

Engraving of two sets of circular waves spreading from points A and B and crossing, with lines of cancellation reaching the right-hand edge at points labeled C, D, E and F — His drawing of waves from **two sources**, A and B

For a century Newton’s view had ruled: light is a stream of particles
In 1801 Young, a London doctor, showed interference with water waves and argued that light does the same. The two-slit experiment is in his lectures of 1807
He was attacked in print and went back to medicine. Fresnel’s work won the argument after 1817

Portrait: after Thomas Lawrence, before 1830, public domain. Drawing: Thomas Young, public domain.

C.3 WAVE PHENOMENA · SL + HL

Interference of light

Young’s double slit.

\(s = \dfrac{\lambda D}{d}\)

\(s\): the separation of neighboring fringes
\(d\): the separation of the slits
\(D\): the distance from slits to screen

ONE SOURCE, TWO SLITSNOT TO SCALE

C.3 WAVE PHENOMENA · SL + HL

Why two slits work when two lamps do not

Both slits are lit by the same wavefront from one source, so the light leaving them is coherent. Each slit is narrow enough to diffract the light into a wide fan, and where the two fans overlap on the screen they interfere.

Where the equation comes from

The first bright fringe away from the center is where the path difference is one wavelength. With the screen far away, that happens at a small angle \(\theta\) where \(d \sin \theta = \lambda\), so \(\theta \approx \frac{\lambda}{d}\). On a screen at distance \(D\) that angle corresponds to a distance \(D \theta = \frac{\lambda D}{d}\) from the center. Each further fringe needs one more wavelength, so they are evenly spaced.

Reading the equation

A longer wavelength spreads the fringes: red fringes are wider apart than blue. Slits closer together spread them too, which surprises people. A more distant screen gives the pattern more room to open out.

Measure across many fringes and divide: ten fringe spacings can be measured ten times more precisely than one.

Mixing up \(d\) (slit separation, a fraction of a millimeter) and \(D\) (distance to the screen, meters).

Worked example

Fringes from a laser.

\(\lambda = 630 \,\mathrm{nm}\) · \(d = 0.50 \,\mathrm{mm}\) · \(D = 2.0 \,\mathrm{m}\) \(s = \dfrac{\lambda D}{d}\) \(s = \dfrac{630 \times 10^{-9} \times 2.0}{0.50 \times 10^{-3}}\) \(s = \class{res}{2.5 \times 10^{-3} \,\mathrm{m}}\)

Ten fringe spacings measure 21 mm with another laser \(\lambda = \dfrac{s d}{D} = \dfrac{2.1 \times 10^{-3} \times 0.50 \times 10^{-3}}{2.0}\) \(\lambda = \class{res}{5.3 \times 10^{-7} \,\mathrm{m}}\) 525 nm: green

C.3 WAVE PHENOMENA · SL + HL

A ruler for light

The second column is how the wavelength of light was first measured. A wavelength of half a thousandth of a millimeter cannot be measured directly, but the experiment magnifies it by the factor \(\frac{D}{d}\), here 4000, into a fringe spacing of millimeters that an ordinary ruler can handle.

With white light

Each color makes its own set of fringes with its own spacing. All colors agree at the center, which is white. Further out they fall out of step: the first bright fringe on each side has a blue inner edge and a red outer edge, and after a few fringes the colors overlap into a blur.

Three different prefixes in one line: nm, mm and m. Convert everything to meters before substituting.

What the screen shows

The double-slit pattern.

Bright fringes where the path difference is \(n \lambda\)
Dark fringes half way between
Equally spaced
For very narrow slits: all about equally bright

INTENSITY ACROSS THE SCREENIDEAL SLITS

C.3 WAVE PHENOMENA · SL + HL

The shape between the peaks

The intensity does not jump from bright to dark. As you move across the screen the phase difference between the two waves changes smoothly, so the resultant amplitude does too, and the intensity follows a smooth curve from a maximum down to zero and back.

How bright is bright

At a maximum the amplitude is twice that from one slit, so the intensity is four times as great. At a minimum it is zero. Averaged across the screen, that is twice the intensity of one slit: exactly the energy that two slits let through.

Evidence for the wave nature of light

Opening a second slit makes some places on the screen darker than they were with one slit alone. Adding light and getting darkness is something particles could not do, and superposition explains it at once.

Real slits have some width, and the fringes then fade away from the center. Part 4 (HL) explains why.

Quick check

Interference.

01What does “coherent” mean?

A constant phase difference

02Path difference \(2.5 \lambda\), sources in phase. Bright or dark?

Dark: destructive

03The slits are moved closer together. The fringes?

Further apart

04Red light is replaced by blue. The fringes?

Closer together

05\(\lambda\) = 500 nm, \(D\) = 1.0 m, \(d\) = 0.25 mm. Fringe spacing?

\(\frac{500 \times 10^{-9} \times 1.0}{0.25 \times 10^{-3}} = 2.0 \,\mathrm{mm}\)

06Why do two separate lamps not give fringes?

They are not coherent

PRESS → TO REVEAL EACH ANSWER

Part 04 · Additional higher level

Slits and gratings.

The single-slit pattern · how it shapes the double-slit pattern
Multiple slits and diffraction gratings

C.3 WAVE PHENOMENA · ADDITIONAL HL

One slit has a pattern of its own

Single-slit diffraction.

\(\theta = \dfrac{\lambda}{b}\)

\(\theta\): the angle of the first minimum, in radians
\(b\): the width of the slit
Central maximum: twice as wide as the others, and far brighter

ONE SLIT, MONOCHROMATIC LIGHT

C.3 WAVE PHENOMENA · ADDITIONAL HL

Why one slit gives dark bands

Think of the slit as a row of many tiny sources, all in phase. Straight ahead, their waves all travel the same distance and add up: the bright center. Now look in the direction where the wave from one edge of the slit travels exactly one wavelength further than the wave from the other edge. Then the wave from the top edge is half a wavelength out of step with the wave from the middle, and they cancel. The same is true of every source in the top half and its partner in the bottom half. Everything cancels in pairs: the first minimum. That direction is where \(b \sin \theta = \lambda\), and for small angles \(\theta = \frac{\lambda}{b}\).

The other maxima are faint

The first one on each side has less than 5% of the intensity of the central maximum, and they fade quickly after that.

Using this for the double slit. Here \(b\) is the width of one slit; in \(s = \frac{\lambda D}{d}\), \(d\) is the distance between two slits.

The angle comes out in radians. To convert a small angle to a distance on a screen at distance \(D\), multiply: \(y = D \theta\).

Worked example

The width of the central maximum.

\(\lambda = 600 \,\mathrm{nm}\) · slit width \(b = 0.10 \,\mathrm{mm}\) \(\theta = \dfrac{\lambda}{b} = \dfrac{600 \times 10^{-9}}{0.10 \times 10^{-3}}\) \(\theta = \class{res}{6.0 \times 10^{-3} \,\mathrm{rad}}\)

On a screen 2.0 m away First minimum: \(y = D \theta = 2.0 \times 6.0 \times 10^{-3} = 12 \,\mathrm{mm}\) Width of central maximum \(= 2y = \class{res}{24 \,\mathrm{mm}}\) The slit itself is only 0.10 mm wide

C.3 WAVE PHENOMENA · ADDITIONAL HL

Geometry would predict a thin line

If light traveled strictly in straight lines, a 0.10 mm slit would cast a 0.10 mm stripe of light on the screen. The actual patch is 24 mm across, 240 times wider. Diffraction is not a small correction here: it decides the whole pattern.

Why telescopes are large

The same spreading happens when light passes through any opening, including the front of a telescope or the pupil of your eye. A point of light is imaged as a small blur whose angular size is about \(\frac{\lambda}{b}\). A wider opening gives a smaller blur and finer detail, which is one reason astronomers build mirrors many meters across.

Giving 12 mm as the width. The central maximum reaches from the first minimum on one side to the first minimum on the other: twice the distance.

Qualitative

The effect of slit width.

Narrower slit: \(\theta = \dfrac{\lambda}{b}\) is larger, so the pattern is wider
Less light gets through, spread over more screen: much fainter
Wider slit: a narrow, bright pattern

TWO SLIT WIDTHS, THE SAME LIGHT

C.3 WAVE PHENOMENA · ADDITIONAL HL

Squeeze the slit and the light spreads

It runs against intuition: closing a slit down makes the patch of light on the screen wider, not narrower. This is the same rule as in Part 2, the narrower the gap compared with the wavelength the more the wave spreads, now with a number attached.

Why the brightness falls so fast

Halving the width halves the light that gets through, and spreads it over twice the width of screen. The intensity at the center therefore falls to a quarter.

Wavelength does the opposite

For a given slit, red light gives a wider pattern than blue, since \(\theta\) is proportional to \(\lambda\). The course treats the single slit with monochromatic light only.

The slit is rectangular: long and narrow. The pattern spreads in the direction of the narrow dimension.

Two effects at once

The real double-slit pattern.

Fringe spacing: set by the slit separation \(d\)
Fringe brightness: set by the single-slit pattern of width \(b\)
A fringe that falls on a single-slit minimum is missing

The single-slit pattern modulates the double-slit pattern.

SLIT SEPARATION FOUR TIMES THE SLIT WIDTH

C.3 WAVE PHENOMENA · ADDITIONAL HL

Why they multiply

Two-slit interference needs light from both slits to reach a point on the screen. How much light each slit sends in a given direction is decided by its own diffraction pattern. In a direction where a single slit sends nothing, there is nothing to interfere, whatever the path difference. So the intensity is the ideal fringe pattern scaled, point by point, by the single-slit envelope.

Counting fringes

On the slide \(d = 4b\). Bright fringes are at angles \(\frac{n \lambda}{d}\); the first minimum of the envelope is at \(\frac{\lambda}{b} = \frac{4 \lambda}{d}\). So the fourth fringe on each side lands on the minimum and is missing. Inside the central maximum of the envelope there are fringes \(n = 0, \pm 1, \pm 2, \pm 3\): seven in all. In general, with \(\frac{d}{b}\) a whole number, there are \(2 \frac{d}{b} - 1\).

In a sketch, draw the dashed envelope first, then fit equally spaced fringes underneath it.

From two slits to many

Multiple slits.

The same slit spacing: main maxima at the same angles
More slits: the maxima are narrower and brighter
Between them: almost nothing

TWO, FIVE AND TWENTY SLITS

C.3 WAVE PHENOMENA · ADDITIONAL HL

Why the positions do not move

A main maximum is a direction in which light from each slit is exactly a whole number of wavelengths ahead of the light from its neighbor. If that is true for one pair of neighboring slits it is true for every pair, however many there are. The condition involves only the spacing \(d\).

Why they sharpen

With two slits, moving slightly off the maximum changes the phase difference a little and the intensity falls gently. With a thousand slits, a tiny phase difference between neighbors builds up along the row: the first and the five-hundredth slit are soon half a cycle apart and cancel, as do the second and the five-hundred-and-first, and so on. A very small change of angle is enough to destroy the maximum.

Why they brighten

At a maximum, \(N\) slits give an amplitude \(N\) times that of one, and an intensity \(N^{2}\) times as great.

Between two main maxima there are \(N - 2\) faint extra maxima. With many slits they are too weak to see.

The story · Munich, 1814–1821

Background story · not assessed

The glassmaker’s apprentice.

Portrait of Joseph von Fraunhofer, a young man with curly hair in a dark coat with a high collar — **Joseph von Fraunhofer**

Orphaned at 11 and apprenticed to a harsh glassmaker
In 1801 the workshop collapsed on him. The prince who led the rescue gave him books and time to study
1814: he mapped 574 dark lines in the spectrum of the Sun
1821: he made a diffraction grating and was the first to measure the wavelengths of spectral lines with one

He died of tuberculosis in 1826, aged only 39.

Portrait: unknown artist, 19th century, public domain.

C.3 WAVE PHENOMENA · ADDITIONAL HL

Thousands of slits

The diffraction grating.

\(n \lambda = d \sin \theta\)

\(d\): the spacing of neighboring slits
\(n\): the order of the maximum: 0, 1, 2 …
\(\theta\): its angle from the straight-through direction

LIGHT AT NORMAL INCIDENCE

C.3 WAVE PHENOMENA · ADDITIONAL HL

The equation is a path difference

Light leaving two neighboring slits at an angle \(\theta\) differs in path by \(d \sin \theta\). A maximum needs that to be a whole number of wavelengths: \(d \sin \theta = n \lambda\). It is the constructive interference condition of Part 3, written with the angle.

Finding \(d\)

Gratings are labeled with the number of lines in each millimeter. With 300 lines per millimeter, \(d = \frac{1}{300}\) mm \(= 3.33 \times 10^{-6}\) m.

No small-angle shortcut

The beams from a grating leave at large angles, tens of degrees, so \(\sin \theta\) must be kept as it is. That is the difference from the double slit, where the fringes are close together and \(s = \frac{\lambda D}{d}\) is good enough.

How many orders

\(\sin \theta\) cannot exceed 1, so \(n\) cannot exceed \(\frac{d}{\lambda}\). The highest order is that number rounded down.

Using the number of lines per millimeter as \(d\). \(d\) is its reciprocal, converted to meters.

Worked example

Sodium light on a grating.

300 lines per mm · \(\lambda = 589 \,\mathrm{nm}\) \(d = \dfrac{1.0 \times 10^{-3}}{300} = 3.33 \times 10^{-6} \,\mathrm{m}\) \(\sin \theta = \dfrac{n \lambda}{d} = \dfrac{1 \times 589 \times 10^{-9}}{3.33 \times 10^{-6}}\) \(\sin \theta = 0.177\) \(\theta = \class{res}{10^{\circ}}\) for the first order

The highest order \(n \le \dfrac{d}{\lambda} = \dfrac{3.33 \times 10^{-6}}{589 \times 10^{-9}} = 5.7\) Highest order: \(\class{res}{n = 5}\) Eleven beams in all: 5 each side and the center

C.3 WAVE PHENOMENA · ADDITIONAL HL

The orders are not evenly spaced in angle

The second order has \(\sin \theta = 0.354\), so \(\theta = 20.7^{\circ}\), about double the first. But the fifth has \(\sin \theta = 0.884\) and \(\theta = 62^{\circ}\), far more than five times 10°. Equal steps in \(\sin \theta\) are not equal steps in \(\theta\).

Measuring a wavelength

The grating does this far better than the double slit. Its maxima are sharp and bright, and the angles are large enough to measure accurately with a spectrometer. Measure the angle between the two first-order beams, halve it, and use \(\lambda = d \sin \theta\). The yellow light of sodium turns out to be two lines, at 589.0 nm and 589.6 nm, and a good grating separates them.

Always round the highest order down. An order of 5.7 does not exist, and 6 would need \(\sin \theta > 1\).

Spectra

White light through a grating.

ON THE SCREENNORMAL INCIDENCE

\(n = 0\): every wavelength goes straight on, so the center is white
Other orders: \(\sin \theta \propto \lambda\), so red is deviated more than violet, and each order is a spectrum

C.3 WAVE PHENOMENA · ADDITIONAL HL

Why the center is white

For \(n = 0\) the equation gives \(\theta = 0\) whatever the wavelength: all the colors land in the same place and recombine.

Higher orders are wider, and overlap

In each order the spread in \(\sin \theta\) between violet and red is proportional to \(n\), so the second-order spectrum is at least twice as wide as the first, and more than that at large angles. Third-order violet (\(3 \times 400 = 1200\) nm) lies at a smaller angle than second-order red (\(2 \times 700 = 1400\) nm), so from there on the spectra run into each other.

A grating and a prism

Both split white light, but in opposite orders: a prism bends violet most, a grating bends red most. A grating’s spreading follows a simple equation, so it can measure wavelengths; a prism’s depends on the properties of the glass.

Several separate wavelengths

Light from a gas lamp contains only certain wavelengths. Each forms its own sharp line in every order: a line spectrum, the fingerprint of the element (E.1).

The surface of a CD or DVD is a reflecting grating: its tracks are about a micrometer apart, which is why it shows colors.

Quick check · HL

Slits and gratings.

01A single slit is made narrower. The central maximum?

Wider and fainter

02\(\lambda\) = 500 nm, \(b\) = 0.050 mm. Angle of the first minimum?

\(\frac{500 \times 10^{-9}}{5.0 \times 10^{-5}} = 0.010 \,\mathrm{rad}\)

03What sets the spacing of the fringes in a real double-slit pattern?

The slit separation \(d\)

04More slits, same spacing. What happens to the maxima?

Same angles; sharper and brighter

05A grating has 500 lines per mm. Slit spacing?

\(2.0 \times 10^{-6} \,\mathrm{m}\)

06In a grating spectrum, which color is deviated most?

Red

PRESS → TO REVEAL EACH ANSWER

Summary

Wave phenomena on one slide.

Refraction

\(\frac{n_{1}}{n_{2}} = \frac{\sin \theta_{2}}{\sin \theta_{1}} = \frac{v_{2}}{v_{1}}\)

Frequency unchanged
Critical angle: \(\sin \theta_\mathrm{c} = \frac{n_{2}}{n_{1}}\)

Diffraction

Spreading at a gap or edge

Greatest when the gap is about one wavelength

Interference

\(n \lambda\) · \(\left(n + \frac{1}{2}\right) \lambda\)

Path difference for constructive and destructive
Coherent sources

Young’s double slit

\(s = \frac{\lambda D}{d}\)

Equally spaced fringes

Single slit

\(\theta = \frac{\lambda}{b}\)

First minimum
It modulates the double-slit pattern

Grating

\(n \lambda = d \sin \theta\)

Sharp, bright maxima
White light: a spectrum in each order

C.3 WAVE PHENOMENA · SL + HL

Summary

Key words.

WavefrontA line joining points in phase

Refractive index\(n = \frac{c}{v}\)

Critical angleAngle of incidence that gives refraction at 90°

DiffractionSpreading of a wave at a gap or edge

SuperpositionDisplacements add where waves overlap

CoherentConstant phase difference

Path differenceExtra distance traveled by one wave

ModulationThe single-slit envelope shaping the fringes

OrderThe whole number \(n\) of a grating maximum

C.3 WAVE PHENOMENA · SL + HL