C.3 Wave phenomena

A single slit gives a broad central maximum with its first minimum at $\theta = \lambda/b$ (HL).

Waves do more than travel in straight lines. A beam of light bends when it enters water, sound reaches a listener round the corner of a building, and two sets of ripples on a pond pass through each other, building tall peaks in some places and leaving calm water in others. These behaviours are wave phenomena: reflection, refraction, diffraction and interference. They occur for every kind of wave, from water ripples and sound to light and radio, which is why they are used to test whether something behaves as a wave.

The same ideas run through modern technology. Optical fibres carry internet traffic across oceans using total internal reflection, noise-cancelling headphones use destructive interference, and spectrometers use diffraction gratings to measure the wavelengths of light from distant stars.

Wavefronts and rays

A wave on a string travels in one dimension, but ripples on water spread across a two-dimensional surface, and sound and light spread through three-dimensional space. Two simple pictures describe these waves: wavefronts and rays.

Wavefronts are usually drawn along successive crests, so neighbouring wavefronts are one wavelength apart. A wavefront moves forward at the wave speed, and in one period it advances by one wavelength.

The two standard shapes appear in the wavefront diagrams. A small source, such as a stone dropped into a pond or a small loudspeaker, produces circular wavefronts in two dimensions (spherical in three), and the rays spread out radially. Far from the source, a small part of a circular wavefront is almost straight, so a distant source produces plane wavefronts, and the rays are parallel. Sunlight reaching the Earth is a good example: the Sun is so far away that its wavefronts are effectively plane.

Both pictures describe the same wave. A ray diagram is often quicker for following a beam through a series of boundaries, while a wavefront diagram shows why the beam changes direction and what happens to the wavelength.

Waves at a boundary: reflection and transmission

When a wave reaches a boundary between two media, such as air and glass or shallow and deep water, some or all of its energy can be sent back (reflected) and some can carry on into the second medium (transmitted). How the energy divides depends on how different the two media are. Light meeting a clean window is mostly transmitted, with a few per cent reflected at each surface, which is why a faint reflection of a room can be seen in a window at night. Light meeting a polished metal surface is almost entirely reflected.

The law of reflection

Angles at a boundary are always measured from the normal, a line drawn perpendicular to the boundary at the point where the ray meets it. The angle between the incident ray and the normal is the angle of incidence $i$; the angle between the reflected ray and the normal is the angle of reflection $r$.

In the reflection diagram the reflected wavefronts are as far apart as the incident ones. The wave stays in the same medium, so its speed is unchanged, and its frequency is set by the source, so its wavelength cannot change either. Reflection changes only the direction of travel.

Reflection is used to locate objects. Bats, sonar and ultrasound scanners send out a pulse and time the echo: if the pulse returns after time $t$, the reflecting surface is a distance $vt/2$ away, the factor of 2 arising because the pulse travels there and back.

Transmission: what changes and what does not

When a wave is transmitted into a new medium, its speed changes, because the speed is set by the properties of the medium. Its frequency stays the same: each crest arriving at the boundary produces exactly one crest on the other side, so the number of crests per second cannot change. Since $v = f\lambda$, a change of speed at constant frequency means the wavelength changes in proportion to the speed:

A wave that slows down therefore has a shorter wavelength in the new medium. Light slows from $3.00\times10^{8}\ \text{m s}^{-1}$ in a vacuum to about $2.0\times10^{8}\ \text{m s}^{-1}$ in glass, so its wavelength in glass is about two-thirds of its wavelength in a vacuum, while its colour, which depends on frequency, does not change. Water waves slow down as they move into shallower water, which is why their crests crowd together near a beach.

Refraction

If a wave crosses a boundary along the normal, it simply slows down (or speeds up) and carries straight on. If it arrives at an angle, it also changes direction. This change of direction caused by a change of speed is refraction.

Why a change of speed bends the wave

Consider a plane wavefront that meets the boundary at an angle. One end of the wavefront enters the slower medium first and slows down, while the rest of the wavefront is still moving at the original speed. The part already across falls behind, so the wavefront swings round, like a line of marching people where those who step onto mud first are slowed while the others keep going. In the slower medium the wavefronts are closer together (shorter wavelength) and the direction of travel is turned towards the normal.

The wavefronts are continuous across the boundary: each one simply bends where it crosses. Going the other way, from a slow medium into a fast one, the wave speeds up, the wavelength increases and the ray bends away from the normal. Refraction does not happen at all if the wave arrives along the normal (angle of incidence 0°): the whole wavefront slows at the same moment, so there is a change of speed and wavelength but no change of direction.

Refractive index

For light, the speed in a material is described by its refractive index, a number that compares it with the speed of light in a vacuum.

Light travels more slowly in every material than in a vacuum, so $n \geq 1$ for all materials. A larger refractive index means slower light; such a medium is often called optically denser. Air has $n = 1.0003$, so close to 1 that it is treated as 1 in calculations. The refractive index of a material depends slightly on the wavelength: violet light travels a little more slowly in glass than red light, which is why a prism spreads white light into a spectrum.

Values are for yellow light at room temperature; they vary by a few parts in a thousand across the visible spectrum.

Snell’s law

The amount of bending is fixed by the ratio of the speeds. For a ray passing from medium 1, where its angle to the normal is $\theta_1$, into medium 2, where its angle is $\theta_2$, the result found experimentally by Willebrord Snell in 1621 is

This is Snell’s law. It is often more convenient in the form $n_1\sin\theta_1 = n_2\sin\theta_2$: the quantity $n\sin\theta$ is the same on both sides of the boundary. The speed form, $\sin\theta_1/\sin\theta_2 = v_1/v_2$, applies to every kind of wave, including sound and water waves, for which refractive index is rarely used.

Refraction through a glass block

A rectangular glass block shows refraction at two boundaries. Light bends towards the normal as it enters and away from the normal as it leaves. Because the two faces are parallel, the angle inside the glass is the same at both faces, so by Snell’s law the light leaves at the same angle to the normal as it entered.

The emergent ray is parallel to the incident ray but shifted sideways. The thicker the block and the larger the angle of incidence, the greater the shift. Some light is also reflected at each face, so a little of the energy never reaches the far side. Measuring pairs of angles in and out of a block like this, and plotting $\sin\theta_1$ against $\sin\theta_2$, gives a straight line through the origin whose gradient is the refractive index of the glass.

Refraction of other waves

Refraction is a property of all waves. Sound travels faster in warm air than in cold air, so on a still evening, when the air near the ground is cooler than the air above it, sound bends back down towards the ground and can be heard much further away than usual. Water waves slow down in shallow water; as waves approach a beach at an angle, the end in shallower water slows first, so the crests swing round until they arrive almost parallel to the shore.

Seismologists and geologists use the refraction of sound and seismic waves at the boundaries between rock layers. Geophones placed along the ground record when waves from a small explosion or a dropped weight arrive, and the pattern of arrival times shows the depth of each layer and the wave speed in it.

Critical angle and total internal reflection

When light travels from a medium with a higher refractive index into one with a lower refractive index, for example from glass into air, it bends away from the normal: the angle of refraction is larger than the angle of incidence. As the angle of incidence increases, the angle of refraction reaches 90° first, and the refracted ray then runs along the boundary. The angle of incidence at which this happens is the critical angle.

For any angle of incidence larger than $\theta_\mathrm{c}$, Snell’s law would need $\sin\theta_2 > 1$, which is impossible, so no wave is transmitted. All the energy is reflected back into the first medium. This is total internal reflection.

Two conditions must both be met for total internal reflection: the wave must be travelling towards a medium in which it is faster (lower refractive index), and the angle of incidence must be greater than the critical angle. Even below the critical angle some light is reflected, but the reflected fraction rises steeply as the critical angle is approached, and beyond it the reflection is total: nearly 100% of the energy returns, more than from the best silvered mirror.

Calculating the critical angle

Setting $\theta_2 = 90^\circ$ (so $\sin\theta_2 = 1$) in Snell’s law, $n_1\sin\theta_\mathrm{c} = n_2\sin 90^\circ$, gives

where $n_1$ is the refractive index of the medium the light is in and $n_2$ that of the medium beyond the boundary ($n_2 < n_1$). When the second medium is air, $\sin\theta_\mathrm{c} = 1/n$. The larger the refractive index, the smaller the critical angle.

Uses of total internal reflection

Diamond has a refractive index of 2.42 and therefore a critical angle of only 24.4°. A cut diamond is shaped so that light entering the top strikes the back faces at more than this angle and is totally reflected back out of the top, which is why diamonds sparkle. Prisms with 45° angles are used in binoculars and periscopes to turn light through 90° or 180° by total internal reflection, because the critical angle of glass (about 42°) is less than 45°.

Optical fibres

An optical fibre is a thin strand of very pure glass, a central core surrounded by a cladding of slightly lower refractive index. Light sent into one end strikes the core–cladding boundary at angles greater than the critical angle, so it is totally internally reflected again and again and travels along the core, even when the fibre bends gently.

Optical fibres carry telephone, television and internet signals as pulses of infrared light. Almost all intercontinental internet traffic travels through optical fibres in undersea cables. Fibres can carry far more information than copper wires and lose much less signal over long distances. In medicine, bundles of fibres in an endoscope carry light into the body and an image back out, so a surgeon can see inside the stomach or a joint without a large incision.

The cladding has three jobs: it keeps the reflections at the core boundary total, it protects the core surface from scratches and dirt that would let light escape, and it stops light crossing between fibres that touch in a bundle.

Diffraction

When a wave passes the edge of an obstacle or goes through a gap (an aperture), it spreads out into the region behind. This spreading is diffraction. A person standing beside an open doorway hears the conversation in the next room even though they cannot see the speakers: the sound diffracts through the doorway, while the light, with a far shorter wavelength, does not noticeably spread.

Diffraction through an aperture

The amount of diffraction depends on the size of the gap compared with the wavelength. When the gap is many wavelengths wide, the wave passes through almost unchanged, with only slight curving at the edges. When the gap is about one wavelength wide, the wave spreads out strongly, and the emerging wavefronts are almost semicircular, as if the gap were a new point source.

In a ripple tank, where water waves of a few centimetres' wavelength pass through gaps between barriers, both cases can be seen side by side by moving the barriers. In every case the wavelength and the frequency are the same after the gap as before: the wave is in the same medium, so its speed has not changed.

A gap that is much smaller than the wavelength also produces strong spreading, but very little of the wave gets through, so the transmitted wave is weak. The strongest noticeable spreading with a useful amount of energy occurs when the gap is comparable with the wavelength.

Diffraction around an obstacle

A wave passing the edge of an obstacle bends round into the region behind it. Behind a sea wall, waves curve round the end of the wall into the harbour. Sound, with wavelengths from a few centimetres to several metres, diffracts readily round buildings, people and doorways.

Radio waves show the role of wavelength clearly. Long-wave radio, at frequencies around 200 kHz, has wavelengths of over a kilometre and diffracts round hills, so it can be received in valleys. FM radio, at about 100 MHz, has wavelengths of about 3 m and diffracts much less, so reception in a deep valley can be poor without a nearby transmitter.

Light does diffract, but only noticeably through very narrow gaps or round very small objects, a fraction of a millimetre across. Light from a laser shone through a slit in a piece of foil, or round a single hair, produces a visible spreading pattern. The fact that light diffracts at all is strong evidence that it is a wave.

Superposition of waves and pulses

Two waves that meet on a string or on a water surface pass through each other and continue unchanged. While they overlap, their displacements add.

The pulse diagrams show the principle of superposition in action. Two upward pulses of amplitude $A$ meeting on a string produce, at the moment they coincide, a single pulse of amplitude $2A$. An upward pulse and an equal downward pulse produce, at the instant they coincide, a flat string. After overlapping, each pulse continues with its original shape, amplitude and speed.

When the displacements add to give a larger resultant, the superposition is called constructive. When they partly or completely cancel, it is destructive. Displacements are added with their signs, so an upward displacement of +3 mm and a downward displacement of −2 mm give a resultant of +1 mm.

Interference from two sources

When two sets of waves overlap continuously, superposition produces a pattern of places where the waves always reinforce and places where they always cancel. This is interference. In a ripple tank, two dippers vibrating together create lines of calm water fanning out between lines of large ripples; two loudspeakers playing the same note create loud and quiet regions in a room.

Coherent sources

For a steady pattern, each point must always receive the two waves with the same phase relationship. If the phase difference between the sources kept changing, a point of reinforcement would become a point of cancellation a moment later, and the pattern would shift too quickly to see.

Coherent waves are easy to make for sound and water: two loudspeakers driven by the same signal generator, or two dippers attached to the same vibrating bar. For light it is harder. An ordinary lamp emits light in short, random bursts from countless atoms, so two separate lamps are never coherent: their phase difference changes randomly millions of times a second and the interference pattern averages out to uniform illumination. Coherent light sources are made by splitting the light from one source into two, as in Young’s double slit, or by using a laser, whose light keeps a steady phase over long distances.

Clear interference also needs waves of roughly equal amplitude, so that the minima are close to zero, and the waves must be of the same type (two light waves, or two sound waves).

Path difference and phase difference

Whether two coherent waves meet in phase depends on how much further one has travelled than the other. For sources that oscillate in phase, the difference in the distances from the two sources to a point is the path difference $\Delta x$.

A path difference of one whole wavelength delays one wave by exactly one cycle, so it arrives back in step with the other. A path difference of half a wavelength puts one wave half a cycle behind, so a crest meets a trough. For two sources oscillating in phase:

Here $n$ is any whole number, 0, 1, 2, …. In general, a path difference $\Delta x$ corresponds to a phase difference $\Delta\phi = \dfrac{2\pi\,\Delta x}{\lambda}$: one wavelength of path is one complete cycle, $2\pi$ radians.

Points with other path differences have intermediate amplitudes. If the two sources were in antiphase instead, every condition would swap: a path difference of $n\lambda$ would then give destructive interference and $(n + \tfrac12)\lambda$ constructive.

The two-source pattern shows how these conditions map out in space. The line exactly midway between the sources has zero path difference, so it is a line of maxima. Moving sideways, the path difference grows, passing through $\tfrac12\lambda$ (a minimum), $\lambda$ (the next maximum), and so on. The lines of maxima and minima fan out from between the sources, and they are closer together when the wavelength is shorter or the sources are further apart.

The same reasoning applies to any coherent waves. In a school laboratory, two slits in a metal plate placed in front of a single microwave transmitter act as two coherent sources, and a detector moved across in front of them finds a series of strong and weak signals.

Noise-cancelling headphones

Noise-cancelling headphones use destructive interference. A small microphone on each ear cup picks up the background noise, and electronics generate a sound wave with the same amplitude but opposite displacement (in antiphase). The speaker plays this “anti-noise” together with the music, and at the ear the two waves cancel. The method works best for steady, low-frequency sounds such as the drone of an aircraft engine, whose long wavelengths make the timing easier to match; sudden high-pitched sounds are reduced mainly by the padding of the ear cups.

Young’s double-slit experiment

Light passing through two narrow, closely spaced slits produces a pattern of bright and dark bands, or fringes, on a screen, because the light from the two slits interferes. Thomas Young first demonstrated this in the early 1800s. The experiment was strong evidence that light is a wave, at a time when many scientists followed Newton’s view that light was a stream of particles, and it allowed the wavelength of light to be measured for the first time.

The set-up

Monochromatic light (light of a single wavelength) from a laser falls on two parallel slits a small distance $d$ apart. Each slit diffracts the light, so the light from the two slits spreads out and overlaps. Because both slits are lit by the same beam, they act as coherent sources in phase. On a screen a distance $D$ away, much larger than $d$, a row of equally spaced bright and dark fringes appears.

Before lasers were available, the slits were lit by a lamp shining through a single narrow slit placed in front of the double slit. Light from the single slit diffracts and reaches both of the double slits from the same point, so the two slits receive light with a fixed phase relationship.

At the centre of the screen the two paths are equal, so there is a bright central fringe. Moving out from the centre, the path difference increases. A bright fringe appears wherever it equals a whole number of wavelengths, and a dark fringe wherever it equals a whole number plus a half.

The fringe spacing

The distance between the centres of neighbouring bright fringes (or neighbouring dark fringes) is the fringe spacing $s$. For small angles, which is always the case when $D \gg d$,

where $\lambda$ is the wavelength, $D$ the distance from the slits to the screen and $d$ the separation of the slit centres. The fringe spacing is the same across the pattern, so the fringes are evenly spaced.

The equation shows how to change the pattern. The fringes spread further apart if the wavelength is longer (red light gives wider fringes than blue), if the screen is moved further away, or if the slits are moved closer together. Because $\lambda$ is so small, $d$ must be under a millimetre and $D$ around a metre or more for fringes a few millimetres apart.

White light

With white light, each wavelength produces its own set of fringes with its own spacing. All the colours have a bright fringe at the centre, where the path difference is zero for every wavelength, so the central fringe is white. On either side the fringes are coloured, with blue on the inner edge (shortest $\lambda$, smallest $s$) and red on the outer edge. After a few fringes the different colours overlap so much that the pattern blurs into uniform white.

Single-slit diffraction (HL)

When monochromatic light passes through a single narrow slit of width $b$ and falls on a distant screen, the pattern is not just a blurred patch. There is a broad, bright central maximum, with a series of much fainter, narrower maxima on either side, separated by dark minima.

Why there are dark minima

Each point across the width of the slit acts as a source of waves, and all these waves are in phase as they leave the slit, because a plane wavefront arrives at normal incidence. Straight ahead ($\theta = 0$) all the waves travel the same distance, so they arrive in phase and give the bright central maximum. At an angle $\theta$, waves from different parts of the slit travel different distances.

Consider the direction in which the wave from the bottom edge B travels exactly one wavelength further than the wave from the top edge A, so that $b\sin\theta = \lambda$. Divide the slit into a top half and a bottom half. The wave from A and the wave from the middle point M then differ in path by $\lambda/2$, so they cancel. The same is true for every pair of points a distance $b/2$ apart: the point just below A pairs with the point just below M, and so on across the slit. Every wave is cancelled by a partner, giving zero intensity: the first minimum.

For small angles, $\sin\theta \approx \theta$ in radians, so the first minimum is at

where $\theta$ is measured from the straight-through direction to the first minimum, in radians. Further minima occur at $2\lambda/b$, $3\lambda/b$, … (dividing the slit into 4, 6, … parts that cancel in pairs). The central maximum extends from $-\lambda/b$ to $+\lambda/b$, so its angular width is $2\lambda/b$, twice the width of each side maximum.

The effect of slit width and wavelength

Because $\theta = \lambda/b$, a narrower slit gives a wider pattern, and a longer wavelength gives a wider pattern. Narrowing the slit also lets less light through, so the pattern becomes dimmer as well as wider. With a wide slit ($b \gg \lambda$) the angle $\lambda/b$ is tiny and the light passes almost straight through: this is the limit in which light seems to travel in straight lines.

With white light, each colour has its central maximum centred in the same place, so the centre is white, but red light ($\lambda \approx 700$ nm) spreads about 1.75 times further than violet ($\lambda \approx 400$ nm). The edges of the central maximum are tinged with red, and the side maxima are coloured, blue inside and red outside.

Double-slit fringes under the single-slit envelope (HL)

The simple double-slit picture assumes slits so narrow that each spreads light equally in all directions, giving fringes of equal brightness. Real slits have a width $b$, so the light from each slit is concentrated within its own single-slit pattern. The two slits produce the same single-slit pattern, on top of each other, and the interference fringes are formed only where this light falls.

The result is a double-slit pattern modulated by the single-slit pattern: the fringe positions are set by the slit separation $d$, with spacing $\lambda D/d$, while their brightness follows the single-slit curve, which acts as an envelope. The fringes are brightest near the centre and fade towards the first minimum of the envelope at $\theta = \lambda/b$, then reappear, much fainter, in the side maxima of the envelope.

Missing orders

If an interference maximum falls exactly where the envelope is zero, that fringe vanishes and is called a missing order. The $n$th interference maximum is at $d\sin\theta = n\lambda$, and the first envelope minimum is at $b\sin\theta = \lambda$. These coincide when $n = d/b$. So if $d = 4b$, the fourth-order fringes are missing; if $d = 3b$, the third-order fringes are missing, together with the sixth and ninth, which lie on the second and third envelope minima.

The photograph shows the effect with slits 0.4 mm apart and 0.1 mm wide, so $d/b = 4$: in the central maximum of the envelope there is a group of bright fringes (the central one and three on each side, seven in all), then a gap where the fourth-order fringe should be, then fainter groups.

Multiple slits and diffraction gratings (HL)

A diffraction grating is a plate with a very large number of equally spaced parallel slits, usually ruled or printed lines, often several hundred per millimetre. The distance between the centres of neighbouring slits is the grating spacing $d$.

The grating equation

Monochromatic light arrives at normal incidence and every slit diffracts it widely. In a direction at angle $\theta$, light from each slit travels $d\sin\theta$ further than light from the neighbouring slit. If this extra path is a whole number of wavelengths, light from every slit arrives in phase, and there is a bright maximum:

where $n = 0, 1, 2, \dots$ is the order of the maximum. The zero-order maximum is straight ahead. Unlike the double-slit equation $s = \lambda D/d$, the grating equation uses $\sin\theta$ directly and needs no small-angle approximation: grating maxima are usually at large angles.

Why the maxima are so sharp

With two slits, moving slightly away from a maximum makes the two waves only slightly out of step, so the brightness falls off gradually and the fringes are broad. With many slits, moving even a little away from a maximum makes the extra path add up across the grating: slit 1 and a slit halfway across the grating soon differ by half a wavelength, and the light from all the slits cancels in pairs. The bright maxima therefore become very narrow, with almost complete darkness between them.

The angles of the principal maxima are the same for 2 slits as for thousands, provided $d$ is the same; what changes is their sharpness. More slits also let through more light, so the maxima are much brighter. Sharp, bright, widely separated maxima at precisely measurable angles make the grating far more useful than a double slit for measuring wavelength.

The number of orders

Since $\sin\theta$ cannot exceed 1, the highest order that can be seen satisfies $n \leq d/\lambda$. The number of maxima on either side of the centre is the largest whole number not greater than $d/\lambda$. If $d/\lambda$ is exactly a whole number, that order would be at 90°, along the grating itself, and cannot actually be observed.

White light and several wavelengths

For white light, every wavelength has its zero-order maximum straight ahead, so the central maximum is white. In each higher order the angle depends on the wavelength, so the light is spread into a spectrum, violet (shortest wavelength, smallest angle) nearest the centre and red furthest out. Higher orders are spread over a wider range of angles, so they are more dispersed, but they are also dimmer.

The spectra of different orders can overlap. For a grating with 600 lines per mm, the second-order spectrum runs from $\sin\theta = 0.48$ to 0.84 and the third-order from 0.72, so the red end of the second order overlaps the violet end of the third. The first-order spectrum never overlaps the second, because the red end of the first order is at $700/d$ in $\sin\theta$ and the violet end of the second is at $800/d$.

If the source emits a few separate wavelengths, as a gas discharge lamp does, each order contains a set of sharp lines, one for each wavelength, in the same order of wavelength in every order but spread further apart in the higher orders. This is the basis of the grating spectrometer, used to identify elements by their spectral lines and to measure the composition, temperature and motion of stars. The closely spaced tracks of a CD, about 1.6 µm apart, act as a reflection grating, which is why a CD shows bands of colour.

Whether two orders overlap can be checked without drawing the spectra: compare $\sin\theta$ for the longest wavelength in the lower order with $\sin\theta$ for the shortest wavelength in the higher order. Since both are divided by the same $d$, the comparison does not depend on the grating.

Applying the ideas in exam questions

Paper 1A: refraction into a slower medium

Paper 1A: changing a double-slit experiment

Paper 1B: measuring the wavelength of a laser

A group of students measures the fringe spacing of a double-slit pattern for different slit-to-screen distances $D$. The slit separation is $d = 0.300 \pm 0.005$ mm. For each $D$ they measure the width $w$ of 10 fringe spacings and record it in the table.

Paper 2: an optical fibre

Paper 2: two-source interference of sound

Paper 2: a diffraction grating (HL)

Self-check

• To do:

  I can draw and interpret wavefront and ray diagrams for plane and circular waves.

• To do:

  I can state the law of reflection and measure angles from the normal.

• To do:

  I can explain which quantities change when a wave is transmitted into a new medium, and why the frequency does not.

• To do:

  I can explain refraction using wavefronts and draw wavefront diagrams with a shorter wavelength in the slower medium.

• To do:

  I can define refractive index and use $n = c/v$.

• To do:

  I can use Snell’s law with refractive indices or with wave speeds.

• To do:

  I can derive and use $\sin\theta_\mathrm{c} = n_2/n_1$ and state the two conditions for total internal reflection.

• To do:

  I can describe how an optical fibre guides light.

• To do:

  I can describe diffraction through an aperture and round an obstacle, and say how the gap size compared with the wavelength affects it.

• To do:

  I can apply the principle of superposition to pulses and waves.

• To do:

  I can explain what coherent sources are and why they are needed for a stable interference pattern.

• To do:

  I can use path difference to decide whether interference at a point is constructive or destructive.

• To do:

  I can describe Young’s double-slit experiment and use $s = \lambda D/d$, including the effect of changing $\lambda$, $D$ or $d$.

• To do:

  I can describe the single-slit intensity pattern and use $\theta = \lambda/b$ for the first minimum (HL).

• To do:

  I can explain how slit width and wavelength affect a single-slit pattern (HL).

• To do:

  I can explain how the single-slit envelope modulates a double-slit pattern and find missing orders (HL).

• To do:

  I can use $n\lambda = d\sin\theta$ for a grating, find the highest order and describe grating patterns for white light and for several wavelengths (HL).

1. A

   A

2. B

   B

3. C

   C

4. D

   D

(c) $\dfrac{1}{17}\times100 = 6\%$ ✓ (the same as the percentage uncertainty in $w$, since $s = w/10$)

it has the smallest $w$, so the fixed ±1 mm uncertainty is the largest fraction of the reading ✓