C.3 Wave Phenomena

IB DP PHYSICS Theme C: Wave Behaviour C.3 Wave Phenomena

Revision Notes · Standard and Higher Level Fahad H. Ahmad

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Original notes prepared for the IB Diploma Programme Physics course (first assessment 2025)

What the syllabus requires

C.3 is about what waves do when they meet boundaries, obstacles and each other. Use this list as a final checklist before the exam. Sections flagged (HL) are Higher Level only.

Understanding You should be able to...

Reflection State and apply the law of reflection; describe reflection of pulses at fixed and free ends of strings, including the phase change.

Refraction Explain refraction as a consequence of a change in wave speed; apply Snell's law n1 sin θ1 = n2 sin θ2.

Total internal reflection Calculate critical angles with sin c = n2/n1 and explain applications such as optical fibres.

Diffraction Describe the spreading of waves at apertures and edges, and state when the effect is greatest (aperture ≈ λ).

Superposition and interference Apply the superposition principle; state the coherence condition; use path-difference conditions nλ and (n + ½)λ.

Double-slit interference Describe Young's experiment and apply s = λD/d to fringe patterns.

(HL) Single-slit diffraction Sketch the intensity pattern; use θ = λ/b for the first minimum; explain the modulation of double-slit fringes.

(HL) Diffraction gratings Explain why more slits give sharper maxima; apply nλ = d sin θ; find the maximum order; describe spectroscopy.

(HL) Interference intensity Use I ∝ A2 to compare intensities at maxima and minima.

Exam note: SL candidates need sections 1-6 of these notes. HL candidates need everything, and should expect C.3 to be combined with C.1 (simple harmonic motion), C.2 (wave basics) and C.4 (standing waves) in Paper 2.

1. Reflection

When a wave meets a boundary it cannot cross, it is reflected. The rule is the same for every kind of wave - water, sound, light, pulses on a rope:

angle of incidence = angle of reflection (θi = θr)

Both angles are measured from the normal - the line perpendicular to the surface at the point of incidence

• and the incident ray, reflected ray and normal all lie in the same plane. Reflection changes the direction of travel but leaves the speed, frequency and wavelength unchanged, because the wave stays in the same medium.

Wavefronts and rays

A wavefront joins neighbouring points that oscillate in phase (for example, all the crests); a ray is an arrow drawn perpendicular to the wavefronts showing the direction of energy travel. The two pictures are equivalent: plane wavefronts hitting a flat barrier at some angle leave the barrier at the same angle on the other side of the normal, exactly as the ray diagram predicts. In exam sketches, always keep wavefronts perpendicular to rays and keep the wavefront spacing (the wavelength) constant on reflection.

Echoes

An echo is simply the reflection of sound from a large surface. Because the sound travels to the reflector and back, the distance to the reflector is d = v t / 2. Sonar, ultrasound imaging and radar all use this round-trip timing idea.

Worked example 1 - echo timing

A student stands facing a large cliff, claps once, and hears the echo 0.60 s later. Taking the speed of sound as 340 m s-1, find the distance to the cliff.

The sound covers the distance twice (there and back) in 0.60 s.

d = v t / 2 = (340 × 0.60) / 2 = 102 m ≈ 1.0 × 102 m.

Forgetting the factor of 2 is the classic error - it doubles the answer.

Reflection of pulses: fixed and free ends

Send a pulse along a rope and watch what returns:

• Fixed end (rope tied to a wall): the reflected pulse is inverted - a crest returns as a trough. There is a phase change of π rad (half a wavelength). Physically, the wall pulls back on the rope with an equal and opposite force (Newton's third law), flipping the pulse.
• Free end (rope ending in a light ring on a pole): the reflected pulse comes back upright - no phase change. The free end overshoots to double amplitude as the pulse arrives.

Why it matters: The π phase change at a fixed (denser) boundary explains the node at the closed end of a string in C.4 standing waves, and the same idea reappears for light reflecting off a denser medium in thin-film interference contexts.

2. Refraction and Snell's law

Refraction is the change of direction of a wave when it crosses a boundary into a medium where its speed is different. The speed change is the cause; the bending is the effect. The frequency cannot change at a boundary (the media are driven at the rate the wave arrives), so from v = fλ the wavelength changes in the same ratio as the speed.

Picture plane wavefronts arriving obliquely at the boundary between deep and shallow water in a ripple tank. The end of each wavefront that enters the shallow (slower) region first is slowed down first, so the wavefront pivots - like a marching band wheeling as one end steps onto muddy ground. The wavefronts end up closer together (shorter λ) and turned toward the normal.

• Entering a slower medium (higher n): the ray bends toward the normal.
• Entering a faster medium (lower n): the ray bends away from the normal.
• Along the normal (θ1 = 0): the wave slows or speeds up but does not bend.

Refractive index and Snell's law

The refractive index of a medium compares the wave speed in the medium with the speed in a reference medium (for light, vacuum):

n=c/v

Because v ≤ c, we have n ≥ 1; a larger n means a slower, optically denser medium. Snell's law connects the angles (measured from the normal) on the two sides of the boundary:

n1 sin θ1 = n2 sin θ2 or equivalently sin θ1 / sin θ2 = v1 / v2 = λ1 / λ2

The second form is the more general one: it applies to any wave, including sound and water waves, where 'refractive index' is not usually quoted.

Medium (for visible light) Refractive index n Speed of light v = c/n

Vacuum 1 (exactly) 3.00 × 108 m s-1

Air 1.0003 ≈ 1.00 3.00 × 108 m s-1

Water 1.33 2.26 × 108 m s-1

Crown glass ≈ 1.5 2.0 × 108 m s-1

Diamond 2.42 1.24 × 108 m s-1

Figure 1. Refraction from air (n1 = 1.00) into glass (n2 = 1.50): the ray bends toward the normal, and the dotted line shows the undeviated path. Snell's law n1 sin θ1 = n2 sin θ2 with θ1 = 30° gives θ2 = arcsin(sin 30° / 1.5) ≈ 19.5°.

Worked example 2 - light entering glass

A ray of light in air strikes a glass block (n = 1.52) at 40° to the normal. Find (a) the angle of refraction, (b) the speed of light in the glass.

• n1 sin θ1 = n2 sin θ2: 1.00 × sin 40° = 1.52 × sin θ2 sin θ2 = 0.643 / 1.52 = 0.423, so θ2 = 25°. The ray bends toward the normal, as expected for a slower medium.
• v = c / n = 3.00 × 108 / 1.52 = 1.97 × 108 m s-1.

The frequency is unchanged; the wavelength inside the glass shrinks to λ/1.52.

Wavelength and frequency in a medium

Two statements are worth memorising because they anchor every refraction argument: the frequency never changes at a boundary (each wavefront arriving must produce one wavefront leaving), while the speed and wavelength change in the same ratio, since v = fλ. In a medium of refractive index n, light therefore has speed v = c/n and wavelength λmedium = λvacuum/n.

Worked example 3 - sodium light entering water

Sodium light has wavelength 589 nm in air. It enters water (n = 1.33). Find (a) its frequency, (b) its wavelength in the water.

• f = c/λ = 3.00 × 108 / 589 × 10-9 = 5.09 × 1014 Hz - and it keeps this frequency in the water.
• λwater = λ/n = 589 / 1.33 = 443 nm.

Check: v = fλwater = 5.09 × 1014 × 443 × 10-9 = 2.26 × 108 m s-1 = c/1.33, as required.

Exam technique: Angles in Snell's law are always measured from the normal, never from the surface. If a question quotes the angle to the surface, subtract from 90° before doing anything else.

3. Total internal reflection and optical fibres

When light travels from a denser medium toward a less dense one (n1 > n2), it bends away from the normal, so the refracted angle is larger than the incident angle. As the incident angle grows, the refracted ray flattens toward the boundary. At the critical angle c the refracted ray grazes along the boundary (θ2 = 90°). Putting θ2 = 90° into Snell's law:

sin c = n2 / n1 (for a boundary with air: sin c = 1/n)

For incidence beyond the critical angle no refracted ray is possible: 100% of the light is reflected back into the dense medium, obeying the ordinary law of reflection. This is total internal reflection (TIR). Two conditions must both hold:

• the light must be travelling from higher n toward lower n, and
• the angle of incidence must exceed the critical angle: θ > c.

Worked example 4 - critical angles

Calculate the critical angle for (a) water (n = 1.33) and (b) glass (n = 1.50), each against air.

• sin c = 1/1.33 = 0.752, so c = 48.8°.
• sin c = 1/1.50 = 0.667, so c = 41.8°.

Because 45° > 41.8°, a 45°-45°-90° glass prism totally internally reflects light hitting its hypotenuse - the principle behind prism binoculars and bicycle reflectors.

Optical fibres

An optical fibre is a hair-thin thread with a glass core of high refractive index surrounded by cladding of slightly lower refractive index. Light entering the core at a shallow angle strikes the core-cladding boundary beyond the critical angle and is totally internally reflected, again and again, zig-zagging along the fibre with almost no loss even around gentle bends. Digital data travels as pulses of infrared laser light.

Advantages over copper cable: enormous bandwidth (many signals multiplexed), low attenuation (fewer repeater amplifiers), no electromagnetic interference or crosstalk, hard to tap, thin and light. The cladding also protects the reflecting surface and stops light leaking between adjacent fibres in a bundle - the same idea is used in medical endoscopes to see inside the body.

Worked example 5 - critical angle in a fibre

An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.40. Find the critical angle at the core-cladding boundary.

sin c = n2 / n1 = 1.40 / 1.50 = 0.933, so c = 69.0°.

Only rays hitting the wall at more than 69° from the normal (i.e. within about 21° of the fibre axis) are guided - which is why light must be launched nearly parallel to the axis.

Common misconception: TIR cannot happen going from air into glass. Going into a denser medium the ray bends toward the normal, so a refracted ray always exists. Check the direction of travel before using sin c = n2/n1.

4. Diffraction

Diffraction is the spreading of a wave when it passes through an aperture or past the edge of an obstacle. The wave bends into the geometric shadow region where a stream of particles could never go. Like reflection, diffraction changes only the direction of travel: the speed, frequency and wavelength are all unchanged (same medium).

The amount of spreading depends on how the aperture width b compares with the wavelength:

Aperture vs wavelength What emerges b much larger than λ Almost straight-through beam; slight curving only at the edges of the wavefronts.

b comparable to λ (b ≈ λ) Strong spreading - maximum diffraction; wavefronts emerge as near-semicircles.

b smaller than λ The gap acts like a point source; circular wavefronts spread into the whole region beyond, but less energy gets through.

Everyday examples

• You can hear a conversation through an open door without seeing the speakers: audible sound has λ ≈ 0.1-10 m, comparable to the doorway, so sound diffracts strongly around it - but light, with λ ≈ 5 × 10-7 m, passes through in effectively straight lines.
• Long-wavelength radio waves (kilometres) diffract over hills and around buildings; short-wavelength TV and satellite signals need line of sight.
• Water waves entering a harbour mouth spread out in semicircular arcs when the entrance is about one wavelength wide.

Why waves spread: Huygens' picture

A helpful model (and the one HL uses for the single slit in section 7) is Huygens' principle: every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the envelope of those wavelets. A wide gap transmits many wavelets whose envelope is nearly flat, so the beam continues straight; a gap about one wavelength wide transmits essentially a single wavelet, which spreads as a semicircle.

Reflection, refraction and diffraction compared

Phenomenon Speed Wavelength Frequency Direction

Reflection unchanged unchanged unchanged reversed about the normal (θi = θr)

Refraction changes changes (same ratio as speed) unchanged bends toward/away from normal (unless along the normal)

Diffraction unchanged unchanged unchanged spreads into the shadow region

Exam technique: When sketching diffraction at a gap, keep the wavelength (spacing of wavefronts) the same before and after the gap - changing it is one of the most common ways to lose the mark.

5. Superposition and interference

The principle of superposition

When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements. The waves then pass through each other completely unchanged. Displacements in the same direction reinforce; opposite displacements cancel.

Interference and coherence

Interference is the stable pattern of reinforcement and cancellation produced when waves from two sources overlap. A stable pattern needs coherent sources: sources with the same frequency and a constant phase difference. (Incoherent sources still superpose, but the pattern shifts randomly billions of times per second and averages away.) In practice coherence is achieved by driving two sources from one oscillator, or by splitting one wavefront in two, as Young did.

Whether two coherent waves reinforce or cancel at a point depends on the path difference - the difference between the distances from the two sources to that point (for sources oscillating in phase):

Constructive: path difference = nλ Destructive: path difference = (n + ½)λ n = 0, 1, 2, ...

A whole number of wavelengths means the waves arrive crest-on-crest (in phase, phase difference 0, 2π, 4π...); an extra half wavelength means crest-on-trough (antiphase, phase difference π).

Path difference Phase difference Result at that point

0, λ, 2λ, ... = nλ 0, 2π, 4π, ... rad Constructive: maximum, double amplitude

½λ, 1½λ, ... = (n + ½)λ π, 3π, ... rad Destructive: minimum, zero amplitude (equal sources) anything in between between 0 and π Intermediate amplitude

Two-source patterns: water and sound

Two dippers in a ripple tank, driven by the same beam, send out circular waves that overlap in a fan of alternating maxima (lines of double- amplitude water, called antinodal lines) and minima (calm nodal lines). The central line, equidistant from both sources, is always a maximum (path difference zero). Exactly the same experiment works with two loudspeakers fed by one signal generator: walking across the room you hear loud-quiet-loud alternation.

Worked example 6 - two loudspeakers

Two loudspeakers connected in phase to the same 680 Hz signal generator face a listener. The listener stands 3.75 m from one speaker and 2.50 m from the other. The speed of sound is 340 m s-1. Is the sound at this point loud or quiet?

λ = v / f = 340 / 680 = 0.500 m.

Path difference = 3.75 - 2.50 = 1.25 m = 1.25 / 0.500 = 2.5 wavelengths = (2 + ½)λ.

A half-integer number of wavelengths means destructive interference - a quiet point (in a real room, only nearly silent, because the amplitudes from the two speakers are not exactly equal there and reflections intrude).

Link: Superposition is the master idea of Theme C: interference (this topic), beats and standing waves (C.4) are all superposition in different geometries.

6. Young's double-slit experiment

In 1801 Thomas Young produced interference with light - the decisive evidence that light is a wave. Monochromatic light falls on two very narrow slits, separated by a small distance d. Each slit diffracts the light into the region beyond, and because both slits are fed by the same incident wavefront, they behave as two coherent sources. On a screen a distance D away, the overlapping light forms fringes: evenly spaced bright and dark bands.

A bright fringe forms where the path difference from the two slits is nλ; a dark fringe where it is (n + ½)λ. Since the slit separation is tiny compared with the screen distance (d << D), the geometry gives a simple result for the spacing s between the centres of adjacent bright fringes (or adjacent dark fringes - it is the same):

s = λD / d

Symbol Meaning Typical size (light) s fringe spacing: centre of one bright fringe to the next ≈ 1 mm

λ wavelength of the light 4-7 × 10-7 m

D distance from slits to screen ≈ 1-3 m d separation of the two slit centres (NOT the slit width) ≈ 0.1-1 mm

Where the formula comes from

For a point on the screen at distance x from the centre, the path difference between the two slits is approximately d sin θ, and for small angles sin θ ≈ tan θ = x/D. Bright fringes need path difference nλ, so xn = nλD/d: the maxima are equally spaced, one fringe every λD/d. The same reasoning with (n + ½)λ puts the dark fringes exactly halfway between the bright ones.

Worked example 7 - laser fringes

A laser of wavelength 633 nm illuminates two slits 0.25 mm apart. Fringes are observed on a screen 2.0 m away. Find the fringe spacing.

s = λD / d = (633 × 10-9 × 2.0) / (0.25 × 10-3) s = 5.1 × 10-3 m ≈ 5 mm - comfortably visible by eye, which is why laser double-slit demos work so well.

Worked example 8 - measuring a wavelength

In a double-slit experiment with d = 0.40 mm and D = 1.5 m, a student measures the distance across 10 fringe spacings (from fringe 0 to fringe 10) as 18.0 mm. Find the wavelength of the light.

Measuring across many fringes reduces the percentage uncertainty: s = 18.0 / 10 = 1.80 mm.

λ = s d / D = (1.80 × 10-3 × 0.40 × 10-3) / 1.5 = 4.8 × 10-7 m = 480 nm (blue-green light).

Worked example 9 - bright or dark?

Light of wavelength 600 nm passes through slits of separation 0.40 mm; the screen is 2.0 m away. What is observed at a point P on the screen 7.5 mm from the centre of the pattern?

Fringe spacing: s = λD/d = (600 × 10-9 × 2.0)/(0.40 × 10-3) = 3.0 mm.

P lies at 7.5 / 3.0 = 2.5 fringe spacings, so the path difference there is 2.5λ = (2 + ½)λ.

Half-integer path difference means a dark fringe - the third minimum out from the central maximum.

Figure 2. Young's double-slit intensity: evenly spaced bright cos2 fringes with spacing Δy = λD/d. For λ = 633 nm, D = 2.0 m and d = 0.25 mm the spacing is ≈ 5.06 mm.

What happens when you change λ, D or d?

Change Effect on fringe spacing s = λD/d

Increase the wavelength (red instead of blue) s increases - red fringes are wider apart than blue.

Move the screen further away (larger D) s increases, but the fringes become dimmer.

Move the slits closer together (smaller d) s increases - narrower slit separation, wider pattern.

Use water instead of air between slits and screen

λ is shorter in water, so s decreases.

Cover one slit Fringes vanish - only the broad single-slit diffraction patch remains.

Widen both slits (same separation) Brighter fringes, but fewer visible (see HL section 7).

White-light fringes

With white light, every wavelength makes its own fringe pattern with its own spacing. All colours have a maximum at the centre (path difference zero), so the central fringe is white. Either side, the fringes become coloured: blue (shortest λ, smallest s) appears on the inner edge of each fringe and red on the outer edge. After a few fringes the overlapping colours wash out into general illumination - so only a handful of white-light fringes are visible.

Common pitfall: In s = λD/d the two distances on top and bottom are easy to swap. Remember d is the tiny one (slit separation, under a millimetre) and D is the big one (metres to the screen). A fringe spacing that comes out bigger than the screen is a sure sign they were swapped.

7. (HL) Single-slit diffraction

A single slit of finite width b does not produce uniform illumination: it produces a diffraction pattern. Every point across the width of the slit acts as a source of secondary wavelets (Huygens' idea), and these wavelets interfere with each other.

The intensity pattern

• A broad, bright central maximum centred on the straight-through direction.
• Dark minima either side, the first at angle θ = λ/b.
• Weak secondary maxima beyond, roughly half the width of the central maximum and much fainter (the first secondary maximum has only about 5% of the central intensity; the next about 2%).

The first minimum occurs where wavelets from the top half of the slit cancel pairwise with wavelets from the bottom half - each pair differing in path by λ/2. For the small angles typical of light this gives:

first minimum: θ = λ / b (θ in radians; b = slit width)

The central maximum therefore has angular width 2λ/b - twice the width of every other bright band. Two consequences follow directly from θ = λ/b:

• Narrower slit (smaller b): wider pattern, dimmer overall - less energy gets through.
• Longer wavelength (larger λ): wider pattern - red light spreads more than blue.

Worked example 10 - width of the central maximum

Light of wavelength 550 nm passes through a slit of width 0.10 mm and falls on a screen 2.0 m away. Find (a) the angle of the first minimum, (b) the linear width of the central maximum on the screen.

• θ = λ/b = (550 × 10-9) / (0.10 × 10-3) = 5.5 × 10-3 rad (about 0.32°).
• Each first minimum sits a distance Dθ = 2.0 × 5.5 × 10-3 = 11 mm from the centre, so the central maximum is 2Dθ = 22 mm wide.

Note the answer to (a) is in radians directly - no conversion needed for the small-angle formula, but convert to degrees only if asked.

Figure 3. Single-slit diffraction: a sinc2 envelope with a broad central maximum whose first minima lie where sin θ = λ/b. For λ

= 550 nm, b = 0.10 mm and D = 2.0 m the first minima fall ≈ 11 mm from the centre, so the central maximum is ≈ 22 mm wide.

How the single slit modulates the double-slit pattern

Real double slits have finite width, so two effects act at once: the interference of light from the two slits (fine fringes, spacing set by the separation d) and the diffraction of light at each slit (broad envelope, width set by the slit width b). The result is the fine Young fringes drawn inside the single-slit intensity envelope: fringes near the centre are bright, those further out fade, and any fringe that lands exactly on a diffraction minimum is missing. Since d > b always, the fringes are always finer than the envelope. For example, if d = 3b, every third interference maximum is suppressed.

Sketching tip: Draw the broad single-slit envelope first (central hump twice the width of the side humps), then fill it with evenly spaced interference fringes whose heights follow the envelope. Examiners look for: equal fringe spacing, envelope symmetry, and weak secondary humps.

8. (HL) Multiple slits and diffraction gratings

From 2 slits to N slits

Keep the slit separation d fixed and add more slits. The condition for a principal maximum is unchanged - all slits in phase - so the bright fringes stay in the same positions. But between them, N slits give N - 2 tiny secondary maxima and N - 1 minima, and as N grows these mop up the light between principal maxima. The principal maxima become:

• sharper - their width is proportional to 1/N, and
• brighter - peak intensity proportional to N2 (see section 9).

A diffraction grating takes this to the extreme: hundreds of slits ('lines') per millimetre, N in the tens of thousands, giving needle-sharp bright lines on a dark background - ideal for precise wavelength measurement.

The grating equation

Light from neighbouring slits travels an extra path d sin θ to reach a distant point at angle θ from the straight-through direction. All N slits reinforce when this equals a whole number of wavelengths:

nλ = d sin θ n = 0, 1, 2, ... (order of the maximum)

Here d is the grating spacing: for a grating ruled with k lines per metre, d = 1/k. The n = 0 maximum is straight through for every wavelength; higher orders appear symmetrically either side. Note that grating angles are usually large, so the small-angle approximation must NOT be used - work with sin θ itself.

Maximum order

Since sin θ cannot exceed 1, the highest visible order is the largest integer n satisfying n ≤ d/λ. Beyond that, the geometry simply cannot supply the path difference.

Worked example 11 - sodium light on a grating

Sodium light (λ = 589 nm) falls normally on a grating with 600 lines per millimetre. Find (a) the angle of the first-order maximum, (b) the angle of the second-order maximum, (c) the total number of orders visible.

d = 1 / 600 mm = 1.667 × 10-6 m.

• sin θ1 = λ/d = 589 × 10-9 / 1.667 × 10-6 = 0.353, so θ1 = 20.7°.
• sin θ2 = 2 × 0.353 = 0.707, so θ2 = 45.0°.
• d/λ = 2.83, so nmax = 2: orders n = 0, 1, 2 on each side - 5 maxima in total (the pattern is symmetric about n = 0).

Spectroscopy

Because the angle of each maximum depends on λ, a grating spreads light into its component wavelengths: each order n > 0 is a complete spectrum, with red deviated more than violet (opposite to a prism). Measuring θ for a spectral line gives λ directly from nλ = d sin θ. This is how the compositions of stars are read from their emission and absorption lines, how gas discharge lamps are identified, and how instruments from CD-player pickups to astronomical spectrographs select wavelengths. Higher orders are more spread out (better resolution), but dimmer, and adjacent orders may overlap: the red end of the second order can overlap the violet end of the third.

Worked example 12 - do the orders overlap?

White light (400-700 nm) falls normally on a grating with 400 lines per millimetre. Show that the second-order and third-order spectra overlap.

d = 1/400 mm = 2.50 × 10-6 m.

Red end of 2nd order: sin θ = 2 × 700 × 10-9 / 2.50 × 10-6 = 0.560, θ = 34.1°.

Violet end of 3rd order: sin θ = 3 × 400 × 10-9 / 2.50 × 10-6 = 0.480, θ = 28.7°.

The violet of the 3rd order (28.7°) appears at a smaller angle than the red of the 2nd order (34.1°), so the two spectra overlap between these angles.

Check yourself: A grating question that yields sin θ > 1 means that order does not exist - state this rather than forcing a calculator answer. It is a favourite way of testing understanding of nmax.

9. (HL) Intensity in interference patterns

The intensity of a wave (power per unit area, W m-2) is proportional to the square of the amplitude:

I ∝ A2

Apply this to two coherent sources, each alone producing amplitude A and intensity I0 at the screen:

Location Resultant amplitude Intensity

Constructive maximum (path difference nλ) A + A = 2A (2A)2 ∝ 4 I0

Destructive minimum (path difference (n + ½)λ) A-A=0

Average over the pattern - 2 I0 (energy is conserved)

The maxima are four times - not twice - as intense as a single source. No energy is destroyed at the dark fringes; it is redistributed to the bright ones, so the average across the pattern is exactly the 2I0 that two independent sources would give.

With N slits, the amplitudes of all N waves add in phase at a principal maximum, giving amplitude NA and intensity N2I0. Since total energy scales only as N, the maxima must simultaneously become narrower (width ∝ 1/N) - which is precisely why a grating gives sharp, bright lines.

If the two amplitudes are unequal (A1 > A2), the minima are no longer completely dark: the resultant amplitude there is A1 - A2, so Imin ∝ (A1 - A2)2 > 0. This is why real fringe patterns lose contrast away from the centre.

Exam technique: Amplitudes add; intensities do not. Always convert intensity to amplitude (square root), add or subtract the amplitudes, then square to get back to intensity.

10. Common pitfalls

• Measuring angles from the surface instead of the normal in reflection and Snell's law.
• Saying frequency changes on refraction - it never does; speed and wavelength change together.
• Applying total internal reflection when light travels into a denser medium - TIR needs n1 > n2 AND θ > c.
• Mixing up the symbols in s = λD/d: d is the slit separation (small), D the screen distance (large), s the fringe spacing - and confusing d with the slit width b of single-slit work.
• Treating θ = λ/b as if it gave degrees - it gives radians, and it is a small-angle formula; the grating equation nλ = d sin θ is exact and must be used at large angles.
• Forgetting the phase change of π when a pulse reflects from a fixed end (none at a free end).
• Claiming maxima with two sources are twice as intense as one source - amplitude doubles, so intensity quadruples.
• Converting lines per mm to grating spacing incorrectly: 600 lines per mm means d = (1/600) mm = 1.67 × 10-6 m, not 600 × 10-3 m.
• Forgetting the factor of 2 in echo problems - the sound travels there and back.

11. Quick reference

Result Statement

Law of reflection θi = θr, measured from the normal

Pulse reflection fixed end: inverted (π phase change); free end: upright (no phase change)

Refractive index n = c/v; sin θ1/sin θ2 = v1/v2 = λ1/λ2

Snell's law n1 sin θ1 = n2 sin θ2

Critical angle sin c = n2/n1 (= 1/n against air); TIR when θ > c, dense to less dense only

Diffraction spreading at gaps and edges; greatest when aperture ≈ λ

Interference conditions constructive: path difference = nλ; destructive: (n + ½)λ; sources must be coherent

Double slit s = λD/d (d = slit separation)

(HL) Single slit first minimum at θ = λ/b (radians); central max has angular width 2λ/b

(HL) Grating nλ = d sin θ; d = 1/(lines per metre); nmax = largest integer ≤ d/λ

(HL) Intensity I ∝ A2; two equal coherent sources: maxima 4I0, minima 0; N slits: N2I0

12. Test yourself

Attempt these without notes; full answers below. Questions 7-9 are HL only.

• Light strikes the surface of a glass block (n = 1.50) at 55° to the normal. Find the angle of refraction, and state which way the ray bends.
• The refractive index of diamond is 2.42. Calculate the speed of light in diamond and its critical angle against air.
• A pulse travels along a rope toward a wall to which the rope is firmly tied. Describe the reflected pulse, and contrast it with reflection from a free end.
• Two loudspeakers in phase emit sound of frequency 850 Hz (speed of sound 340 m s-1). At a point where the path difference is 0.60 m, is the interference constructive or destructive?
• In a double-slit experiment, light of wavelength 486 nm passes through slits 0.30 mm apart onto a screen 2.4 m away. Find the fringe spacing.
• Double-slit fringes of spacing 2.1 mm are formed with d = 0.50 mm and D = 1.8 m. Find the wavelength. Describe the pattern if the source were white light instead.
• (HL) Light of wavelength 640 nm passes through a single slit of width 0.080 mm. Find the angular position of the first minimum and the width of the central maximum on a screen 1.5 m away.
• (HL) A grating has 300 lines per millimetre. For light of wavelength 650 nm find the angle of the first-order maximum and the highest order observable.
• (HL) Each of two coherent sources alone produces intensity I0 at a point P. What is the intensity at P if the waves arrive (a) in phase, (b) in antiphase, (c) if one source's amplitude is doubled and they arrive in phase?
• Explain why you can hear someone speaking around an open doorway but cannot see them.

Answers

• sin θ2 = sin 55° / 1.50 = 0.819 / 1.50 = 0.546, so θ2 = 33°. The ray bends toward the normal (air to glass = slower medium).
• v = c/n = 3.00 × 108 / 2.42 = 1.24 × 108 m s-1. sin c = 1/2.42 = 0.413, so c = 24.4°. This tiny critical angle traps light inside a cut diamond, causing its sparkle.
• Fixed end: the pulse returns inverted (crest becomes trough) - a phase change of π rad - with the same speed, length and (ideally) amplitude. Free end: the pulse returns upright with no phase change.
• λ = 340/850 = 0.40 m. Path difference = 0.60 m = 1.5λ = (1 + ½)λ: destructive - a quiet point.
• s = λD/d = (486 × 10-9 × 2.4)/(0.30 × 10-3) = 3.9 × 10-3 m ≈ 3.9 mm.
• λ = sd/D = (2.1 × 10-3 × 0.50 × 10-3)/1.8 = 5.8 × 10-7 m ≈ 580 nm. With white light: a white central fringe, then coloured fringes with blue edges nearer the centre and red edges further out, fading into overlap after a few fringes.
• θ = λ/b = 640 × 10-9 / 8.0 × 10-5 = 8.0 × 10-3 rad. Central maximum width = 2Dθ = 2 × 1.5 × 8.0 × 10-3 = 2.4 × 10-2 m = 24 mm.
• d = 1/300 mm = 3.33 × 10-6 m. sin θ1 = λ/d = 650 × 10-9/3.33 × 10-6 = 0.195, so θ1 = 11.2°. d/λ = 5.13, so the highest order is n = 5.
• (a) Amplitudes add: 2A, so I = 4I0. (b) Amplitudes cancel: I = 0. (c) Amplitudes A + 2A = 3A, so I = 9I0.
• Audible sound has wavelengths (roughly 0.1-10 m) comparable to the width of a doorway, so it diffracts strongly and spreads around the opening. Visible light has a wavelength about a factor of 106 smaller than the doorway, so its diffraction is negligible and it travels in effectively straight lines - no line of sight, no image.