Summary to follow. 14 syllabus statements (3 HL) · 37 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 14 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Wavefront
Wavefront
A line (in two dimensions) or surface (in three dimensions) joining neighbouring points of a wave that oscillate in phase, for example all the points on one crest. Wavefronts drawn at successive crests are one wavelength apart, and they move forward at right angles to themselves as the wave travels.
Ray
A line drawn to show the direction in which a wave, and the energy it carries, travels. A ray is perpendicular to the wavefronts at every point it crosses them. A ray is a construction for describing the wave, not a physical object: the wave also exists between the rays that happen to be drawn.
Circular, spherical and plane wavefronts
A point source on a surface, such as a dipper in a ripple tank, produces circular wavefronts; a point source in three dimensions produces spherical wavefronts. In both cases the rays point radially outwards from the source. Far from the source, or from a long straight source, the wavefronts are straight (plane) and the rays are parallel.
Students often think Rays are physical beams that make up the wave, so there is no wave between the rays drawn on a diagram. In fact No. A ray is a line drawn to show the direction in which the wave's energy travels. The wave fills the whole region, including the spaces between the rays that happen to be drawn.
Students often think Waves carry the medium along with them, so a ray traces the path of particles travelling outwards from the source. In fact No. A wave transfers energy, not matter. The particles of the medium oscillate about fixed positions; the ray shows the direction in which the energy travels.
Reflection
Reflection
The change in direction of a wave at a boundary so that it returns into the medium it came from. The angle of reflection equals the angle of incidence, both measured between the ray and the normal to the boundary. The reflected wave has the same speed, frequency and wavelength as the incident wave.
Refraction
The change in direction of a wave when it crosses a boundary obliquely into a medium in which its speed is different. The frequency, set by the source, does not change, so the wavelength changes in proportion to the speed (v = fλ). A wave that slows down bends towards the normal; one that speeds up bends away from it. At normal incidence the speed and wavelength change but the direction does not.
Transmission and partial reflection
At a boundary between two media an incident wave is in general partly reflected and partly transmitted into the second medium, and its energy is shared between the reflected and transmitted waves. The transmitted wave has the same frequency as the incident wave. Total internal reflection is the case in which nothing is transmitted.
Students often think At a boundary a wave is either reflected or transmitted, so light meeting clear glass all passes through and none is reflected. In fact In general it is both. Part of the wave is reflected and part is transmitted; total internal reflection is the only case in which nothing is transmitted.
Students often think Refraction means bending, so light always changes direction on entering a different medium, even when it travels along the normal. In fact No. Along the normal (normal incidence) the speed and wavelength change but the direction does not.
Diffraction
Diffraction
The spreading of a wave into the region behind an obstacle, or beyond the edges of an aperture, where a straight-line (ray) model would predict a shadow. Diffraction changes the shape of the wavefronts but not the wavelength, frequency or speed of the wave.
Conditions for significant diffraction
Diffraction occurs at every edge, but it is significant only when the wavelength is comparable with, or larger than, the size of the aperture or obstacle. For a given wavelength, the narrower the aperture the greater the spreading. Everyday sounds such as speech and music (wavelengths of roughly 0.1 m to 3 m) diffracts noticeably round doorways, pillars and people; visible light (wavelengths of about 4 × 10⁻⁷ m to 7 × 10⁻⁷ m) diffracts noticeably only at very narrow slits and fine edges.
Students often think Light travels only in straight lines and does not diffract; only mechanical waves such as sound and water waves bend round obstacles. In fact Yes. All waves diffract. Visible light has such a short wavelength (about 5 × 10⁻⁷ m) that its diffraction is noticeable only at very narrow openings and fine edges.
Students often think Diffraction depends only on the size of the gap or obstacle, so all waves meeting the same obstacle diffract by the same amount. In fact No. It depends on the size of the gap or obstacle compared with the wavelength.
Wavefront–ray diagram for refraction
Wavefront–ray diagram for refraction
When plane wavefronts meet a boundary obliquely, the part of each wavefront that enters the slower medium first travels less far in the same time, so the wavefront turns. In the slower medium the wavefronts are closer together (shorter wavelength) and the rays are closer to the normal. The angle between a wavefront and the boundary equals the angle between its ray and the normal, because the wavefront is perpendicular to the ray and the boundary is perpendicular to the normal.
Wavefront diagram for diffraction
Plane wavefronts passing through an aperture several wavelengths wide emerge straight in the middle and curved only at the edges. When the aperture is about one wavelength wide or narrower, the emerging wavefronts are almost semicircular. The spacing of the wavefronts, the wavelength, is the same beyond the aperture as before it.
Students often think Waves that squeeze through a gap have their wavelength changed, usually shortened, as they pass through. In fact No. The wave stays in the same medium, with the same speed and frequency, so its wavelength is unchanged.
Students often think A wave that slows down bends away from the normal, and one that speeds up bends towards it. In fact Towards the normal. A wave that speeds up bends away from the normal.
Critical angle
Critical angle
The angle of incidence c, in the medium of higher refractive index n₁, for which the angle of refraction in the medium of lower refractive index n₂ is 90°. From Snell's law, sin c = n₂/n₁. For glass of refractive index 1.50, c = 41.8° at a boundary with air (n = 1.00) and 62.5° at a boundary with water (n = 1.33).
Total internal reflection
Complete reflection of a wave at a boundary, with no transmitted wave. It needs two conditions: the wave travels towards a medium of lower refractive index (lower wave speed to higher wave speed), and the angle of incidence is greater than the critical angle. Optical fibres use it to trap light in a core of higher refractive index than the surrounding cladding.
Students often think Total internal reflection happens when light meets a medium of higher refractive index at a large angle, because the light cannot get into the denser medium. In fact No. It can happen only when light travels towards a medium of lower refractive index, such as from glass towards air.
Students often think The angle of incidence (or of refraction) is the angle between the ray and the surface of the boundary. In fact No. Every angle is measured between the ray and the normal to the boundary.
Refractive index, n
Refractive index, n
The ratio of the speed of light in a vacuum to the speed of light in a medium, n = c/v. It has no unit. The larger the refractive index, the lower the speed of light in the medium. For air, n = 1.00 to three significant figures.
Snell's law
n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁, where θ₁ and θ₂ are the angles between the ray and the normal in media 1 and 2 and v₁ and v₂ are the wave speeds; equivalently n₁ sin θ₁ = n₂ sin θ₂. Because the frequency does not change, the wavelengths are in the same ratio as the speeds: λ₂/λ₁ = v₂/v₁. The speed form applies to all waves, including water and sound waves.
Students often think The refractive index of the medium the light enters decides the change on its own, as if the light always came from air. In fact No. It is the ratio n₁/n₂ that matters; n₁ drops out only when the first medium is air.
Students often think Light travels faster in an optically denser medium (larger refractive index), so the ratios in Snell's law are used the wrong way up. In fact No. n = c/v, so the larger the refractive index, the lower the speed of light.
Principle of superposition
Principle of superposition
When two or more waves or pulses meet at a point, the resultant displacement is the sum of the individual displacements, taking account of their signs (directions). The waves or pulses then pass through each other and continue unchanged, as if the other had not been there.
Wave pulse
A single short disturbance, such as one hump sent along a stretched string, that travels through a medium. When two pulses on a string overlap, a pulse displaced above the rest position and one displaced below it partly or completely cancel while they overlap; afterwards both emerge unchanged.
Students often think Pulses collide and bounce off each other like balls, returning the way they came. In fact They pass through each other. While they overlap their displacements add; afterwards each continues with its original shape and direction.
Students often think Superposition means adding the sizes of the displacements, whatever their directions. In fact No. Displacements add with their signs, so an upward and a downward pulse partly or completely cancel.
Coherent sources
Coherent sources
Sources that emit waves with a constant phase difference, which requires them to have the same frequency. Only coherent sources give a stable interference pattern, whose maxima and minima stay in fixed places. Two independent lamps are not coherent, because the phase of the light from each changes randomly many millions of times a second. Coherent light is obtained by dividing the light from one source, for example with two slits, or by using a laser.
Students often think Waves from sources that are not coherent do not superpose at all; they pass through each other without interfering. In fact Yes. All waves superpose. The light from the lamps does interfere at each instant, but the phase difference changes so rapidly that any pattern shifts far too fast to be seen.
Students often think Sources are coherent if they have the same frequency (the same colour), whatever happens to their phases. In fact No. Coherent sources must have a constant phase difference; the same frequency is necessary but not sufficient.
Path difference
Path difference
The difference between the distances travelled by two waves from their sources to a given point, measured in metres (m). For two sources that oscillate in phase, a path difference of one whole wavelength corresponds to a phase difference of 2π, so the path difference decides whether the waves arrive in phase.
Constructive interference
Superposition of two waves that arrive in phase, giving a resultant amplitude equal to the sum of the individual amplitudes. For two coherent sources oscillating in phase it occurs where path difference = nλ, with n = 0, 1, 2, … . Two waves of equal amplitude A give amplitude 2A and, since intensity is proportional to amplitude squared, four times the intensity of one wave.
Students often think Each wave must travel a whole number of wavelengths to the point for constructive interference, so each path is compared with λ separately. In fact No. It is the path difference that matters: constructive interference occurs where path difference = nλ, whatever the individual path lengths.
Students often think A path difference of an odd number of wavelengths (3λ, 5λ, …) gives destructive interference. In fact No. Any whole number of wavelengths, odd or even, gives constructive interference. Destructive interference needs an odd number of half-wavelengths, (n + ½)λ.
Destructive interference
Destructive interference
Superposition of two waves that arrive in antiphase, giving a resultant amplitude equal to the difference of the individual amplitudes, zero if they are equal. For two coherent sources oscillating in phase it occurs where path difference = (n + ½)λ, with n = 0, 1, 2, … . Energy is not destroyed at points of destructive interference; it is redistributed to the points of constructive interference.
Students often think A path difference of λ corresponds to a phase difference of π (half a cycle), so it puts the waves in antiphase. In fact No. A path difference of λ is a phase difference of 2π; a phase difference of π needs a path difference of λ/2.
Students often think n starts at 1, so the smallest path difference for destructive interference is 1½λ. In fact n = 0, which gives the smallest path difference for destructive interference, λ/2.
Young's double-slit interference
Young's double-slit interference
Monochromatic light passes through two narrow parallel slits whose centres are a distance d apart; the slits act as coherent sources. On a screen a distance D away, with D much larger than d, equally spaced bright and dark fringes appear. The fringe separation s, the distance between the centres of adjacent bright (or adjacent dark) fringes, is s = λD/d, with every length in metres.
Measuring the fringe separation
Because s is only a few millimetres, it is found by measuring across many fringes and dividing by the number of spaces: from the centre of the 1st to the centre of the (N + 1)th bright fringe there are N fringe separations. Increasing D or λ, or decreasing d, increases s.
Students often think The slit width b and the slit separation d play the same role, so the fringe separation can be found from, or changed by, the width of the slits. In fact d, the distance between the centres of the two slits. The width of each slit does not appear in s = λD/d and does not change the fringe separation.
Students often think The fringe separation is the measured distance divided by the number of fringes counted. In fact No. Eleven bright fringes have ten spaces between them, so s = x/10.
Single-slit diffraction pattern HL
Single-slit diffraction pattern
When monochromatic light passes through a single rectangular slit of width b, a distant screen shows a bright central maximum with much weaker subsidiary maxima on either side, separated by dark minima. The central maximum is twice as wide as each subsidiary maximum, and the subsidiary maxima become fainter the farther they are from the centre.
Angle to the first minimum, θ = λ/b
θ is the angle, in radians, between the centre of the central maximum and the first minimum on either side, for a slit of width b (m) and wavelength λ (m). The central maximum spans 2λ/b, so on a screen a distance D away its width is about 2λD/b. The relation uses the small-angle approximation, so θ is in radians.
Effect of slit width on the single-slit pattern
Narrowing the slit makes the central maximum wider (θ = λ/b increases) and less intense, because less light passes through the slit and it is spread over a wider angle. Widening the slit gives a narrower, more intense central maximum. The effect on intensity is treated qualitatively.
Students often think The central maximum of the single-slit pattern is the geometric image of the slit, so its width on the screen equals the slit width b whatever the wavelength or screen distance. In fact No. Its width is set by diffraction: it extends to θ = λ/b either side of the straight-through direction, so on a screen at distance D it is about 2λD/b wide, which depends on λ and D and is far wider than the slit.
Students often think θ = λ/b is the whole angular width of the central maximum, from the first minimum on one side to the first minimum on the other. In fact No. θ is the angle from the centre of the central maximum to the first minimum on one side; the full angular width is 2λ/b.
Modulation of the double-slit pattern HL
Modulation of the double-slit pattern
Each slit of a real double slit has a finite width b, so the light leaving each slit forms its own single-slit diffraction pattern. The observed pattern is the double-slit interference fringes, spaced according to d, with intensities that follow the single-slit pattern (the envelope), set by b: the fringes are brightest near the centre and fade towards the minima of the envelope.
Missing orders
An interference maximum that falls at the same angle as a minimum of the single-slit envelope receives no light and is missing from the pattern. Interference maxima are at sin θ = nλ/d and the first envelope minimum is at sin θ = λ/b, so when d = 3b the third-order fringes are missing and five fringes (n = 0, ±1, ±2) lie within the central maximum of the envelope.
Students often think Double-slit fringes are all equally bright; the finite width of the slits cannot change their relative brightness or remove any fringe. In fact No. The fringe intensities follow the single-slit envelope: bright near the centre, fading towards the envelope's minima, where a fringe can be missing altogether.
Students often think The slit width b sets the spacing of the fringes and the slit separation d sets the width of the envelope. In fact The slit separation d sets the fringe spacing (sin θ = nλ/d); the slit width b sets the envelope (first minimum at sin θ = λ/b).
Diffraction grating and grating spacing d HL
Diffraction grating and grating spacing d
A diffraction grating is a large number of equally spaced parallel slits. The grating spacing d is the distance between the centres of adjacent slits, in metres; for a grating with N lines per millimetre, d = 1/N mm = 1/(N × 10³) m.
Grating equation, nλ = d sin θ
Principal maxima occur at angles θ for which the path difference between waves from adjacent slits is a whole number of wavelengths: nλ = d sin θ, where n = 0, 1, 2, … is the order and n = 0 is the central maximum. Since sin θ cannot exceed 1, the highest order observed is the largest whole number not greater than d/λ. Grating angles are often large, so the small-angle approximation is not used.
Effect of the number of slits
Increasing the number of equally illuminated slits while keeping d the same does not move the principal maxima, whose angles are fixed by nλ = d sin θ, but makes them much sharper (narrower) and brighter. Multiple-slit and grating patterns therefore have sharp, bright maxima separated by wide, almost dark regions, which is why gratings are used to measure wavelengths.
White light and several wavelengths through a grating
Each wavelength produces maxima at its own angles. The central maximum (n = 0) is white, because the path difference there is zero for every wavelength. In each order n ≥ 1 the light is spread into a spectrum with violet (shortest wavelength, smallest θ) nearest the centre and red farthest out. Higher-order spectra are wider and can overlap; for visible light (about 400–700 nm) the second- and third-order spectra overlap. Longer wavelengths give fewer orders, because n cannot exceed d/λ.
Students often think Violet (shorter-wavelength) light is always deviated more than red, so in grating and double-slit patterns red is nearest the centre and shorter wavelengths spread out more. In fact No. A grating deviates red light more than violet, because sin θ = nλ/d increases with wavelength. A prism deviates violet more because glass has a larger refractive index for violet light.
Students often think The central maximum is the first order, so the second-order maximum is the first bright line beside the centre. In fact No. The central maximum is the zero-order maximum, n = 0; the first-order maxima are the next ones on either side.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Circular ripples spread out across a ripple tank from a vibrating dipper. A student sketches a wavefront–ray diagram of the waves. Which statement correctly describes the rays on the diagram?
Answer and reasoning
They are drawn along the crests, parallel to the wavefronts, joining points that are in phase. — A student who treats rays and wavefronts as the same kind of line picks this. A line joining points in phase along a crest is a wavefront. A ray crosses the wavefronts at 90° and shows which way they move.
They trace the paths along which water moves outwards from the dipper as the wave travels. — A student who thinks waves carry the medium with them picks this. The water oscillates about fixed positions, as a floating cork shows; it is the energy that travels outwards along the rays.
They are thin beams that make up the wave, so there is no wave between two drawn rays. — A student who thinks rays are physical beams picks this. Rays are lines drawn to show direction; the ripples fill the whole tank, including the spaces between the few rays that happen to be drawn.
They are perpendicular to the wavefronts and show the direction in which the wave's energy travels. — Rays are lines drawn at right angles to the wavefronts, pointing the way the wave, and its energy, moves. For circular ripples the wavefronts are circles centred on the dipper and the rays point radially outwards from it.
2 A narrow beam of monochromatic light in air strikes the flat surface of a glass block along the normal, so that the angle of incidence is 0°. Which statement about what happens at the surface is correct?
Answer and reasoning
None of the light is reflected; it all passes into the glass. — A student who thinks a wave is either reflected or transmitted picks this. Clear glass still reflects a few per cent of the light at its surface, which is why you can see reflections in a window; the rest is transmitted.
The light entering the glass changes direction as it is refracted. — A student who equates refraction with bending picks this. Refraction is caused by the change of speed; the direction changes only for oblique incidence. With θ₁ = 0, Snell's law gives θ₂ = 0, so the light continues along the normal.
The light entering the glass keeps the frequency it had in air. — The frequency is fixed by the source and does not change at a boundary. The light slows down in glass, so its wavelength decreases (λ = v/f); arriving along the normal, it does not change direction, and a few per cent of it is reflected back along the normal.
The light entering the glass keeps the wavelength it had in air. — A student who thinks only the speed changes at a boundary picks this. The frequency is fixed by the source, so when the speed falls in glass the wavelength must fall in the same proportion (v = fλ).
3 A person standing behind a tree trunk about 1 m wide can clearly hear a street musician on the other side of the trunk but cannot see her. Which statement best explains this?
Answer and reasoning
Sound wavelengths, roughly 0.1 m to 3 m, are comparable with the trunk's width; light's are far smaller. — Diffraction round an obstacle is significant when the wavelength is comparable with the size of the obstacle. Sound of wavelength about a metre bends well into the region behind a 1 m trunk; visible light, of wavelength about 5 × 10⁻⁷ m, hardly spreads at all, so the trunk casts a sharp shadow.
Light travels only in straight lines and cannot diffract; only mechanical waves bend round obstacles. — A student who thinks light cannot diffract picks this. Light does diffract, as a laser shone through a very narrow slit shows. Behind a 1 m trunk the effect is negligible only because light's wavelength is millions of times smaller than the trunk.
Sound and light diffract equally round the same trunk, but the diffracted light is too dim to see. — A student who thinks diffraction depends only on the size of the obstacle picks this. The amount of spreading depends on the wavelength compared with the obstacle, so sound (λ about 1 m) spreads far more round the trunk than light (λ about 5 × 10⁻⁷ m).
Sound diffracts round the trunk only because one of its wavelengths is exactly equal to the trunk's width. — A student who thinks diffraction happens only when the size equals the wavelength picks this. Diffraction is significant for any wavelength comparable with, or larger than, the obstacle; the musician's sounds cover a wide range of wavelengths and all of them bend round the trunk.
Working λ = v/f with v ≈ 340 m s⁻¹: sounds of about 100 Hz to 3 kHz have λ ≈ 3.4 m to 0.11 m, comparable with the 1 m trunk. Visible light has λ ≈ 4 × 10⁻⁷ m to 7 × 10⁻⁷ m, about a million times smaller than the trunk, so its diffraction is negligible.
4 Light meets the boundary between two transparent media. Which statement gives a condition that must be met for total internal reflection to occur?
Answer and reasoning
The light must be travelling towards a medium of higher refractive index. — A student who thinks light is reflected because it cannot get into a denser medium picks this. Light entering a medium of higher refractive index bends towards the normal and is always partly transmitted, whatever the angle; there is no critical angle in that direction.
The light must be travelling towards a medium of lower refractive index. — A critical angle exists only when light travels towards a medium of lower refractive index (for example glass to air), because only then can the angle of refraction reach 90°. This is one of the two conditions; the other is that the angle of incidence, measured from the normal, is greater than the critical angle.
The ray's angle to the surface must be greater than the critical angle. — A student who measures angles from the surface picks this. The critical angle, like every angle in Snell's law, is measured between the ray and the normal. A ray at a large angle to the surface is close to the normal and is mostly transmitted.
The angle of incidence must be equal to the critical angle for the boundary. — A student who takes the critical angle itself as the condition picks this. At exactly the critical angle the refracted ray grazes along the boundary; total internal reflection needs an angle of incidence greater than the critical angle.
5 A ray of light passes from air into a transparent liquid. The incident ray makes an angle of 40.0° with the surface of the liquid and the refracted ray makes an angle of 55.0° with the surface. The speed of light in air is 3.00 × 10⁸ m s⁻¹. What is the speed of light in the liquid?
Answer and reasoning
4.01 × 10⁸ m s⁻¹ — A student who expects light to speed up in the denser liquid inverts the ratio: 3.00 × 10⁸ × sin 50.0°/sin 35.0° = 4.01 × 10⁸ m s⁻¹, faster than light in air. The refracted ray is closer to the normal, so the light has slowed down: v₂/v₁ = sin θ₂/sin θ₁.
3.82 × 10⁸ m s⁻¹ — A student who uses the angles measured from the surface calculates 3.00 × 10⁸ × sin 55.0°/sin 40.0° = 3.82 × 10⁸ m s⁻¹. Snell's law uses angles from the normal, 50.0° and 35.0°.
2.25 × 10⁸ m s⁻¹ — Angles are measured from the normal: θ₁ = 90.0° − 40.0° = 50.0° and θ₂ = 90.0° − 55.0° = 35.0°. From sin θ₂/sin θ₁ = v₂/v₁, v₂ = 3.00 × 10⁸ × sin 35.0°/sin 50.0° = 2.25 × 10⁸ m s⁻¹. The ray bends towards the normal, consistent with slowing down.
2.10 × 10⁸ m s⁻¹ — A student who cancels the sines uses the ratio of the angles: 3.00 × 10⁸ × 35.0/50.0 = 2.10 × 10⁸ m s⁻¹. sin 35.0°/sin 50.0° = 0.749, not 0.700, so v₂ = 2.25 × 10⁸ m s⁻¹.
Working θ₁ = 90.0° − 40.0° = 50.0°; θ₂ = 90.0° − 55.0° = 35.0°. v₂/v₁ = sin θ₂/sin θ₁ ⇒ v₂ = (3.00 × 10⁸ m s⁻¹)(sin 35.0°/sin 50.0°) = (3.00 × 10⁸)(0.574/0.766) = 2.25 × 10⁸ m s⁻¹.
6 In a Young's double-slit experiment using a filament lamp, the light passes through a single narrow slit before it reaches the double slit. What is the purpose of the single slit?
Answer and reasoning
It makes the light monochromatic, because sources of a single frequency are automatically coherent. — A slit does not select a frequency; a filter does that. Equal frequency is necessary but not sufficient for coherence: the sources must also keep a constant phase difference, which is what illuminating both slits from the same narrow single slit achieves.
It turns the lamp's light into laser-like light, without which stable fringes cannot be produced. — A laser is not needed. Young produced fringes with sunlight, and a lamp with a single slit works because the double slit divides one wavefront into two coherent parts. A slit cannot change the nature of the light, only where it comes from.
It lights both slits with the same wavefronts, so the two slits act as coherent sources. — Light from different points of a lamp filament has random, rapidly changing phases. A single narrow slit selects light from a very small region, so each wavefront that spreads from it reaches both slits of the double slit. The two slits then emit waves with a constant phase difference: they are coherent, and a stable fringe pattern forms.
It forms a sharp image of the lamp on the double slit, so that each slit casts a bright image on the screen. — The fringes are not images of the slits: light from each slit spreads by diffraction over the whole screen, and the bright and dark fringes are where the two sets of waves superpose constructively and destructively. The single slit's role is to make the two slits coherent.
7 Two dippers vibrating in phase act as coherent sources S₁ and S₂ on a ripple tank, producing waves of wavelength 2.0 cm. Point P is 31.0 cm from S₁ and 37.0 cm from S₂. What kind of interference occurs at P?
Answer and reasoning
Destructive: the paths are 15.5λ and 18.5λ, not whole numbers of λ — A student who compares each path with the wavelength separately picks this. What matters is the difference between the paths, 18.5λ − 15.5λ = 3λ, a whole number of wavelengths, so the waves arrive in phase.
Destructive: the path difference, 3λ, is an odd number of λ — A student who remembers 'odd multiple' but drops 'of half a wavelength' picks this. Any whole number of wavelengths gives constructive interference; destructive interference needs (n + ½)λ, such as 2.5λ or 3.5λ.
Neither: maxima occur only where P is equidistant from S₁ and S₂ — A student who thinks waves are in phase only after equal distances picks this. A path difference of 3λ delays one wave by exactly three cycles, which leaves the two waves in phase: P lies on the third-order maximum.
Constructive: the path difference is 6.0 cm, which is 3λ — Path difference = 37.0 cm − 31.0 cm = 6.0 cm = 3 × 2.0 cm = 3λ. A whole number of wavelengths means the waves from the in-phase sources arrive in phase, so the interference at P is constructive.
Working Path difference = S₂P − S₁P = 37.0 cm − 31.0 cm = 6.0 cm. 6.0 cm/2.0 cm = 3, so path difference = 3λ = nλ with n = 3 ⇒ constructive interference.
8 In a double-slit experiment, monochromatic light is incident normally on the slits. The amplitude at the centre of each bright fringe is twice the amplitude that one slit alone would produce there, and the amplitude at each dark fringe is zero. What has happened to the energy of the light that does not arrive at the dark fringes?
Answer and reasoning
It has been redistributed to the bright fringes, where the intensity is four times that from one slit. — Intensity is proportional to amplitude squared, so doubling the amplitude gives four times the intensity of one slit alone. The dark fringes receive nothing, and averaged across the pattern the intensity is twice that from one slit, exactly the energy supplied by the two slits. Interference redistributes energy; it does not create or destroy it.
It has been destroyed where the waves cancel, so less energy reaches the screen than passes through the slits. — A student who thinks cancellation destroys energy picks this. Energy is conserved: what is missing from the dark fringes appears at the bright fringes, whose intensity is four times, not twice, that from one slit.
It did not set out towards the dark fringes, which are shadows of the barrier between the two slits. — A student who uses the straight-line model of light picks this. Light from both slits reaches every point on the screen. Covering one slit makes the dark fringes light up, which shows that they are dark because the two waves cancel there.
It has been converted into thermal energy in the screen at the dark fringes, where the two waves cancel. — A student who assumes 'missing' energy must be lost as heat picks this. A zero amplitude means no energy is arriving at a dark fringe, so there is none to be absorbed there; the energy flow is directed to the bright fringes.
9 Monochromatic light of wavelength 600 nm is incident normally on a single rectangular slit of width 0.12 mm. A screen is 2.0 m from the slit. What is the width of the central maximum on the screen, measured between the first minima on either side? HL
Answer and reasoning
1.0 × 10⁻² m — A student who takes θ = λ/b as the full angular width of the central maximum stops at Dθ = 1.0 × 10⁻² m. θ is measured from the centre to the first minimum on ONE side, so the full width is twice this.
2.0 × 10⁻² m — θ = λ/b = (600 × 10⁻⁹)/(0.12 × 10⁻³) = 5.0 × 10⁻³ rad is the angle from the centre to the first minimum on one side. On the screen that is Dθ = 2.0 × 5.0 × 10⁻³ = 1.0 × 10⁻² m, so the full width between the two first minima is 2.0 × 10⁻² m.
3.5 × 10⁻⁴ m — A student who treats θ = 5.0 × 10⁻³ as an angle in degrees calculates 2 × 2.0 × tan(0.0050°) = 3.5 × 10⁻⁴ m. λ/b is a ratio of lengths and gives the angle in radians.
2.0 × 10⁻⁵ m — A student who substitutes b = 0.12 without converting from millimetres gets θ = 5.0 × 10⁻⁶ rad and a width of 2.0 × 10⁻⁵ m. With b = 0.12 × 10⁻³ m the width is 2.0 × 10⁻² m.
Working θ = λ/b = (600 × 10⁻⁹ m)/(0.12 × 10⁻³ m) = 5.0 × 10⁻³ rad (centre to first minimum). Width = 2Dθ = 2 × 2.0 m × 5.0 × 10⁻³ = 2.0 × 10⁻² m.
10 Monochromatic light is incident normally on two slits, each of width b, whose centres are a distance d apart, where d is several times b. Which describes the pattern on a distant screen? HL
Answer and reasoning
Equally spaced fringes of equal brightness across the whole screen, as in the ideal Young's pattern — A student who thinks the finite slit width cannot change the relative brightness of the fringes picks this. Equal fringes are an idealisation for very narrow slits; with slits of width b the fringe brightness follows the single-slit envelope and falls to zero at its minima.
Fringes whose spacing is set by the slit width b, within an envelope whose width is set by d alone — A student who swaps the roles of b and d picks this. The fringe spacing depends on the separation d (s = λD/d); the envelope depends on the slit width b (first minimum at θ = λ/b).
Two bright bands, one opposite each slit and each as wide as a slit, with a dark gap between them — A student who uses the straight-line model of light picks this. Light spreads out from each slit by diffraction and the two waves interfere, giving many fringes whose spacing depends on the wavelength, not two images of the slits.
Equally spaced fringes whose brightness follows the single-slit pattern, fading at its minima — The fringe spacing is set by d, as in Young's experiment, but each slit's light is itself a single-slit diffraction pattern set by b. The single-slit pattern modulates the fringes: they are brightest near the centre and fade to nothing at the minima of the single-slit envelope.
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27 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A wavefront diagram shows straight water waves of wavelength 2.0 cm, with wavefronts parallel to a barrier, passing through a gap 6.0 cm wide in the barrier; beyond the gap the wavefronts are straight in the middle and curved only at the edges. The gap is narrowed to 0.40 cm. How do the wavefronts beyond the gap change?
Answer and reasoning
They spread out less, as spreading is greatest when the gap equals λ. — A student who thinks diffraction peaks when the gap equals the wavelength picks this. Spreading keeps increasing as the gap is narrowed; through a gap a fifth of a wavelength wide the waves spread out more than they did through the 6.0 cm gap, not less.
They spread into almost semicircular wavefronts, still 2.0 cm apart. — The gap has gone from three wavelengths wide to a fifth of a wavelength wide, so the waves now spread into almost semicircular wavefronts. Diffraction does not change the speed or frequency, so the wavefronts are still 2.0 cm apart.
They form a narrower band of straight wavefronts, 0.40 cm wide, 2.0 cm apart. — A student who thinks a gap simply trims a beam picks this. A gap narrower than the wavelength makes the waves spread out beyond it, into almost semicircles; the narrower the gap, the more curved the emerging wavefronts.
They spread into almost semicircular wavefronts, but closer than 2.0 cm apart. — A student who thinks diffraction changes the wavelength picks this. The water depth, and so the wave speed, is the same on both sides of the barrier and the frequency is unchanged, so the wavelength stays 2.0 cm.
Working Before: gap/λ = 6.0 cm/2.0 cm = 3, so the spreading is modest. After: gap/λ = 0.40 cm/2.0 cm = 0.20, so the waves spread into almost semicircular wavefronts. Speed and frequency are unchanged, so λ = 2.0 cm on both sides of the gap.
2 On a wavefront diagram, straight wavefronts in deep water make an angle of 40° with the straight boundary of a region of shallow water. The wave speed in the shallow water is 0.75 times the wave speed in the deep water. What angle do the wavefronts in the shallow water make with the boundary?
Answer and reasoning
29° — The angle between a wavefront and the boundary equals the angle between its ray and the normal, so θ₁ = 40°. Then sin θ₂ = (v₂/v₁) sin θ₁ = 0.75 × sin 40° = 0.482, so θ₂ = 29°, which is also the new wavefront–boundary angle. The wave slows down, so it turns towards the normal and the angle decreases.
55° — A student who converts every wavefront angle to a ray angle by subtracting from 90° takes θ₁ = 50°, finds θ₂ = 35.1° and gives 90° − 35.1° = 55°. The wavefront–boundary angle already equals the ray–normal angle, because both pairs of lines are perpendicular: θ₁ = 40° and θ₂ = 29°.
59° — A student who thinks a wave that slows down bends away from the normal uses sin θ₂ = sin 40°/0.75, giving 59°. A wave that slows down bends towards the normal: sin θ₂ = 0.75 × sin 40°, so θ₂ = 29°.
30° — A student who cancels the sines takes θ₂/θ₁ = v₂/v₁ and gets θ₂ = 0.75 × 40° = 30°. Sine is a function of the angle, not a factor: sin θ₂ = 0.75 × sin 40° = 0.482, so θ₂ = 29°.
Working The angle between a wavefront and the boundary equals the angle between its ray and the normal, so θ₁ = 40°. v₂/v₁ = sin θ₂/sin θ₁ ⇒ sin θ₂ = 0.75 × sin 40° = 0.75 × 0.643 = 0.482 ⇒ θ₂ = 29°, which is also the angle the refracted wavefronts make with the boundary.
3 A ray of light travelling in glass of refractive index 1.50 meets a flat boundary with water of refractive index 1.33. The ray makes an angle of 35.0° with the boundary. What happens to the ray at the boundary?
Answer and reasoning
It refracts into the water; the angle of refraction is 40.3° — A student who takes the 35.0° angle with the boundary as the angle of incidence calculates sin θ₂ = 1.50 × sin 35.0°/1.33, giving 40.3°. Angles are measured from the normal, so θ₁ = 55.0° and θ₂ = 67.5°.
It refracts into the water; the angle of refraction is 46.6° — A student who thinks light is faster in the medium of larger refractive index expects it to slow down in water and bend towards the normal, using sin θ₂ = 1.33 × sin 55.0°/1.50 to get 46.6°. Light is slower in glass (n = 1.50) than in water (n = 1.33), so it speeds up and bends away from the normal.
It is totally internally reflected back into the glass block — A student who uses sin c = 1/1.50 finds c = 41.8° and, since 55.0° > 41.8°, predicts total internal reflection. That is the critical angle for glass to air. For glass to water, sin c = 1.33/1.50 and c = 62.5°, which is larger than 55.0°, so the ray refracts.
It refracts into the water; the angle of refraction is 67.5° — The angle of incidence is measured from the normal: θ₁ = 90° − 35.0° = 55.0°. The critical angle is given by sin c = 1.33/1.50, so c = 62.5°. Since 55.0° < 62.5°, the ray refracts (with some partial reflection): sin θ₂ = 1.50 × sin 55.0°/1.33 = 0.924, θ₂ = 67.5°, bending away from the normal as it speeds up.
Working θ₁ = 90.0° − 35.0° = 55.0°. Critical angle: sin c = n₂/n₁ = 1.33/1.50 = 0.887 ⇒ c = 62.5°. θ₁ < c, so the ray is refracted (and partly reflected). n₁ sin θ₁ = n₂ sin θ₂ ⇒ sin θ₂ = (1.50 × sin 55.0°)/1.33 = (1.50 × 0.819)/1.33 = 0.924 ⇒ θ₂ = 67.5°.
4 Monochromatic light has a wavelength of 450 nm in water, of refractive index 1.33. The light passes from the water into glass of refractive index 1.50. What is its wavelength in the glass?
Answer and reasoning
508 nm — A student who thinks light is faster in the medium of larger refractive index uses the ratio upside down: 450 × 1.50/1.33 = 508 nm. In glass (n = 1.50) light is slower than in water (n = 1.33), so the wavelength must decrease.
399 nm — The frequency is unchanged, so λ₂/λ₁ = v₂/v₁ = n₁/n₂. λ₂ = 450 nm × 1.33/1.50 = 399 nm. The glass has the larger refractive index, so the light is slower and its wavelength shorter there.
450 nm — A student who thinks only the speed changes at a boundary keeps the wavelength at 450 nm. The frequency stays the same, so the wavelength falls in proportion to the speed.
300 nm — A student who uses the glass's refractive index alone, as if the light came from air, calculates 450/1.50 = 300 nm. The light comes from water, so the ratio n₁/n₂ = 1.33/1.50 must be used.
Working f is unchanged ⇒ λ₂/λ₁ = v₂/v₁ = n₁/n₂. λ₂ = 450 nm × (1.33/1.50) = 399 nm.
5 Two pulses of the same shape and width travel towards each other along a stretched string. Pulse P has a maximum displacement of +3.0 cm and pulse Q a maximum displacement of −1.0 cm. Which statement about the pulses is correct?
Answer and reasoning
While the peaks coincide, the maximum displacement of the string is +4.0 cm. — A student who adds the sizes of the displacements gets 3.0 + 1.0 = 4.0 cm. Displacements add with their signs; Q is below the rest position, so it reduces the displacement: +3.0 − 1.0 = +2.0 cm.
After they have overlapped, the pulses rebound and travel back the way they came. — A student who pictures pulses as colliding objects picks this. Pulses are disturbances of the string, not objects; after overlapping they pass through each other and keep moving in their original directions.
While their peaks coincide, the string's maximum displacement is +2.0 cm. — By the principle of superposition, displacements add with their signs: +3.0 cm + (−1.0 cm) = +2.0 cm. Because the pulses have the same shape, the combined displacement is largest where the peaks meet. Afterwards the pulses pass through each other and each continues with its original shape and direction.
After they have overlapped, they stay combined as one pulse of +2.0 cm. — A student who thinks overlapping pulses merge permanently picks this. The +2.0 cm shape exists only during the overlap; afterwards pulse P (+3.0 cm) and pulse Q (−1.0 cm) emerge unchanged.
Working Superposition: y = y_P + y_Q = (+3.0 cm) + (−1.0 cm) = +2.0 cm while the peaks coincide; afterwards the pulses separate with displacements +3.0 cm and −1.0 cm again.
6 Two identical sodium lamps placed side by side illuminate a white screen. No interference fringes can be seen anywhere on the screen. Which statement explains this?
Answer and reasoning
The phase difference between the light from the two lamps changes rapidly and randomly, so they are not coherent. — Each lamp emits light in short, random bursts from many atoms, so the phase difference between the two lamps' light changes randomly millions of times a second. The light does superpose, but any pattern moves so fast that the screen appears evenly lit. A stable pattern needs coherent sources, with a constant phase difference.
Light waves from two separate lamps cannot superpose, so they pass through each other without interfering. — A student who thinks non-coherent waves do not superpose picks this. All waves superpose: at each instant the light from the lamps interferes, but the pattern changes too quickly to be seen because the phase difference is not constant.
The lamps have the same frequency, so they are coherent and fringes form, but they are too fine to be seen. — A student who thinks equal frequency makes sources coherent picks this. Coherence also needs a constant phase difference, and the phase of each lamp's light changes randomly. No stable fringes form at all, however fine.
Stable interference fringes can be produced only with laser light, and sodium lamps do not emit laser light. — A student who thinks only lasers give interference picks this. Light from a single sodium lamp, passed through a single slit and then a double slit, gives clear fringes because both slits are lit by the same wavefronts. The problem here is two independent sources.
7 Two loudspeakers connected to the same signal generator emit sound in phase at a frequency of 680 Hz. The speed of sound in air is 340 m s⁻¹. What is the smallest path difference that gives destructive interference?
Answer and reasoning
0.50 m — A student who equates a path difference of λ with a phase difference of π (half a cycle) gives λ = 0.50 m. A path difference of λ is a phase difference of 2π, a whole cycle, so the waves arrive in phase and interfere constructively. Antiphase needs a path difference of λ/2 = 0.25 m.
1.00 m — A student who rearranges the wave equation as λ = f/v gets 680/340 = 2.0 and halves it to give 1.00 m. The wavelength is λ = v/f = 0.50 m, so half a wavelength is 0.25 m.
0.25 m — λ = v/f = 340/680 = 0.50 m. Destructive interference needs path difference = (n + ½)λ, and the smallest value is for n = 0: ½ × 0.50 m = 0.25 m.
0.75 m — A student who starts n at 1 uses (1 + ½)λ = 1.5 × 0.50 = 0.75 m. That path difference does give a minimum, but not the smallest one: n = 0 gives 0.25 m.
Working λ = v/f = (340 m s⁻¹)/(680 Hz) = 0.50 m. Destructive: path difference = (n + ½)λ; smallest for n = 0 ⇒ 0.5 × 0.50 m = 0.25 m.
8 In a Young's double-slit experiment, monochromatic light is incident normally on two slits whose centres are 0.40 mm apart; each slit is 0.030 mm wide. The screen is 1.60 m from the slits. The distance from the centre of the 1st bright fringe to the centre of the 11th bright fringe is 24.0 mm. What is the wavelength of the light?
Answer and reasoning
5.5 × 10⁻⁷ m — A student who divides by the number of fringes takes s = 24.0 mm/11 = 2.18 mm and gets λ = 5.5 × 10⁻⁷ m. Eleven bright fringes have only ten spaces between them, so s = 2.40 mm.
6.0 × 10⁻⁷ m — From the 1st to the 11th bright fringe there are 10 fringe separations, so s = 24.0 mm/10 = 2.40 × 10⁻³ m. Then λ = sd/D = (2.40 × 10⁻³ × 0.40 × 10⁻³)/1.60 = 6.0 × 10⁻⁷ m, which is orange light.
6.0 × 10⁻⁴ m — A student who substitutes d = 0.40 without converting from millimetres gets 6.0 × 10⁻⁴ m, a length far too large for light. With d = 0.40 × 10⁻³ m, λ = 6.0 × 10⁻⁷ m.
4.5 × 10⁻⁸ m — A student who uses the slit width, 0.030 mm, in place of the slit separation gets 4.5 × 10⁻⁸ m. The fringe separation is set by the distance between the slit centres, d = 0.40 mm; the slit width does not appear in s = λD/d.
Working Number of fringe separations = 11 − 1 = 10 ⇒ s = 24.0 mm/10 = 2.40 mm = 2.40 × 10⁻³ m. s = λD/d ⇒ λ = sd/D = (2.40 × 10⁻³ m)(0.40 × 10⁻³ m)/(1.60 m) = 6.0 × 10⁻⁷ m.
9 In a Young's double-slit experiment, monochromatic red light is incident normally on the slits and the fringe separation on the screen is 2.0 mm. Which single change would make the fringe separation 4.0 mm?
Answer and reasoning
Doubling the distance between the centres of the slits — A student who thinks moving the slits apart spreads the fringes picks this. s = λD/d, so doubling d halves the fringe separation to 1.0 mm.
Using blue light, of shorter wavelength, in place of the red — A student who thinks shorter wavelengths spread out more, as violet does in a prism, picks this. s = λD/d is proportional to λ, and blue light has a shorter wavelength than red, so the fringes move closer together.
Halving the width of each slit, keeping their separation — A student who confuses slit width with slit separation picks this. The fringe separation depends on d, the distance between the slit centres, which is unchanged; the slit width does not appear in s = λD/d. Narrower slits let less light through, so the fringes are dimmer, but s stays 2.0 mm.
Halving the distance between the centres of the two slits — s = λD/d, so s is inversely proportional to d: halving the slit separation doubles s from 2.0 mm to 4.0 mm.
Working s = λD/d ⇒ s ∝ 1/d. Halving d: s = 2 × 2.0 mm = 4.0 mm.
10 Monochromatic light of wavelength 600 nm is incident normally on a single slit, and the diffraction pattern is viewed on a distant screen. The slit width is then reduced from 0.20 mm to 0.10 mm. How does the central maximum change? HL
Answer and reasoning
It becomes half as wide and less intense, as it is an image of a slit that is now half as wide. — A student who thinks the central maximum is a geometric image of the slit picks this. For slits this narrow the width of the pattern is set by diffraction, θ = λ/b, so a narrower slit gives a wider central maximum, not a narrower one.
It is unchanged, as both slits are far wider than 600 nm, so neither diffracts significantly. — A student who thinks diffraction happens only when the slit width equals the wavelength picks this. Diffraction still occurs when b is a few hundred wavelengths: θ = λ/b = 3.0 × 10⁻³ rad and then 6.0 × 10⁻³ rad, which on a screen 2 m away is a few millimetres, easily visible, and it doubles.
It becomes twice as wide and less intense, as less light passes and is spread over a wider angle. — θ = λ/b, so halving b doubles the angle to the first minimum and the central maximum becomes twice as wide. Less light gets through the narrower slit, and it is spread over a wider angle, so the central maximum is less intense.
It becomes twice as wide, with unchanged peak intensity, which depends only on the source. — A student who thinks the brightness depends only on the source picks this. The width does double, but less light passes through the narrower slit and it is spread more widely, so the central maximum is less intense.
11 A double slit has slits of width b = 0.10 mm whose centres are d = 0.30 mm apart. It is illuminated normally with monochromatic light. How many bright interference fringes lie within the central maximum of the single-slit diffraction envelope? HL
Answer and reasoning
5 — Interference maxima are at sin θ = nλ/d and the first minima of the envelope at sin θ = λ/b. These coincide when n = d/b = 3, so the third-order fringes fall on the envelope minima and are missing. Inside the central maximum are the orders n = 0, ±1 and ±2: five fringes.
7 — A student who thinks the envelope only dims fringes and never removes one counts orders 0, ±1, ±2 and ±3. The third-order fringes lie exactly at the envelope's first minima, where neither slit sends any light, so they are missing.
3 — A student who takes λ/b as the full angular width of the envelope puts its edges at λ/(2b), which admits only n < d/(2b) = 1.5, that is n = 0 and ±1. The first minima are at λ/b from the centre, so orders up to ±2 fit inside.
1 — A student who swaps the roles of b and d puts the fringes λ/b apart inside an envelope reaching only λ/d, so only the central fringe fits. The fringes are λ/d apart and the envelope reaches λ/b, three times farther.
Working Interference maxima: sin θ = nλ/d. First envelope minimum: sin θ = λ/b. Fringes inside the central maximum satisfy n < d/b = 0.30 mm/0.10 mm = 3, so n = 0, ±1, ±2 (n = ±3 coincides with the minima and is missing): 2 × 2 + 1 = 5 fringes.
12 Light of wavelength 650 nm is incident normally on a diffraction grating that has 600 lines per millimetre. At what angle to the straight-through direction is the second-order maximum observed? HL
Answer and reasoning
23.0° — A student who counts the central maximum as the first order uses n = 1 for the second bright line: sin θ = 0.390 and θ = 23.0°. The central maximum is n = 0, so the second-order maximum has n = 2 and θ = 51.3°.
44.7° — A student who uses the small-angle form θ = nλ/d gets 0.780 rad = 44.7°. Grating angles are large, so sin θ = nλ/d must be used: θ = sin⁻¹ 0.780 = 51.3°.
51.3° — d = 1/(600 × 10³) m = 1.67 × 10⁻⁶ m. For n = 2: sin θ = nλ/d = 2 × 650 × 10⁻⁹ × 600 × 10³ = 0.780, so θ = 51.3°.
0.04° — A student who takes d = 1/600 m instead of 1/600 mm gets sin θ = 7.8 × 10⁻⁴ and θ = 0.04°. With 600 lines per millimetre, d = 1.67 × 10⁻⁶ m.
Working d = 1/(600 mm⁻¹) = 1/(6.00 × 10⁵ m⁻¹) = 1.667 × 10⁻⁶ m. nλ = d sin θ ⇒ sin θ = 2 × 650 × 10⁻⁹ m × 6.00 × 10⁵ m⁻¹ = 0.780 ⇒ θ = 51.3°.
13 Monochromatic light is incident normally first on a double slit and then on a diffraction grating whose adjacent slits have the same separation d as the double slit. Every slit is equally illuminated. Compared with the double-slit fringes, how do the grating's principal maxima appear? HL
Answer and reasoning
Farther apart than the fringes, because the grating has many more slits — A student who confuses more slits with more lines per millimetre picks this. Here the slit spacing d is the same, so the maxima are at the same angles; more slits only make them sharper and brighter.
At the same angles, but wider and dimmer, as the light is shared among more slits — A student who thinks light is shared out among more slits picks this. Each slit is equally illuminated, so more slits pass more light, and the extra slits make the maxima narrower, not wider.
Identical in position, width and brightness, because only d and λ matter — A student who judges that the number of slits cannot matter because it is not in nλ = d sin θ picks this. The equation gives only the positions; the width and brightness of the maxima depend strongly on the number of slits.
At the same angles as the double-slit fringes, but much sharper and brighter — The angles of the maxima are fixed by nλ = d sin θ, which is the same for both because d and λ are the same. With many slits, waves reinforce only very close to those angles and cancel elsewhere, so the maxima become narrow, and more slits pass more light, so they are brighter.
14 White light is incident normally on a diffraction grating. Which describes the pattern seen on a screen? HL
Answer and reasoning
A white central maximum, with a spectrum in each order on either side and violet nearest the centre — At the central maximum (n = 0) the path difference is zero for every wavelength, so all colours coincide and it is white. In each higher order sin θ = nλ/d is larger for longer wavelengths, so each spectrum runs from violet (nearest the centre) to red (farthest out).
A white central maximum, with a spectrum in each order on either side and red nearest the centre — A student who transfers the colour order of a prism picks this. A prism deviates violet most, but a grating deviates the longest wavelength most: sin θ = nλ/d, so red is farthest from the centre in each order.
A spectrum at the centre, as well as in each order on either side, with violet nearest to the centre — A student who thinks a grating splits colours at every maximum picks this. At n = 0 the path difference is zero for every wavelength, so all colours arrive together and the central maximum is white.
A white central maximum and a white maximum in each order, as all colours share the same angles — A student who thinks the grating alone fixes the positions of the maxima picks this. sin θ = nλ/d depends on λ, so for n ≥ 1 each colour has its own angle and the light is spread into a spectrum.
15 Light containing just two wavelengths, 450 nm and 700 nm, is incident normally on a diffraction grating with slit spacing d = 2.00 × 10⁻⁶ m. Which statement about the maxima produced is correct? HL
Answer and reasoning
On each side, the 450 nm light gives maxima up to order 4, and the 700 nm light up to order 3. — A student who rounds d/λ = 2.86 to the nearest whole number gives 3 for the 700 nm light. The third order would need sin θ = 3 × 700 × 10⁻⁹/2.00 × 10⁻⁶ = 1.05, which is impossible, so the highest order is 2.
On each side, the 450 nm light gives maxima up to order 4, but the 700 nm light only up to order 2. — sin θ = nλ/d cannot exceed 1, so n ≤ d/λ. For 450 nm, d/λ = 4.44, so the highest order is 4; for 700 nm, d/λ = 2.86, so the highest order is 2. The longer wavelength is diffracted through larger angles, so it runs out of orders sooner.
The 700 nm light gives more orders than the 450 nm light, because red light is deviated less. — A student who transfers the prism's colour order to a grating picks this. In a grating sin θ = nλ/d, so red (700 nm) is deviated more than blue (450 nm), and it gives fewer orders, 2 compared with 4.
Both wavelengths give maxima up to the same order, as the grating alone fixes the maxima. — A student who thinks the grating alone fixes the maxima picks this. The number of orders depends on d/λ: 4.44 for 450 nm and 2.86 for 700 nm, so the two wavelengths reach different highest orders.
Working n_max = largest whole number ≤ d/λ. 450 nm: d/λ = (2.00 × 10⁻⁶)/(450 × 10⁻⁹) = 4.44 ⇒ n_max = 4. 700 nm: d/λ = (2.00 × 10⁻⁶)/(700 × 10⁻⁹) = 2.86 ⇒ n_max = 2 (n = 3 would need sin θ = 1.05).
16 A small loudspeaker in open air emits sound of frequency 170 Hz. The speed of sound in air is 340 m s⁻¹. A student draws a wavefront–ray diagram of the sound in a horizontal plane through the speaker. Which statement about the diagram is correct?
Answer and reasoning
Every compression and every rarefaction has a wavefront, so neighbouring ones are 1.0 m apart. — A wavefront is drawn through points that are in phase, so it passes through compressions only (or rarefactions only), not both. A compression and the next rarefaction are λ/2 = 1.0 m apart, but adjacent wavefronts on the diagram are a full wavelength, 2.0 m, apart.
The rays show the paths along which air is carried outward from the speaker. — Sound is a wave: the air oscillates about fixed positions and is not carried away from the speaker. A ray shows the direction in which the wave's energy travels, perpendicular to the wavefronts, not a path taken by the medium.
Adjacent wavefronts are 2.0 m apart, both close to the speaker and far from it. — λ = v/f = 340/170 = 2.0 m. Wavefronts join points that are in phase (successive compressions), so neighbours are one wavelength apart, and since neither v nor f changes with distance the spacing is the same everywhere. The wavefronts are circles centred on the speaker (sections of spheres in three dimensions) and the rays are radial lines perpendicular to them.
There is no sound in the gaps between the rays that are drawn on the diagram. — Rays are construction lines showing the direction of energy travel; any number could be drawn. The sound wave fills the whole region around the speaker, including the spaces between whichever rays the student chose to draw.
Working λ = v/f = 340 m s⁻¹ / 170 Hz = 2.0 m. Wavefronts are drawn through points in phase (successive compressions), so adjacent wavefronts are λ = 2.0 m apart, and this spacing is the same at every distance from the speaker because v and f are unchanged. The wavefronts are circles (sections of spheres) centred on the speaker; rays are radial, perpendicular to them.
17 Straight water waves of frequency 5.0 Hz travel at 0.40 m s⁻¹ across the deep part of a ripple tank towards a boundary with a shallow region, in which the wave speed is 0.20 m s⁻¹. The wavefronts are parallel to the boundary. Which statement about the waves at the boundary is correct?
Answer and reasoning
The reflected waves have a wavelength of 8.0 cm and the transmitted waves 4.0 cm. — The frequency, 5.0 Hz, is the same for the incident, reflected and transmitted waves. Reflected waves travel in the deep water at 0.40 m s⁻¹, so λ = 0.40/5.0 = 8.0 cm, the same as the incident waves. Transmitted waves travel at 0.20 m s⁻¹, so λ = 0.20/5.0 = 4.0 cm: the speed halves, and so does the wavelength.
The transmitted waves have a wavelength of 8.0 cm, the same as that of the incident waves. — At a boundary the speed and the wavelength change together, because the frequency is fixed by the source. With v = fλ and f = 5.0 Hz, halving the speed to 0.20 m s⁻¹ halves the wavelength to 4.0 cm. Only the reflected waves, which stay in the deep water, keep the 8.0 cm wavelength.
The transmitted waves have a frequency of 2.5 Hz, half that of the incident waves. — Every wavefront arriving at the boundary produces one transmitted wavefront, so the number of wavefronts per second, the frequency, is the same on both sides: 5.0 Hz. It is the wavelength, not the frequency, that halves when the speed halves.
No waves are reflected at the boundary; all of the wave energy passes into the shallow water. — Wherever the wave speed changes, some of the wave is reflected and some transmitted, even at normal incidence. A weak reflected wave travels back through the deep water with the incident wavelength of 8.0 cm; the transmitted wave carries the rest of the energy at 4.0 cm wavelength.
Working Incident wavelength λ₁ = v₁/f = 0.40/5.0 = 0.080 m = 8.0 cm. The reflected wave travels back in the deep water at 0.40 m s⁻¹ with the same frequency, so its wavelength is also 8.0 cm. The transmitted wave keeps f = 5.0 Hz, so λ₂ = v₂/f = 0.20/5.0 = 0.040 m = 4.0 cm. Part of the wave is reflected and part transmitted, even at normal incidence.
18 Two radio stations broadcast from the same mast: station A at 1.0 MHz, wavelength 300 m, and station B at 100 MHz, wavelength 3.0 m. A house stands in a valley behind a hill about 200 m high, with no straight-line path from the mast to the house. Which station is more likely to be received at the house, and why?
Answer and reasoning
Station B, because its 100 MHz waves carry more energy and so pass over the hill more easily. — How far a wave spreads into the region behind an obstacle depends on its wavelength compared with the obstacle, not on the energy it carries. Station B's short 3.0 m wavelength is much smaller than the hill, so it diffracts very little; the hill blocks it much more effectively than it blocks station A.
Both equally, because the hill is the same size for both, and it alone fixes how much a wave diffracts. — The size of the obstacle only matters in comparison with the wavelength. The same 200 m hill is small relative to a 300 m wavelength but enormous relative to a 3.0 m wavelength, so the two stations' signals diffract by very different amounts.
Neither, because diffraction only occurs when the wavelength is exactly equal to the size of the hill. — There is no requirement for an exact match. Diffraction is noticeable whenever the wavelength is comparable with or larger than the obstacle, and it becomes stronger as the wavelength increases, so the 300 m signal spreads readily into the valley.
Station A, because its 300 m wavelength is comparable with the hill's size, so it diffracts round it. — Diffraction around a body is significant when the wavelength is comparable with, or larger than, the size of the body. Station A's 300 m waves are of the same order as the 200 m hill and spread strongly into the valley behind it; station B's 3.0 m waves are far smaller than the hill and diffract very little, so the hill casts a radio 'shadow' for them.
Working Diffraction around a body is significant when λ is comparable with or larger than the body. Station A: λ = 300 m, of the same order as the 200 m hill, so the waves spread strongly into the valley behind it. Station B: λ = 3.0 m, about 70 times smaller than the hill, so it diffracts very little and the hill shadows the house. Station A is the one more likely to be received.
19 Two pulses of identical shape and size but opposite displacement travel towards each other along a stretched string. At one instant they overlap exactly and the whole string is momentarily straight. Which statement about this instant, and what follows it, is correct?
Answer and reasoning
The pulses have cancelled each other out; the string stays straight from now on. — Superposition is temporary. The string is straight only at this one instant, but its particles are moving, so it does not remain straight: the two pulses emerge on the far side of each other and travel on unchanged.
The string is straight but moving; the pulses then emerge and continue past each other. — At this instant the displacements of the two pulses cancel everywhere, but their velocities add: the parts of the string where the pulses' leading and trailing edges overlap are moving, so the energy is momentarily all kinetic. Each pulse is unaffected by the other, and a moment later both reappear and continue in their original directions with their original shapes.
The string is straight and at rest; the pulses' energy has become internal energy of the string. — The energy has not been converted to internal energy. When the displacement is zero everywhere, the string's particles are moving fastest, so all the pulses' energy is kinetic at that instant. It reappears as pulse energy as the two pulses separate.
The pulses have collided; each then travels back the way it came, with its shape restored. — Pulses are not objects and do not bounce off each other. They pass through one another: after the instant of complete cancellation the positive pulse continues in its original direction and so does the negative pulse.
20 Two loudspeakers, driven in phase by the same signal generator, face a line along which a microphone is moved. Starting from the central maximum of loudness, the microphone is moved along the line to the third maximum away from the centre. At that point its distances from the two speakers differ by 0.60 m. What is the wavelength of the sound?
Answer and reasoning
0.20 m — For sources in phase, maxima occur where the path difference is nλ, and the central maximum is n = 0. The third maximum from the centre is n = 3, so 3λ = 0.60 m and λ = 0.20 m.
0.30 m — This treats the central maximum as the first and so uses n = 2 for the third maximum, giving 0.60/2. The central maximum has path difference zero (n = 0); the third maximum away from it has n = 3, so λ = 0.60/3 = 0.20 m.
0.40 m — This takes the condition for maxima to be a whole number of half-wavelengths, 0.60 m = 3 × λ/2. Constructive interference needs a path difference of a whole number of wavelengths, so 0.60 m = 3λ and λ = 0.20 m.
0.10 m — This pairs a path difference of one wavelength with a phase difference of π, so that the waves are in phase only every 2λ and the third maximum sits at 6λ. One wavelength of path difference is a full cycle, 2π, so the third maximum is at 3λ and λ = 0.20 m.
Working Sources in phase: maxima where the path difference is nλ, with n = 0 at the central maximum. The third maximum away from the centre has n = 3, so 3λ = 0.60 m and λ = 0.20 m.
21 The diagram shows straight water waves in a ripple tank crossing the boundary between region X and region Y. The wavefronts, a ray and the normal at the point where the ray crosses the boundary are drawn. Which statement is correct?
Answer and reasoning
The waves travel more slowly in Y, so Y is the deeper water. — The reading of the diagram is right: the shorter wavelength in Y at the same frequency means a lower speed. But water waves are slower in shallow water, not deep water, so Y is the shallower region. Breaking waves near a beach look faster because they grow taller, but they are slowing down.
The waves travel faster in Y, as the ray bends towards the normal there. — Bending towards the normal is the sign of a wave that has slowed down: the part of each wavefront that enters Y first is held back, swinging the wavefront round. The closer spacing of the wavefronts in Y, at unchanged frequency, also shows a lower speed there.
The waves have a lower frequency in Y, because they arrive there more slowly. — Slowing down does not lower the frequency. Each wavefront arriving at the boundary produces one wavefront in Y, so the number of wavefronts per second is the same on both sides. With f fixed, v = fλ shows that the shorter wavelength in Y corresponds to a lower speed, not a lower frequency.
The waves travel more slowly in Y, so Y is the shallower water. — The frequency is the same on both sides, so the closer wavefront spacing (shorter wavelength) in Y means a lower speed there, v = fλ. The ray bending towards the normal confirms this. Water waves travel more slowly in shallower water, so Y is the shallow region.
22 A student's sketch of water waves passing through a gap in a barrier is shown. The sketch is drawn to the scale marked, and the water is of the same depth everywhere. Which feature of the sketch is incorrect?
Answer and reasoning
The wavefronts beyond the gap are drawn curved instead of as a straight beam. — The curvature is correct. A gap about one wavelength wide acts almost like a point source, and the waves spread out into the region behind the barrier as semicircular wavefronts. Waves do not continue as a straight-sided beam.
The wavefronts beyond the gap are drawn closer together than those approaching it. — Diffraction changes the shape and direction of the wavefronts but not the wavelength: the water is the same depth on both sides, so the speed is unchanged, and the frequency is fixed by the source. The semicircles should be 2.0 cm apart, the same as the incident wavefronts, not 1.0 cm.
The rays beyond the gap are drawn at right angles to the wavefronts, not along them. — Rays are always drawn perpendicular to the wavefronts, in the direction the wave travels. For semicircular wavefronts spreading from the gap, the rays are radial lines from the gap, exactly as the sketch shows.
The wavefronts are drawn crossing the spaces between the rays, where there is no wave. — Rays are construction lines, not the wave itself. The diffracted wave fills the whole region beyond the gap, so the wavefronts correctly run continuously across the spaces between the three rays that happen to be drawn.
23 The diagram shows a ray of light inside a semicircular glass block of refractive index 1.50, surrounded by air of refractive index 1.00. The ray travels along a radius to the centre C of the flat face, making the angle marked with the flat face. What happens to the ray at C?
Answer and reasoning
It refracts into the air at 22.5° to the normal. — This has the refractive indices the wrong way round, sin θ₂ = sin 35°/1.50. Light travels faster in air than in glass, so on leaving the glass it bends away from the normal: sin θ₂ = 1.50 sin 35°, giving 59.4°.
It refracts into the air, emerging at 35.0° to the normal. — This uses only the refractive index of the air the light enters, as if sin θ₂ = sin θ₁/n_air with n_air = 1.00. Snell's law needs both media: n₁ sin θ₁ = n₂ sin θ₂ with n₁ = 1.50 for the glass, so sin θ₂ = 1.50 sin 35° and θ₂ = 59.4°.
It refracts into the air, leaving at 59.4° to the normal. — The 55° is measured from the flat face, so the angle of incidence is 90° − 55° = 35°. The critical angle for glass to air is sin⁻¹(1.00/1.50) = 41.8°, and 35° is less than this, so the ray leaves the glass: 1.50 sin 35° = 1.00 sin θ₂ gives θ₂ = 59.4°, bending away from the normal. Some light is also partially reflected.
It is totally internally reflected, back into the glass. — Taking the marked 55° as the angle of incidence makes it exceed the 41.8° critical angle. But angles of incidence are measured from the normal, so θ₁ = 90° − 55° = 35°, which is less than the critical angle, and the ray refracts out at 59.4°.
Working The angle marked, 55°, is between the ray and the flat face, so the angle of incidence (from the normal) is θ₁ = 90° − 55° = 35°. Critical angle for glass to air: sin c = 1.00/1.50 ⇒ c = 41.8°. Since 35° < 41.8°, the ray refracts into the air (with some partial reflection): 1.50 sin 35° = 1.00 sin θ₂ ⇒ sin θ₂ = 0.860 ⇒ θ₂ = 59.4°. The ray enters the curved face along a radius, so it is undeviated there.
24 The graph shows the displacement y of a stretched string against position x at time t = 0. Pulse A is travelling to the right and pulse B to the left, each at a speed of 1.0 cm s⁻¹. What is the displacement of the string at x = 6.0 cm when t = 3.0 s?
Answer and reasoning
2.0 cm — This adds the two amplitudes, +3.0 and −1.0 cm, but at t = 3.0 s the peak of A is at x = 7 cm, not 6 cm. Superposition adds the displacements each pulse has at the point in question: A contributes +1.5 cm there and B −1.0 cm, giving +0.5 cm.
2.5 cm — This reads the two contributions correctly, 1.5 cm from A and 1.0 cm from B, but adds their sizes while ignoring the sign. B's displacement at x = 6 cm is negative, so the resultant is +1.5 + (−1.0) = +0.5 cm.
0.0 cm — This assumes the pulses met and bounced back, so that neither reaches x = 6 cm by t = 3.0 s. Pulses pass through each other; at t = 3.0 s both overlap x = 6 cm, contributing +1.5 cm and −1.0 cm, so the displacement is +0.5 cm.
0.5 cm — In 3.0 s each pulse moves 3.0 cm. Pulse A's peak is then at x = 7 cm and its leading edge runs from (5, 0) to (7, +3), so at x = 6 cm its displacement is +1.5 cm. Pulse B's trough, −1.0 cm, has arrived at x = 6 cm. The string's displacement is the sum: +1.5 + (−1.0) = +0.5 cm.
Working At t = 0, A is a triangle from x = 2 to 6 cm with its peak of +3.0 cm at x = 4 cm; B is a triangle from x = 8 to 10 cm with its trough of −1.0 cm at x = 9 cm. In 3.0 s each pulse moves 3.0 cm. At t = 3.0 s, A spans x = 5 to 9 cm with its peak at x = 7 cm: at x = 6 cm it is halfway up the leading edge, displacement +3.0 × (1/2) = +1.5 cm. B spans x = 5 to 7 cm with its trough at x = 6 cm: displacement −1.0 cm. Superposition: y = +1.5 + (−1.0) = +0.5 cm.
25 The diagram shows two point sources S₁ and S₂ in a ripple tank, vibrating in phase, and a point P on the water surface. The waves have a wavelength of 2.0 cm. On which line of the interference pattern does P lie?
Answer and reasoning
On the first minimum to one side of the central maximum line. — This treats a path difference of one wavelength, 2.0 cm, as a phase difference of π. One wavelength is a whole cycle, 2π, so the waves arrive in phase and P is on the first maximum; the first minimum needs a path difference of λ/2 = 1.0 cm.
On the first maximum to one side of the central maximum. — S₂P = √(15.0² + 8.0²) = 17.0 cm, so the path difference is 17.0 − 15.0 = 2.0 cm = λ. For in-phase sources a path difference of nλ gives a maximum; n = 1 is the first maximum beside the central (n = 0) line.
On the second maximum to one side of the central maximum. — This counts the 2.0 cm path difference as two half-wavelengths and calls it the second maximum. Maxima need a whole number of wavelengths: 2.0 cm is 1 × λ, so P is on the first maximum.
On no maximum, because P is not equidistant from S₁ and S₂. — Equal distances give only the central maximum. Any path difference that is a whole number of wavelengths also gives a maximum, and here 17.0 − 15.0 = 2.0 cm = λ, so P is on the first maximum to one side.
Working S₁P = 15.0 cm and S₁S₂ = 8.0 cm meet at a right angle at S₁, so S₂P = √(15.0² + 8.0²) = √289 = 17.0 cm. Path difference = 17.0 − 15.0 = 2.0 cm = 1 × λ, a whole number of wavelengths, so the waves arrive in phase: constructive interference (n = 1).
26 The graph shows how the intensity I of the light on a screen varies with position x in a Young's double-slit experiment. The two slits are 0.40 mm apart and each is 0.020 mm wide; the screen is 2.0 m from the slits, and monochromatic light is incident normally on them. What is the wavelength of the light?
Answer and reasoning
6.0 × 10⁻⁷ m — Adjacent maxima on the graph are 3.0 mm apart, so s = 3.0 × 10⁻³ m. Rearranging s = λD/d gives λ = sd/D = (3.0 × 10⁻³ × 0.40 × 10⁻³)/2.0 = 6.0 × 10⁻⁷ m, orange light of 600 nm.
4.8 × 10⁻⁷ m — This divides the 12 mm between the outermost maxima by the five maxima counted, giving 2.4 mm. There are only four fringe spacings between five maxima, so s = 12/4 = 3.0 mm and λ = 6.0 × 10⁻⁷ m.
3.0 × 10⁻⁸ m — This uses the slit width, 0.020 mm, in place of the slit separation. The fringe separation s = λD/d depends on the distance d between the centres of the slits, 0.40 mm, giving λ = 6.0 × 10⁻⁷ m. The slit width affects only the brightness envelope.
6.0 × 10⁻⁴ m — This substitutes s = 3.0 (millimetres) alongside d in metres and D in metres. All lengths must be in the same unit: s = 3.0 × 10⁻³ m gives λ = 6.0 × 10⁻⁷ m. A wavelength of 0.6 mm would be far into the infrared, not visible light.
Working From the graph, adjacent maxima are at x = 0, 3, 6 mm, so the fringe separation is s = 3.0 mm = 3.0 × 10⁻³ m. s = λD/d ⇒ λ = sd/D = (3.0 × 10⁻³ m × 0.40 × 10⁻³ m)/2.0 m = 6.0 × 10⁻⁷ m (600 nm).
27 The graph shows the intensity pattern produced on a distant screen when monochromatic light is incident normally on two rectangular slits of width b whose centres are a distance d apart. The dashed line is the single-slit diffraction envelope. What is the value of the ratio d/b? HL
Answer and reasoning
8.00 — This takes θ = λ/b to be the whole width of the central envelope, which spans eight fringe spacings (from the missing −4 to the missing +4). λ/b is the half-width, from the centre to the first minimum, which is four fringe spacings: 4λ/d = λ/b and d/b = 4.
3.50 — This counts the seven bright fringes inside the central lobe and takes its half-width as 3.5 fringe spacings. The half-width is set by where the envelope reaches zero, which is where the 4th fringe would be, four fringe spacings from the centre, so d/b = 4.
0.25 — This swaps the roles of b and d, taking the fringe spacing to be λ/b and the envelope half-width λ/d. The narrow fringes come from the slit separation d and the broad envelope from the slit width b, so the ratio is d/b = 4, not 1/4.
4.00 — The envelope's first minimum (θ = λ/b) coincides with the position of the missing 4th-order fringe (θ = 4λ/d). Equating them, λ/b = 4λ/d, so d/b = 4: the slit separation is four times the slit width.
Working Interference maxima occur at sin θ ≈ θ = nλ/d; the single-slit envelope has its first minimum at θ = λ/b. On the graph the envelope falls to zero exactly where the 4th-order fringe would be (the central lobe holds orders 0, ±1, ±2, ±3 and the ±4 fringes are missing), so 4λ/d = λ/b, giving d/b = 4.
That was your twenty minutes. Real practice on C.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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