C.3 Wave phenomena flashcards: IB Physics SL and HL

All 78 C.3 Wave phenomena cards

1. DefinitionSL & HL

   State the law of reflection for a wave meeting a plane boundary.

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   The angle of incidence equals the angle of reflection (θ_i = θ_r), both measured between the ray and the NORMAL, and the incident ray, reflected ray and normal all lie in the same plane. Speed, frequency and wavelength are unchanged on reflection. Exam tip: angles must be stated from the normal, not from the surface — this is the most common lost mark; add the same-plane statement for a full 2-mark definition. Angles in degrees (or rad); angle is a scalar quantity. A wave reflected at a fixed (denser) boundary suffers a π (180°) phase change; reflection at a free end produces no phase change.

2. DefinitionSL & HL

   Define refraction of a wave.

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   Refraction is the change in direction of a wave as it crosses the boundary between two media, caused by a change in the SPEED of the wave. Exam tip: the mark is for change of speed causing the change of direction — writing only that the wave bends scores zero. Frequency is unchanged because it is fixed by the source, so speed and wavelength change in the same ratio (v = fλ). Entering a slower, optically denser medium the ray bends TOWARDS the normal. A ray along the normal (θ = 0) does not change direction but its speed still changes. Angles are measured from the normal.

3. DefinitionSL & HLData booklet: Yes

   Define the absolute refractive index of a medium.

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   The refractive index n of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium: n = c/v. Exam tip: say in a vacuum (or free space), not in air, although n_air ≈ 1.00 is an accepted approximation. n is a dimensionless scalar with n ≥ 1 for all real media (water 1.33, crown glass ≈ 1.50, diamond 2.42). A larger n means a slower wave and a more optically dense medium. Because f is unchanged, n also equals λ_vacuum/λ_medium. n varies slightly with wavelength (dispersion), which is why a prism separates white light.

4. DefinitionSL & HLData booklet: Yes

   State Snell's law and define the relative refractive index ₁n₂.

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   Snell's law: n₁ sinθ₁ = n₂ sinθ₂, with both angles measured from the normal. For light passing from medium 1 into medium 2 the relative refractive index is ₁n₂ = n₂/n₁ = sinθ₁/sinθ₂ = v₁/v₂ = λ₁/λ₂. Exam tip: the booklet prints n₁/n₂ = sinθ₂/sinθ₁ = v₂/v₁ — the subscripts on the n ratio are REVERSED relative to the sine and speed ratios, and blindly matching them is the classic error. Dimensionless scalar. Snell's law applies to any wave (water waves, sound), not only to light.

5. DefinitionSL & HLData booklet: No – derive

   Define the critical angle.

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   The critical angle θ_c is the angle of incidence, in the optically denser medium, for which the angle of refraction in the less dense medium is exactly 90°, so the refracted ray travels along the boundary. From Snell's law sinθ_c = n₂/n₁, and for a medium–air boundary sinθ_c = 1/n. Exam tip: you must state that the light is travelling in the MORE dense medium (n₁ > n₂); θ_c does not exist otherwise. Measured in degrees from the normal; scalar. Numerical anchors: water θ_c = 48.8°, crown glass θ_c = 41.8°, diamond θ_c = 24.4°.

6. DefinitionSL & HL

   Define total internal reflection and state the conditions required for it.

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   Total internal reflection is the complete reflection of a wave back into the denser medium at a boundary, with no transmitted (refracted) ray. Conditions: (1) the wave travels from a medium of higher refractive index towards one of lower refractive index; (2) the angle of incidence is greater than the critical angle. Exam tip: total means 100% of the energy is reflected — say that no light is refracted or transmitted; writing most of the light loses the mark. At exactly θ_c the refracted ray grazes the surface at 90°. Applications: optical fibres, endoscopes, prismatic reflectors in binoculars and bicycle reflectors.

7. DefinitionSL & HL

   Outline the structure of a step-index optical fibre and state the function of the cladding.

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   A step-index fibre has a thin glass core of high refractive index surrounded by cladding of LOWER refractive index, so light entering within the acceptance angle strikes the core–cladding boundary above the critical angle and is totally internally reflected repeatedly along the fibre. Exam tip: the cladding is not merely protection — it guarantees the n₂ < n₁ boundary needed for TIR, keeps the core surface free of scratches and dirt that would let light escape, and prevents cross-talk between neighbouring fibres. Multipath (modal) dispersion arises because rays at different angles travel different path lengths, broadening a pulse; it is reduced by a narrower core or a graded-index profile.

8. DefinitionSL & HL

   Define diffraction and state when it is significant.

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   Diffraction is the spreading out of a wave as it passes through an aperture or around an obstacle. Exam tip: diffraction is significant when the wavelength is comparable to the width of the gap or obstacle, λ ≈ b; for b ≫ λ the spreading is negligible. For the top mark say that the wave spreads INTO the geometrical shadow. Speed, frequency and wavelength are all unchanged by diffraction — only the shape of the wavefronts and the distribution of energy change, and the amplitude falls as the energy spreads over a wider region. This is why sound (λ ~ 1 m) diffracts round a doorway but light (λ ~ 5 × 10⁻⁷ m) does not.

9. DefinitionSL & HL

   State the principle of superposition.

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   When two or more waves meet at a point, the resultant displacement at that point and instant is the vector sum of the displacements each wave would produce there on its own. Exam tip: the mark is for sum of DISPLACEMENTS — writing sum of amplitudes or sum of intensities is wrong and is the single most frequent error; displacements may be negative, so they can cancel. Superposition is instantaneous and the waves continue unchanged after they have passed through each other. Displacement is a vector; the unit is that of the displacement (m for a string, Pa for a sound pressure wave).

10. DefinitionSL & HLData booklet: No – memorise

    Define constructive interference and state its path-difference condition.

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    Constructive interference occurs where waves from two coherent sources arrive IN PHASE, so their displacements add and the resultant amplitude is a maximum, A₁ + A₂. It occurs when the path difference is a whole number of wavelengths, Δ = nλ with n = 0, 1, 2, …, equivalent to a phase difference of 0, 2π, 4π … rad. Exam tip: state both the phase condition and the path-difference condition — quoting only that the waves add scores 1 of 2 marks. Since intensity ∝ amplitude², two equal sources give a maximum intensity 4 times that of one source alone, not twice.

11. DefinitionSL & HLData booklet: No – memorise

    Define destructive interference and state its path-difference condition.

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    Destructive interference occurs where waves from two coherent sources arrive antiphase (phase difference π rad), so their displacements subtract and the resultant amplitude is a minimum, |A₁ − A₂|. Condition: path difference Δ = (n + ½)λ, an odd number of half-wavelengths; phase difference π, 3π, 5π … rad. Exam tip: complete cancellation (zero intensity) also requires EQUAL amplitudes — say amplitudes equal when asked why the dark fringes are completely dark. Energy is not destroyed: it is redistributed from the minima into the maxima, so total energy is conserved. Scalar path difference in metres.

12. DefinitionSL & HL

    Define coherent sources.

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    Two sources are coherent if they emit waves of the same frequency (and hence the same wavelength) with a CONSTANT phase difference between them. Exam tip: same frequency alone is not enough — the phase relationship must be constant in time; the sources need not be in phase, only constantly related. For high-contrast fringes the waves should also have similar amplitude and be monochromatic. In Young's experiment coherence is obtained by illuminating both slits from a single slit or laser so both beams come from the same wavefront. Two separate filament lamps are incoherent, so no stable fringes are seen.

13. DefinitionSL & HLData booklet: No – memorise

    Distinguish between path difference and phase difference for two interfering waves.

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    Path difference is the extra distance one wave travels compared with the other in reaching a point, measured in metres. Phase difference is the fraction of a cycle by which the two oscillations differ, measured in radians (or degrees). They are linked by phase difference = (2π/λ) × path difference. Exam tip: students routinely quote a path difference in radians or omit the unit; remember that a path difference of λ corresponds to 2π rad and λ/2 to π rad. Both quantities are scalars. Waves in phase have Δ = nλ; antiphase corresponds to Δ = (n + ½)λ. λ must be the wavelength in the medium concerned.

14. DefinitionSL & HL

    Define monochromatic light and explain why it is used in the double-slit experiment.

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    Monochromatic light consists of a single frequency, and therefore a single wavelength in a given medium. Exam tip: define by FREQUENCY, since wavelength changes between media; single colour is accepted but is a weaker answer. Monochromatic light is used so that every fringe of a given order falls at one sharp position. With white light the central maximum is white, and the higher orders become coloured spectra because s = λD/d differs for each wavelength — violet lies closest to the centre, red furthest — and the orders overlap and wash out at large n. A laser gives light that is monochromatic, coherent and intense.

15. DefinitionSL & HLData booklet: Yes

    Define the fringe spacing in a double-slit interference pattern.

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    The fringe spacing s is the distance between the centres of two ADJACENT bright fringes (or two adjacent dark fringes) measured on the screen, given by s = λD/d. Unit: metre; scalar. Exam tip: to reduce the percentage uncertainty, measure across many fringes (for example 10 spacings) and divide — this is the standard improvement answer; remember that 10 bright fringes enclose only 9 spacings. The fringes are equally spaced only in the small-angle regime D ≫ d. Increasing λ or D, or decreasing d, increases s; blocking one slit destroys the fringes entirely.

16. EquationSL & HLData booklet: Yes

    State the equation n = c/v and define every symbol with its unit.

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    n = c/v (data booklet). n = absolute refractive index of the medium (no unit); c = speed of light in a vacuum = 3.00 × 10⁸ m s⁻¹; v = speed of light in the medium (m s⁻¹). Valid for a stated frequency, since n depends slightly on wavelength (dispersion). Rearranged forms you must know: v = c/n and λ_medium = λ_vacuum/n, because f is unchanged. Common misuse: substituting the speed in the first medium when the boundary lies between two media — there use n₁v₁ = n₂v₂. Sanity check: for water n = 1.33 gives v = 3.00 × 10⁸/1.33 = 2.26 × 10⁸ m s⁻¹, sensibly less than c.

17. EquationSL & HLData booklet: Yes

    State the refraction equation as printed in the data booklet and define each term.

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    n₁/n₂ = sinθ₂/sinθ₁ = v₂/v₁ (booklet), which is equivalent to n₁ sinθ₁ = n₂ sinθ₂. n₁, n₂ = refractive indices of media 1 and 2 (no unit); θ₁ = angle of incidence in medium 1 and θ₂ = angle of refraction in medium 2, both measured FROM THE NORMAL in degrees; v₁, v₂ = wave speeds in each medium (m s⁻¹). Since f is constant, λ₂/λ₁ = v₂/v₁ as well. Valid for plane boundaries and any type of wave. Misuse: reading the subscripts on the n ratio as though they matched the sine ratio. Check: air to glass (n = 1.50) at θ₁ = 30° gives sinθ₂ = 0.333, θ₂ = 19.5° — bent towards the normal.

18. EquationSL & HLData booklet: No – derive

    Derive the expression for the critical angle from Snell's law.

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    Set θ₂ = 90° in n₁ sinθ₁ = n₂ sinθ₂: n₁ sinθ_c = n₂ × 1, hence sinθ_c = n₂/n₁, and for a medium–air boundary sinθ_c = 1/n. θ_c = critical angle in the denser medium (degrees); n₁ = refractive index of the denser medium, n₂ that of the less dense medium (both dimensionless). Valid only when n₁ > n₂. Not printed in the booklet, so derive it each time. Misuse: writing sinθ_c = n₁/n₂ and obtaining sinθ_c > 1 — a calculator error message is the clue that it has been inverted. Check: glass n = 1.50 gives sinθ_c = 0.667 and θ_c = 41.8°.

19. EquationSL & HLData booklet: No – memorise

    State the path-difference and phase-difference conditions for two-source interference.

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    Constructive: path difference Δ = nλ, phase difference 2πn rad. Destructive: Δ = (n + ½)λ, phase difference (2n + 1)π rad. n = 0, 1, 2, … (integer, no unit); λ = wavelength in the medium in which the waves travel (m); Δ in metres. Valid for coherent sources with no additional phase change; if one wave reflects from a denser medium, add an extra λ/2 and the two conditions swap over. Not printed in the booklet — memorise. Misuse: using the vacuum wavelength when the waves travel in water or glass. Check: two speakers 0.85 m and 1.70 m from a listener with λ = 0.85 m give Δ = λ, so a loud maximum is heard.

20. EquationSL & HLData booklet: Yes

    State the double-slit equation s = λD/d and define every symbol.

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    s = λD/d (booklet). s = fringe spacing, the distance between adjacent bright (or adjacent dark) fringes (m); λ = wavelength of the light (m); D = distance from the slits to the screen (m); d = separation of the centres of the two slits (m). Valid when D ≫ d so that sinθ ≈ tanθ ≈ θ, and for coherent monochromatic illumination. Rearranged: λ = sd/D, the standard experimental route to λ. Misuse: confusing d (slit separation, sub-millimetre) with D (screen distance, metres), and using the width of 10 fringes as s. Check: λ = 600 nm, d = 0.50 mm, D = 2.0 m gives s = 2.4 × 10⁻³ m = 2.4 mm.

21. EquationSL & HLData booklet: No – derive

    Give the angular and linear positions of the nth bright fringe in a double-slit pattern.

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    Constructive interference requires d sinθ = nλ, so sinθ_n = nλ/d; for small angles θ_n ≈ nλ/d in radians and the distance from the centre of the screen is y_n = nλD/d, so successive fringes are separated by s = λD/d. n = order (0, 1, 2, …); d = slit separation (m); D = slit-to-screen distance (m); y_n in metres. Derive this — the booklet prints nλ = d sinθ for a grating, which is the same physics. Misuse: applying the small-angle form at large angles, where fringes cease to be evenly spaced. Check: d = 0.20 mm, λ = 500 nm, n = 1 gives sinθ = 2.5 × 10⁻³, θ = 0.14°.

22. EquationSL & HLData booklet: No – memorise

    How do amplitudes and intensities combine at a double-slit maximum and minimum?

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    At a maximum the displacements add: A_max = A₁ + A₂; at a minimum A_min = |A₁ − A₂|. Since intensity I ∝ A² (booklet), two equal sources each of amplitude A and intensity I₀ give I_max ∝ (2A)² = 4I₀ and I_min = 0. A = amplitude (m); I = intensity (W m⁻²). Valid only for coherent sources; incoherent sources simply give I₁ + I₂ = 2I₀ with no pattern. Misuse: adding intensities rather than amplitudes for coherent light. Check: the mean of the fringe pattern is 2I₀, exactly the incoherent total, so energy is conserved — it is redistributed, not created.

23. EquationSL & HLData booklet: Yes

    How do speed, wavelength and frequency change when a wave refracts into a denser medium?

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    Frequency f is unchanged because it is fixed by the source; speed and wavelength both decrease in the same ratio, since v = fλ (booklet c = fλ). In terms of refractive index, v = c/n and λ_medium = λ_vacuum/n. Symbols: f (Hz), v (m s⁻¹), λ (m), n (no unit). Valid at any refracting boundary. Misuse: claiming the frequency falls so the colour changes — colour is set by frequency, so an object does not change colour under water. Check: 600 nm light entering glass of n = 1.50 keeps f = 5.00 × 10¹⁴ Hz but has λ = 400 nm and v = 2.00 × 10⁸ m s⁻¹ inside the glass.

24. Graph/diagramSL & HL

    Describe the graph used to find the refractive index of a glass block from measured angles.

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    Plot sinθ₁ (angle in air, y-axis, no unit) against sinθ₂ (angle in the glass, x-axis, no unit). Because 1 × sinθ₁ = n sinθ₂, the graph is a straight line through the origin whose gradient is n_glass; find n from a large gradient triangle, never from a single data point. A non-zero intercept indicates a systematic error, such as the normal drawn incorrectly or the block traced out of position. Plotting θ₁ against θ₂ instead gives a curve and loses marks. A denser block or liquid steepens the line; the line ends at sinθ₁ = 1, and that x-value equals sinθ_c.

25. Graph/diagramSL & HL

    Sketch and describe the ray diagram for light in glass meeting a glass–air boundary below, at and above the critical angle.

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    Draw the boundary, the normal and the incident ray inside the glass, with arrows on every ray. Below θ_c: a bright refracted ray bends AWAY from the normal (θ₂ > θ₁) plus a weak partially reflected ray with θ_r = θ_i. At θ_c: the refracted ray grazes along the boundary at 90° to the normal and the reflected ray is bright. Above θ_c: no refracted ray at all — total internal reflection, with all the energy in the reflected ray at θ_r = θ_i. Mark every angle from the normal. As θ_i increases towards θ_c the refracted ray dims while the reflected ray brightens.

26. Graph/diagramSL & HL

    Sketch the wavefront diagrams for plane waves passing through a wide gap and through a gap about one wavelength wide.

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    Plane parallel wavefronts approach from the left, spaced by λ. Wide gap (b ≫ λ): the emerging wavefronts remain essentially plane and parallel with only slight curling at the edges, giving a well-defined beam and sharp shadows. Gap b ≈ λ: the emerging wavefronts are almost semicircular arcs centred on the gap, spreading through nearly 180° into the geometrical shadow. In both cases the wavefront SPACING (λ), the speed and the frequency are unchanged — drawing a different wavelength after the gap is a standard error. Amplitude falls as the energy spreads. Narrowing the gap or increasing λ increases the spreading.

27. Graph/diagramSL & HLData booklet: Yes

    Describe the intensity–position graph for an idealised double-slit interference pattern.

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    Axes: intensity I (W m⁻²) on the y-axis against position y on the screen (m), or against angle θ. The pattern is a series of equally spaced, identical maxima of the same height (I_max ∝ 4I₀ for two equal slits) falling to zero between them, symmetric about the central n = 0 maximum, with spacing s = λD/d. Increasing λ or D, or decreasing d, spreads the fringes; covering one slit removes the fringes and leaves a broad single-slit hump. The equal-height model is accepted at SL; at HL the fringe heights are modulated by the single-slit envelope.

28. Graph/diagramSL & HLData booklet: Yes

    Describe the linear graph used to determine λ from double-slit measurements as the screen distance is varied.

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    For each screen distance D measure the width of 10 fringe spacings and divide by 10 to obtain s. Plot s (y-axis, m) against D (x-axis, m). Since s = λD/d the line is straight through the origin with gradient λ/d, so λ = gradient × d. Draw maximum and minimum gradient lines through the error bars to obtain the uncertainty in λ. A positive y-intercept suggests a zero error in D, for example measuring from the wrong face of the slide. Using a smaller slit separation d steepens the line, as does using red light rather than blue since λ_red > λ_blue.

29. Concept/explainSL & HL

    Explain the law of reflection for a wave at a plane boundary, and state what happens to the phase of a pulse reflected at a fixed end.

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    • The incident ray, the reflected ray and the normal all lie in the same plane
    • the angle of incidence equals the angle of reflection, both measured between the ray and the NORMAL to the surface
    • frequency, wavelength and speed are unchanged because the wave stays in the same medium
    • reflection at a fixed (rigid) boundary, or from an optically denser medium, inverts the pulse: a phase change of π (half a wavelength) occurs
    • reflection at a free end, or from a less dense medium, gives no phase change
    • this π phase change is what makes thin-film and Lloyd's-mirror interference conditions swap over. Exam tip: incomplete answers measure angles from the surface and forget to say the phase change happens only at the denser/fixed boundary.
30. Concept/explainSL & HL

    Explain, in terms of wavefronts and wave speed, why light bends towards the normal on entering glass from air.

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    • Refraction is the change of direction of a wave crossing a boundary because its SPEED changes
    • the frequency is fixed by the source and does not change, so from v = fλ the wavelength decreases when the speed decreases
    • light travels more slowly in glass, v = c/n
    • consider a plane wavefront meeting the boundary obliquely: the edge that enters the glass first slows down while the rest still travels fast, so the wavefront pivots
    • the ray, which is perpendicular to the wavefront, therefore turns towards the normal
    • a ray along the normal (θ₁ = 0) does not change direction, though it still slows and its wavelength still shortens. Exam tip: students often claim the frequency or the colour changes; only speed, wavelength and direction change.
31. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of absolute refractive index and how it links Snell's law to wave speeds and wavelengths.

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    • The absolute refractive index of a medium is n = c/v, the ratio of the speed of light in a vacuum to its speed in that medium, so n ≥ 1 and n has no unit
    • at a boundary Snell's law gives n₁ sinθ₁ = n₂ sinθ₂ with both angles measured from the normal
    • combining these, n₁/n₂ = v₂/v₁ = sinθ₂/sinθ₁ = λ₂/λ₁ (booklet)
    • a larger n means a slower wave and a shorter wavelength in that medium
    • n depends slightly on wavelength (dispersion), which is why a prism separates white light and why violet deviates most
    • the same relations apply to sound and water waves, with n replaced by the speed ratio. Exam tip: the ratio form reverses the subscripts — n₁/n₂ = v₂/v₁, not v₁/v₂.
32. Concept/explainSL & HLData booklet: No – derive

    Explain total internal reflection and derive the condition for the critical angle.

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    • Light passing from a denser to a less dense medium (n₁ > n₂) bends AWAY from the normal, so θ₂ > θ₁
    • as θ₁ increases the refracted ray weakens and the reflected ray strengthens
    • at the critical angle θ_c the refracted ray grazes the boundary, θ₂ = 90°
    • substituting into Snell's law: n₁ sinθ_c = n₂ sin90° = n₂, hence sinθ_c = n₂/n₁
    • for θ₁ > θ_c no refraction is possible and 100% of the light is reflected back into the denser medium, obeying the law of reflection
    • examples: optical fibres, prismatic reflectors in binoculars and bicycle reflectors, mirages, sparkle of diamond (n = 2.42, θ_c = 24°). Exam tip: TIR is impossible going from less dense to more dense — always check which medium has the larger n.
33. Concept/explainSL & HL

    Outline how a step-index optical fibre transmits a signal and why total internal reflection rather than a mirror is used.

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    • A step-index fibre has a high-index core surrounded by a lower-index cladding, so a critical angle exists at the core–cladding boundary
    • light launched within the acceptance cone strikes the boundary at an angle greater than θ_c and is totally internally reflected repeatedly along the fibre
    • TIR reflects essentially 100% of the energy, whereas even a good mirror absorbs a few per cent at each reflection, which would be fatal after thousands of reflections
    • the cladding also protects the core surface, prevents light leaking where fibres touch, and reduces cross-talk between fibres
    • narrow cores reduce multipath (modal) dispersion, which would otherwise smear pulses and limit data rate. Exam tip: state the cladding has the SMALLER refractive index; many answers omit the cladding entirely.
34. Concept/explainSL & HL

    Explain what diffraction is and the condition under which it is significant.

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    • Diffraction is the spreading of a wave as it passes through an aperture or around an obstacle, and the bending of wavefronts into the geometrical shadow
    • it is a property of all waves and is evidence for the wave nature of light
    • the spreading is significant when the wavelength is comparable to the size of the aperture or obstacle, λ ≈ a
    • a wide slit (a ≫ λ) gives an almost straight-through beam; narrowing the slit increases the spreading
    • this is why sound (λ ~ 1 m) diffracts round a doorway but light (λ ~ 5 × 10⁻⁷ m) does not noticeably
    • long-wavelength radio waves diffract over hills, while microwaves need line of sight. Exam tip: diffraction changes the direction and amplitude distribution only — the wavelength, frequency and speed are unchanged.
35. Concept/explainSL & HLData booklet: No – memorise

    State the principle of superposition and explain the path-difference conditions for constructive and destructive interference.

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    • The principle of superposition: when two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements at that instant
    • constructive interference occurs where the waves arrive in phase, giving a large amplitude; destructive interference occurs where they arrive in antiphase, giving a small or zero amplitude
    • for two sources in phase the path difference determines the outcome: constructive when path difference = nλ, destructive when path difference = (n + ½)λ, with n = 0, 1, 2, …
    • if the sources are in antiphase the two conditions swap
    • energy is not destroyed at a minimum, it is redistributed into the maxima. Exam tip: quote the phrase vector sum of displacements — amplitudes only add directly when the waves are in phase.
36. Concept/explainSL & HL

    Explain what coherence means and why a single laser or a single slit before the double slit is needed for Young's experiment.

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    • Two sources are coherent if they have the same frequency and a constant phase difference between them
    • only then is the path-difference condition at each point on the screen fixed, so the interference maxima and minima stay in the same places long enough to be seen
    • two independent lamps emit in random bursts, so the phase difference changes randomly ~10⁸ times per second and the pattern averages out to uniform illumination
    • Young solved this by illuminating both slits with light from a single narrow slit, so the two slits are driven by the same wavefront; a laser is naturally coherent and monochromatic
    • the fringes are also clearest when the two amplitudes are similar, giving complete cancellation at the minima. Exam tip: coherent is not the same as monochromatic — say constant phase difference.
37. Concept/explainSL & HLData booklet: Yes

    Describe the fringe pattern produced by two narrow slits with monochromatic light, and how it changes with white light.

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    • A series of bright and dark fringes appears on the screen, equally spaced near the centre, with separation s = λD/d
    • the central fringe (n = 0, zero path difference) lies on the perpendicular bisector of the slits
    • with two ideal narrow slits the maxima all have about the same intensity and the intensity varies smoothly (cos²) between them
    • increasing D or λ, or decreasing d, widens the fringes
    • with white light the central fringe is white because all wavelengths have zero path difference there; the higher orders spread into spectra with blue nearest the centre (shortest λ) and red furthest out, and the orders soon overlap and wash out. Exam tip: state clearly which quantity d is — the slit SEPARATION, not the slit width.
38. Worked problemSL & HLData booklet: Yes

    A ray of light in air strikes a flat glass surface (n = 1.50) at 42° to the normal. Calculate the angle of refraction.

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    Use Snell's law n₁ sinθ₁ = n₂ sinθ₂ with n₁ = 1.00 (air), θ₁ = 42°, n₂ = 1.50. Substitute: 1.00 × sin42° = 1.50 × sinθ₂. sin42° = 0.6691, so sinθ₂ = 0.6691/1.50 = 0.4461. θ₂ = sin⁻¹(0.4461) = 26.5°. The ray bends TOWARDS the normal, as expected on entering a slower, denser medium. Answer: θ₂ = 26.5° (3 s.f.), measured from the normal inside the glass. Check/Trap: make sure the calculator is in degrees, and never write sinθ₂ = sin42°/1.50 as 42/1.50 = 28° — you must take the sine first and the inverse sine last. Also check the sense of the bend: if your answer were larger than 42° you have inverted the ratio.

39. Worked problemSL & HLData booklet: Yes

    Light of wavelength 590 nm in air enters water of refractive index 1.33. Determine the speed, frequency and wavelength of the light in the water.

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    Speed: n = c/v so v = c/n = 3.00 × 10⁸/1.33 = 2.26 × 10⁸ m s⁻¹. Frequency is set by the source and is UNCHANGED at the boundary: f = c/λ_air = 3.00 × 10⁸/(590 × 10⁻⁹) = 5.08 × 10¹⁴ Hz. Wavelength in water: λ_water = v/f = 2.26 × 10⁸/(5.08 × 10¹⁴) = 4.45 × 10⁻⁷ m, or directly λ_water = λ_air/n = 590/1.33 = 444 nm. Answers: v = 2.26 × 10⁸ m s⁻¹, f = 5.08 × 10¹⁴ Hz, λ = 444 nm (3 s.f.). Check/Trap: the classic error is dividing the frequency by n. Frequency never changes on refraction — if it did, energy per photon would change and the colour would change, which it does not.

40. Worked problemSL & HLData booklet: No – derive

    Calculate the critical angle for a glass–air boundary where the glass has refractive index 1.52, and state what happens at 45°.

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    At the critical angle the refracted ray grazes the surface, θ₂ = 90°. Snell: n₁ sinθ_c = n₂ sin90°, so sinθ_c = n₂/n₁ = 1.00/1.52 = 0.6579. θ_c = sin⁻¹(0.6579) = 41.1° (3 s.f.). Because 45° > 41.1°, a ray inside the glass meeting the boundary at 45° undergoes TOTAL INTERNAL REFLECTION: no light is transmitted, and it reflects at 45° on the other side of the normal. This is exactly why 45°–45°–90° glass prisms are used as reflectors in periscopes and binoculars. Check/Trap: TIR requires the light to start in the DENSER medium; a ray arriving from air at 45° simply refracts into the glass. Also θ_c depends on wavelength, so θ_c differs slightly for red and blue.

41. Worked problemSL & HLData booklet: No – derive

    An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.45. Determine the maximum angle to the fibre axis at which light can enter the flat end face from air and still be totally internally reflected.

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    Step 1 — critical angle at the core–cladding wall: sinθ_c = n_clad/n_core = 1.45/1.50 = 0.9667, θ_c = 75.2°. Step 2 — geometry: if the ray inside the core makes angle β with the axis, it meets the wall at (90° − β) to the normal of the wall. TIR needs 90° − β ≥ 75.2°, so β ≤ 14.8°. Step 3 — refraction at the end face: 1.00 × sinα = 1.50 × sinβ, so sinα = 1.50 × sin14.8° = 1.50 × 0.2554 = 0.3831. α = 22.5°. Answer: acceptance angle α_max = 22.5° (half-angle of the acceptance cone). Check/Trap: the commonest slip is using 75.2° directly in Snell at the end face — the wall normal and the axis are perpendicular, so you must subtract from 90°.

42. Worked problemSL & HLData booklet: Yes

    Coherent light of wavelength 633 nm illuminates two slits 0.25 mm apart. The screen is 1.80 m away. Calculate the fringe separation.

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    Use the booklet relation s = λD/d, valid for D ≫ d and small angles. Convert: λ = 633 × 10⁻⁹ m, D = 1.80 m, d = 0.25 × 10⁻³ m. s = (633 × 10⁻⁹ × 1.80)/(0.25 × 10⁻³) = (1.139 × 10⁻⁶)/(2.5 × 10⁻⁴) = 4.56 × 10⁻³ m. Answer: s = 4.56 mm, or 4.6 mm to 2 s.f. matching the precision of d. Sanity check: fringes of a few millimetres over a couple of metres is exactly what a school laser experiment gives. Check/Trap: d is the slit SEPARATION and D the slit-to-screen distance — swapping them here would give an absurd 4.6 km. Convert mm to m before substituting, and remember s is the distance between ADJACENT bright fringes.

43. Worked problemSL & HLData booklet: Yes

    In a Young's slits experiment the distance across 10 bright fringes is measured as 22.0 mm with slits 0.30 mm apart and a screen 1.05 m away. Determine the wavelength of the light.

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    Fringe separation: 10 fringe spacings span 22.0 mm, so s = 22.0/10 = 2.20 mm = 2.20 × 10⁻³ m. Rearrange s = λD/d to give λ = sd/D. Substitute: λ = (2.20 × 10⁻³ × 0.30 × 10⁻³)/1.05 = (6.60 × 10⁻⁷)/1.05 = 6.29 × 10⁻⁷ m. Answer: λ = 6.3 × 10⁻⁷ m = 629 nm ≈ 630 nm (2 s.f., limited by d given to 2 s.f.) — red light, consistent with a helium–neon laser. Check/Trap: measuring across 10 fringes and dividing reduces the fractional uncertainty tenfold; the trap is forgetting to divide, which gives a wavelength ten times too large. Note that 10 bright fringes counted from first to last span 9 spacings if you count the fringes rather than the gaps — state clearly which you measured.

44. Worked problemSL & HLData booklet: Yes

    Water waves of frequency 5.0 Hz travel at 0.40 m s⁻¹ in deep water and 0.25 m s⁻¹ in shallow water. They cross the boundary at 30° to the normal. Determine the angle of refraction and the wavelength in each region.

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    Wavelengths: frequency is unchanged, so λ_deep = v₁/f = 0.40/5.0 = 0.080 m and λ_shallow = v₂/f = 0.25/5.0 = 0.050 m. Refraction: use sinθ₂/sinθ₁ = v₂/v₁ (booklet). sinθ₂ = sin30° × (0.25/0.40) = 0.500 × 0.625 = 0.3125, so θ₂ = sin⁻¹(0.3125) = 18.2°. Answers: θ₂ = 18°(2 s.f.), λ_deep = 8.0 cm, λ_shallow = 5.0 cm. The wave slows, so it bends towards the normal and the wavefronts move closer together — exactly what a ripple tank with a submerged plate shows. Check/Trap: the ratio form has the subscripts crossed, sinθ₂/sinθ₁ = v₂/v₁; writing v₁/v₂ gives sinθ₂ = 0.800 and θ₂ = 53°, a bend away from the normal, which contradicts the slowing.

45. Worked problemSL & HLData booklet: No – memorise

    Two loudspeakers driven in phase emit sound of frequency 660 Hz (speed of sound 330 m s⁻¹). A listener is 4.20 m from one speaker and 5.45 m from the other. Deduce whether a maximum or a minimum is heard.

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    Wavelength: λ = v/f = 330/660 = 0.500 m. Path difference: Δ = 5.45 − 4.20 = 1.25 m. Express in wavelengths: Δ/λ = 1.25/0.500 = 2.5 = (2 + ½)λ. This matches the destructive condition (n + ½)λ with n = 2, so the waves arrive in antiphase and the listener hears a MINIMUM (quiet point). Check/Trap: the sources must be coherent and in phase for this test — if the speakers were driven in ANTIPHASE the conditions swap and 2.5λ would give a maximum, so always state the source phase. The minimum is not silent in practice because the amplitudes are unequal (different distances) and the room reflects sound. Energy is not destroyed: it is redistributed into the maxima elsewhere.

46. Exam technique/trapSL & HL

    Explain the most common errors made when applying Snell's law and the critical-angle condition in exams.

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    Trap 1: measuring angles from the SURFACE rather than the normal. Snell's law is defined with both angles between the ray and the normal; a diagram angle of 50° to the glass surface is 40° to the normal. Always draw and label the normal. Trap 2: pairing the wrong n with the wrong θ — n₁ always goes with the medium the ray is IN when it makes θ₁. Trap 3: assuming total internal reflection can occur going into a denser medium; it cannot, since sinθ_c = n₂/n₁ needs n₂ < n₁. Trap 4: taking sin⁻¹ of a number greater than 1 and writing a maths error instead of concluding TIR occurs. Approach: label media, write n₁ sinθ₁ = n₂ sinθ₂, substitute, and finally check the bend is towards the normal when entering the slower medium.

47. Exam technique/trapSL & HLData booklet: Yes

    Identify the traps in using s = λD/d and describe how to avoid them.

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    The symbols are routinely confused: s is the fringe SEPARATION (distance between adjacent bright fringes), D is the slit-to-screen distance, d is the SLIT SEPARATION (centre to centre), and none of them is the slit width. Trap 1: substituting the slit width for d — the width appears only in the HL diffraction envelope. Trap 2: forgetting to convert mm to m; a factor of 10³ makes the answer absurd. Trap 3: reading s from the screen as the width of one bright fringe rather than the centre-to-centre spacing. Trap 4: using the equation when D is not ≫ d, or at large angles, where the small-angle approximation fails. Approach: quote the equation, list each symbol with its unit, convert all lengths to metres, and finish with a sanity check — school fringes are a few millimetres wide.

48. Exam technique/trapSL & HL

    Outline how the command terms state, describe, explain, determine, deduce, sketch and suggest should be answered in wave-phenomena questions.

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    State: give the fact only, no reasoning — for example, the angle of incidence equals the angle of reflection. Describe: give the observable features in words — bright and dark fringes, equally spaced, brightest at the centre. Explain: give reasons and mechanism — link the path difference to phase and to superposition. Determine: obtain a numerical answer from data, showing working, units and sensible significant figures. Deduce: reach a conclusion AND justify it with the physics, e.g. path difference = 2.5λ therefore a minimum. Sketch: a labelled freehand graph or diagram with the correct SHAPE, axes labelled and key values marked; no plotting required. Suggest: propose a plausible physical reason for an observation not directly taught. Exam tip: marks are lost mainly by explaining when only a statement is asked, and by describing when explanation is required.

49. Exam technique/trapSL & HLData booklet: Yes

    Describe an experiment to determine the refractive index of a glass block, including the graph used and the treatment of uncertainties.

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    Apparatus: rectangular glass block, plain paper, ray box with a single slit (or optical pins), protractor, sharp pencil. Method: draw round the block, mark the normal, direct a ray at the surface, mark the incoming and emerging rays with pins or pencil crosses, remove the block and measure θ₁ (air) and θ₂ (glass) from the normal. Repeat for θ₁ from 10° to 70° in 10° steps, and repeat each reading. Independent variable θ₁, dependent θ₂, controlled: same block, same face, same wavelength (colour), same normal. Linearisation: plot sinθ₁ (y) against sinθ₂ (x); Snell gives sinθ₁ = n sinθ₂, so the gradient is n and the line passes through the origin. Uncertainty: ±0.5° on each angle, propagate to sinθ, add error bars, and take max/min gradients through the bars to obtain Δn. Limitations: thick pencil lines and ray divergence.

50. Exam technique/trapSL & HLData booklet: Yes

    Describe how to measure the wavelength of laser light using a double slit, including controls, limitations and improvements.

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    Apparatus: laser, double slit of known separation d, metre rule, screen, darkened room. Method: fix the slits, measure D from slits to screen with a metre rule, measure the distance across at least 10 fringe spacings with a travelling microscope or a rule and divide by the number of SPACINGS to get s, then λ = sd/D. Independent variable D (vary it and plot s against D: gradient = λ/d, so λ = gradient × d, and the line through the origin shows systematic errors are small). Controlled: same slits, same laser, room dark. Safety: never look along the beam. Limitations: fringe edges are fuzzy so locating centres is the dominant uncertainty; d quoted by the manufacturer may itself be uncertain; the screen may not be perpendicular. Improvements: larger D for wider fringes, more fringes measured, repeat readings, use a travelling microscope.

51. Exam technique/trapSL & HL

    Explain the difference between random and systematic errors, and between precision and accuracy, using a fringe-spacing measurement as the example.

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    Random errors scatter readings either side of the true value: judging the centre of a fuzzy fringe gives values a little too large or too small each time. They are reduced by repeating and averaging, and by measuring across many fringes so the error is divided. They show as scatter of points about the best-fit line and set the size of the error bars. Systematic errors shift every reading the same way: a metre rule with a worn zero end, or measuring D to the front of the slit holder rather than the slits, makes every D too small and every calculated λ too small. They show as a non-zero intercept on an s against D graph and are NOT reduced by repeating. Precise means the readings are closely grouped (small random error); accurate means the mean is close to the true value (small systematic error). A set can be precise but inaccurate.

52. DefinitionHL onlyData booklet: No – memorise

    Define a diffraction grating and its grating spacing d.

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    A diffraction grating is a plate ruled with a very large number of equally spaced parallel slits (lines) which diffract light and produce sharp, widely separated interference maxima. The grating spacing d is the centre-to-centre distance between adjacent slits, d = 1/N, where N is the number of lines per metre. Unit of d: metre; typically a few micrometres. Exam tip: convert lines per millimetre correctly — 600 lines mm⁻¹ = 6.00 × 10⁵ m⁻¹, so d = 1.67 × 10⁻⁶ m; omitting the mm to m conversion is the commonest grating error. Increasing the number of illuminated slits sharpens and brightens the maxima without moving them.

53. DefinitionHL onlyData booklet: Yes

    Define the order n of a maximum produced by a diffraction grating.

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    The order n is the integer in nλ = d sinθ giving the number of whole wavelengths of path difference between light from ADJACENT slits: n = 0 is the undeviated central maximum, n = 1 the first-order maxima either side, and so on. Exam tip: n must be an integer and sinθ ≤ 1, so the highest observable order is n_max = d/λ rounded DOWN; quoting a fractional order, or one requiring sinθ > 1, is a standard trap. Higher orders are more widely dispersed, giving better separation of wavelengths in a spectrum, but they are dimmer and may be missing where the single-slit envelope has a zero.

54. DefinitionHL only

    State the phase change that occurs when light is reflected at a boundary.

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    Light reflected at a boundary with a medium of HIGHER refractive index undergoes a phase change of π rad (180°), equivalent to an extra path difference of λ/2; reflection at a boundary with a lower-index medium produces no phase change. Exam tip: this is exactly why the thin-film conditions look inverted — with a π change at ONE surface only, constructive interference is 2dn = (m + ½)λ and destructive is 2dn = mλ. Always count how many π changes occur before choosing a condition. Mechanical analogy: a pulse on a string reflected from a fixed end is inverted, while reflection from a free end is not.

55. DefinitionHL onlyData booklet: No – derive

    Explain what an anti-reflection coating is and state its minimum thickness.

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    A thin transparent film (typically magnesium fluoride, n = 1.38) is deposited on a lens so that light reflected from the air–film surface and light reflected from the film–glass surface interfere DESTRUCTIVELY, suppressing the reflection and increasing transmission. Because n_air < n_film < n_glass there is a π phase change at BOTH surfaces, so the two cancel and destructive reflection requires 2dn = (m + ½)λ, giving a minimum (quarter-wave) thickness d = λ/(4n). Exam tip: λ is the wavelength IN VACUUM — the refractive index in the formula already accounts for the film. Coatings are optimised near 550 nm, which is why coated lenses look purple.

56. DefinitionHL onlyData booklet: Yes

    Define the condition for the first minimum in single-slit diffraction.

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    For a slit of width b the first intensity minimum occurs at an angle θ = λ/b from the central axis, with θ in radians. It arises because light from the top half of the slit cancels light from the bottom half in pairs when the path difference between the two edges of the slit is exactly λ. Exam tip: λ/b is the angle to the FIRST MINIMUM, that is the half-width of the central maximum; the central maximum has angular width 2λ/b and linear width 2λD/b on a screen at distance D. Requires λ ≤ b, otherwise no minima exist. Units: b in m, λ in m, θ in rad.

57. EquationHL onlyData booklet: Yes

    State the single-slit diffraction equation θ = λ/b and its conditions of validity.

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    θ = λ/b (booklet). θ = angle from the central axis to the first minimum, in RADIANS; λ = wavelength (m); b = slit width (m). Valid in the small-angle far-field (Fraunhofer) regime with λ ≪ b and screen distance D ≫ b. Useful forms: angular width of the central maximum 2λ/b; linear half-width on a screen λD/b, full width 2λD/b; the nth minimum lies at θ = nλ/b. Misuse: treating θ = λ/b as the full width of the central peak, or working in degrees. Check: λ = 600 nm, b = 0.10 mm, D = 2.0 m gives central width 2 × 600 × 10⁻⁹ × 2.0/(1.0 × 10⁻⁴) = 2.4 × 10⁻² m = 24 mm.

58. EquationHL onlyData booklet: Yes

    State the diffraction-grating equation and define every symbol.

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    nλ = d sinθ (booklet). n = order of the maximum (integer, no unit); λ = wavelength (m); d = grating spacing = 1/(lines per metre) (m); θ = angle of the nth maximum measured from the straight-through zero-order direction, in degrees. Valid for light at normal incidence on a grating of many equally spaced slits. Rearranged: λ = d sinθ/n, and n_max = d/λ rounded down since sinθ ≤ 1. Misuse: applying the small-angle approximation — grating angles are LARGE, so never write θ ≈ nλ/d. Check: 600 lines mm⁻¹ (d = 1.67 × 10⁻⁶ m) with λ = 589 nm at n = 1 gives sinθ = 0.354, θ = 20.7°.

59. EquationHL onlyData booklet: No – memorise

    How is the grating spacing d obtained from the number of lines per millimetre?

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    d = 1/N, where N is the number of lines (slits) per unit length; the units must match, so N in lines per metre gives d in metres. Example: 300 lines mm⁻¹ = 3.00 × 10⁵ lines m⁻¹, hence d = 1/(3.00 × 10⁵) = 3.33 × 10⁻⁶ m. Not printed in the booklet — memorise. Misuse: leaving N in mm⁻¹, which makes d 1000 times too large and sinθ 1000 times too small; the sanity check is that d must be a few micrometres, i.e. a few wavelengths of light. A smaller d gives larger diffraction angles and a more spread-out spectrum but fewer visible orders.

60. EquationHL onlyData booklet: Yes

    State the thin-film condition for constructive interference and define every term.

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    2dn = (m + ½)λ (booklet). d = thickness of the film (m); n = refractive index of the film (no unit); m = 0, 1, 2, … (integer); λ = wavelength IN VACUUM (m). Valid for near-normal incidence on a film with a π phase change at ONE surface only, for example a soap film in air or an oil film on water where n_oil > n_water. The term 2dn is the extra optical path through the film and the ½λ comes from the single phase change. Misuse: substituting the wavelength inside the film — the factor n already accounts for it. Check: soap film n = 1.33, d = 100 nm gives strong reflection at λ = 4dn = 532 nm (green).

61. EquationHL onlyData booklet: Yes

    State the thin-film condition for destructive interference and give an application.

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    2dn = mλ (booklet), where d = film thickness (m), n = refractive index of the film (no unit), m = 1, 2, 3, … and λ = vacuum wavelength (m). Same validity: near-normal incidence with a phase change at one surface only. Application: an anti-reflection coating has a phase change at BOTH surfaces (n_air < n_coating < n_glass), so the two π changes cancel and destructive reflection instead requires 2dn = (m + ½)λ, giving a minimum thickness d = λ/(4n). Misuse: quoting the booklet lines without first counting the phase changes. Check: MgF₂ with n = 1.38 at λ = 550 nm gives d = 550/(4 × 1.38) = 99.6 nm ≈ 100 nm.

62. Graph/diagramHL onlyData booklet: Yes

    Describe the intensity graph for single-slit diffraction and how it changes with slit width and wavelength.

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    Axes: intensity I (W m⁻²) against angle θ (rad) or screen position (m). There is a tall central maximum centred on θ = 0 with zeros at θ = ±λ/b, ±2λ/b, ±3λ/b …, and much weaker secondary maxima between them, the first side maximum being only about 4.5% of the central peak. The central maximum is twice as wide (2λ/b) as each secondary maximum. Narrowing b widens the whole pattern while admitting less light, so the peak is dimmer; increasing λ also widens it. Extract λ or b from the first minimum: λ = bθ₁ ≈ b y₁/D, where y₁ is the distance from the centre to the first dark fringe.

63. Graph/diagramHL onlyData booklet: No – derive

    Describe how the single-slit envelope modulates a double-slit pattern and explain missing orders.

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    Plot I against θ: the closely spaced double-slit fringes, whose spacing is set by the slit separation d, sit inside a broad single-slit envelope whose width is set by the slit WIDTH b, so the fringes fade to zero at the envelope minimum θ = λ/b and reappear, much weaker, in the side lobes. A double-slit maximum coinciding with a single-slit zero is MISSING: this occurs when d sinθ = nλ and b sinθ = mλ simultaneously, that is when n/m = d/b, so for d = 3b the orders n = 3, 6, 9 … are absent. Reducing b widens the envelope so more orders are visible; changing d alters only the fringe spacing.

64. Graph/diagramHL onlyData booklet: Yes

    Compare the intensity patterns of a two-slit and a many-slit arrangement, and describe grating spectra.

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    For the same d and λ the maxima lie at exactly the SAME angles (nλ = d sinθ), but as the number of illuminated slits increases the maxima become much narrower, sharper and brighter with near-total darkness between them, whereas two slits give broad, gently varying fringes. With white light the zero order is white, because θ = 0 for every wavelength, while each higher order spreads into a spectrum with violet nearest the centre and red furthest, since sinθ ∝ λ; higher-order spectra are wider and can overlap. Extract λ by plotting sinθ (y-axis) against n (x-axis): the gradient equals λ/d.

65. Concept/explainHL onlyData booklet: Yes

    Explain the origin of the first minimum in single-slit diffraction and why it occurs at θ = λ/b.

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    • Treat the slit of width b as a large number of secondary (Huygens) sources across the aperture
    • at angle θ, pair each source in the top half with the source a distance b/2 below it; the extra path for the lower one is (b/2) sinθ
    • every pair cancels when (b/2) sinθ = λ/2, that is b sinθ = λ
    • for the small angles used in IB, sinθ ≈ θ in radians, giving the booklet form θ = λ/b for the first minimum
    • further minima occur at θ = nλ/b (n = 1, 2, 3 …), never n = 0
    • the central maximum therefore has angular width 2λ/b, twice that of the subsidiary maxima, and carries most of the energy. Exam tip: b is the slit WIDTH; θ must be in radians, and the formula gives a MINIMUM, not a maximum.
66. Concept/explainHL onlyData booklet: Yes

    Explain how the single-slit diffraction pattern changes when the slit width and the wavelength are changed.

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    • Since θ = λ/b, narrowing the slit increases θ, so the central maximum becomes WIDER and the whole pattern spreads out
    • less light passes through, so the peak intensity falls; the pattern is dimmer as well as broader
    • widening the slit narrows the pattern until, for b ≫ λ, it approaches the geometrical bright band of ray optics
    • increasing the wavelength (red rather than blue) also increases θ and widens the pattern, so red fringes spread more than blue
    • intensity of the first subsidiary maximum is only about 4–5% of the central peak, and successive maxima fall away rapidly
    • total energy through the slit falls as b falls, so peak intensity ∝ b² for a fixed source. Exam tip: sketches must show a central peak twice as wide as the side peaks and rapidly decreasing side-peak heights.
67. Concept/explainHL onlyData booklet: Yes

    Explain why a real two-slit pattern is modulated by a single-slit envelope and what causes missing orders.

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    • Real slits have a finite width b as well as a separation d
    • each slit alone produces a single-slit diffraction pattern (envelope) with minima at sinθ = nλ/b
    • the two slits then interfere, giving fine two-slit maxima at d sinθ = nλ
    • the observed intensity is the two-slit interference pattern multiplied by the single-slit envelope, so the fringes are strongest near the centre and fade towards the edges of the central envelope
    • a fringe is MISSING when an interference maximum falls exactly on a diffraction minimum, i.e. when n_interference/n_diffraction = d/b
    • so if d/b = 3 the orders 3, 6, 9 … are absent
    • the number of visible fringes in the central envelope is 2(d/b) − 1. Exam tip: students draw equal-height fringes; the envelope must be shown.
68. Concept/explainHL onlyData booklet: Yes

    Explain why a diffraction grating produces much sharper and brighter maxima than a double slit.

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    • A grating has N very large (thousands of lines per mm), all illuminated coherently
    • a principal maximum occurs where every slit is in phase, nλ = d sinθ, the same angles as for two slits with the same d
    • moving slightly off that angle, the many small phase differences from N slits sum to zero much more rapidly than for two slits, so the maxima are extremely narrow — the angular width is proportional to 1/N
    • the maxima are also much brighter because N amplitudes add, giving intensity ∝ N² at the peaks
    • the sharpness allows wavelengths that differ by a tiny amount to be separated, which is why gratings are used in spectroscopy rather than double slits. Exam tip: d is the grating spacing = 1/(number of lines per metre); convert lines per mm carefully.
69. Concept/explainHL onlyData booklet: Yes

    Describe the spectra produced by a diffraction grating with white light and compare them with a prism spectrum.

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    • The grating equation nλ = d sinθ shows θ depends on wavelength, so each order (except n = 0) spreads white light into a continuous spectrum
    • the zero order is undeviated and white, since sinθ = 0 for all λ
    • in every other order RED is deviated MOST and violet least, because θ increases with λ
    • several orders appear symmetrically either side of the centre and higher orders may overlap (e.g. the red of order 2 can overlap the violet of order 3)
    • the maximum order is limited by sinθ ≤ 1, so n ≤ d/λ
    • a prism disperses by dispersion of n with wavelength and bends VIOLET most, gives only one spectrum, is brighter but far less sharp and non-linear in wavelength. Exam tip: grating and prism spread the colours in OPPOSITE senses.
70. Concept/explainHL onlyData booklet: Yes

    Explain thin-film interference, including the role of the phase change on reflection.

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    • Light reflects partly at the top surface of the film and partly at the bottom surface; the two reflected beams are coherent and superpose
    • for near-normal incidence the second beam travels an extra optical path 2dn, where d is the film thickness and n its refractive index (the wavelength inside the film is λ/n)
    • reflection at a boundary with a MORE dense medium adds a phase change of π, equivalent to an extra λ/2; reflection at a less dense medium adds none
    • when there is a phase change at one surface only, the conditions become 2dn = (m + ½)λ for constructive and 2dn = mλ for destructive reflection (booklet)
    • hence soap bubbles and oil films show colours that change with thickness and viewing angle. Exam tip: use 2dn, not 2d, and always check how many π phase changes occur.
71. Worked problemHL onlyData booklet: Yes

    Light of wavelength 589 nm passes through a single slit of width 0.040 mm. Calculate the width of the central maximum on a screen 2.00 m away.

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    Angular position of the first minimum: θ = λ/b = (589 × 10⁻⁹)/(4.0 × 10⁻⁵) = 1.47 × 10⁻² rad (small, so the approximation is safe). Linear distance from the centre to the first minimum: y = Dθ = 2.00 × 1.47 × 10⁻² = 2.94 × 10⁻² m. The central maximum runs from the first minimum on one side to the first on the other, so its width = 2y = 5.9 × 10⁻² m. Answer: width ≈ 5.9 cm (2 s.f.). Check/Trap: θ = λ/b is in RADIANS — do not press sin⁻¹ or convert to degrees before multiplying by D. The other classic error is quoting y instead of 2y; the question asked for the full width of the central maximum, which is twice the half-width and twice the width of each side maximum.

72. Worked problemHL onlyData booklet: Yes

    A diffraction grating has 600 lines per millimetre and is illuminated normally by laser light of wavelength 633 nm. Determine the angles of the first and second order maxima and the highest order observable.

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    Grating spacing: d = 1/(600 lines mm⁻¹) = 1/(6.00 × 10⁵ m⁻¹) = 1.67 × 10⁻⁶ m. Use nλ = d sinθ, so sinθ = nλ/d. First order (n = 1): sinθ = (633 × 10⁻⁹)/(1.67 × 10⁻⁶) = 0.380, θ₁ = 22.3°. Second order (n = 2): sinθ = 0.760, θ₂ = 49.4°. Highest order: sinθ ≤ 1 requires n ≤ d/λ = 1.67 × 10⁻⁶/6.33 × 10⁻⁷ = 2.63, and n must be an integer, so n_max = 2. Answers: 22.3°, 49.4°, maximum order 2 (plus the zero order). Check/Trap: convert lines per mm to a spacing in metres first — using 600 as d, or forgetting the reciprocal, is the most frequent error. Note the orders are NOT equally spaced in angle because sinθ, not θ, is proportional to n.

73. Worked problemHL onlyData booklet: Yes

    A grating with 500 lines per mm gives a second-order maximum at 39.0°. Determine the wavelength of the light.

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    Grating spacing: d = 1/(5.00 × 10⁵ m⁻¹) = 2.00 × 10⁻⁶ m. Rearranging nλ = d sinθ gives λ = d sinθ/n. Substitute n = 2, θ = 39.0°: sin39.0° = 0.6293, so λ = (2.00 × 10⁻⁶ × 0.6293)/2 = 6.29 × 10⁻⁷ m. Answer: λ = 6.29 × 10⁻⁷ m = 629 nm (3 s.f.), in the orange–red. Check/Trap: using a higher order improves precision — an uncertainty of ±0.2° in θ gives a fractional uncertainty of only about 0.4% here, and dividing by n reduces the effect of the angular uncertainty further. Do not forget to divide by n; leaving it out doubles the answer to 1259 nm, which is not visible light and should be spotted immediately as a sanity-check failure.

74. Worked problemHL onlyData booklet: Yes

    A soap film of refractive index 1.33 in air is 320 nm thick. Determine which visible wavelength is most strongly reflected at normal incidence.

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    There is a π phase change at the top (air → soap, denser) but none at the bottom (soap → air, less dense), so the phase change occurs at ONE surface and the booklet constructive condition is 2dn = (m + ½)λ. Compute 2dn = 2 × 320 × 10⁻⁹ × 1.33 = 8.51 × 10⁻⁷ m. Then λ = 2dn/(m + ½). For m = 0: λ = 8.51 × 10⁻⁷/0.5 = 1702 nm (infrared, not visible). For m = 1: λ = 8.51 × 10⁻⁷/1.5 = 5.67 × 10⁻⁷ m = 567 nm — green, visible. For m = 2: λ = 8.51 × 10⁻⁷/2.5 = 340 nm (ultraviolet). Answer: the film reflects green light of about 570 nm strongly. Check/Trap: use the optical path 2dn, not 2d, and always test several values of m and reject those outside 400–700 nm.

75. Worked problemHL onlyData booklet: Yes

    Determine the minimum thickness of a magnesium fluoride coating (n = 1.38) on glass (n = 1.52) that minimises reflection of light of wavelength 550 nm at normal incidence.

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    Here the refractive index increases at BOTH surfaces (1.00 → 1.38 → 1.52), so there is a π phase change at each reflection. The two phase changes cancel, leaving only the optical path difference 2dn. Destructive interference of the reflected beams therefore requires 2dn = (m + ½)λ. The minimum non-zero thickness is m = 0: d = λ/(4n) = (550 × 10⁻⁹)/(4 × 1.38) = 9.96 × 10⁻⁸ m ≈ 1.0 × 10⁻⁷ m = 100 nm. Answer: d ≈ 1.0 × 10⁻⁷ m, a quarter-wavelength coating. Check/Trap: count the phase changes BEFORE choosing the formula — with a phase change at only one surface, destructive reflection would instead need 2dn = mλ. Note 550 nm (green) is chosen as the middle of the visible range, so red and blue are reflected slightly, giving coated lenses their purple sheen.

76. Exam technique/trapHL onlyData booklet: Yes

    Explain the traps in using θ = λ/b and distinguishing it from nλ = d sinθ.

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    Trap 1: mixing the two equations. θ = λ/b describes single-slit DIFFRACTION and gives the first MINIMUM, where b is the slit WIDTH. nλ = d sinθ describes a grating and gives MAXIMA, where d is the spacing between adjacent slits. Trap 2: units of angle — θ = λ/b is a small-angle result and returns radians; multiply straight by D to get a distance, and do not take sin⁻¹. Trap 3: quoting the half-width when the question asks for the width of the central maximum (which is 2λD/b). Trap 4: using θ = λ/b for a double slit and expecting the whole pattern; it gives only the envelope. Approach: identify whether the question is about the envelope (width b) or the fringes (separation d), state which equation and why, then substitute in metres.

77. Exam technique/trapHL onlyData booklet: Yes

    Describe how a diffraction grating is used to determine a wavelength and explain why it is more precise than a double slit.

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    Apparatus: laser or spectrometer with collimator and telescope, grating of known lines per mm on a rotating table, protractor scale (vernier). Method: set the grating perpendicular to the beam, locate the maxima either side of the zero order, measure the angle on both sides and halve the difference — this cancels the systematic error of a misaligned zero. Repeat for higher orders and plot sinθ (y) against n (x); the gradient is λ/d, so λ = gradient × d. Controlled: grating spacing, normal incidence, same source. Why more precise: the maxima are extremely sharp (angular width ∝ 1/N), the angles are large (tens of degrees) so a ±0.1° uncertainty is a very small fraction, and higher orders multiply the effect. Limitations: the quoted lines per mm carries its own uncertainty; the grating must be normal to the beam or the two sides differ.

78. Exam technique/trapHL onlyData booklet: Yes

    Explain the traps in thin-film interference questions and the correct method for choosing the condition.

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    Trap 1: using the path difference 2d instead of the OPTICAL path difference 2dn — the wavelength inside the film is λ/n, so the film is optically thicker than it looks. Trap 2: quoting the booklet conditions without checking the phase changes. The forms 2dn = (m + ½)λ constructive and 2dn = mλ destructive apply when a π phase change occurs at ONE surface only (e.g. a soap film in air, or oil on water with n_oil > n_water). If the index increases at both surfaces (an anti-reflection coating) the two π changes cancel and the conditions SWAP. Trap 3: forgetting the question may ask about transmitted rather than reflected light, which is complementary. Method: draw the two reflected rays, mark each boundary as denser or less dense, count the π changes, then choose the condition and test integer values of m for a visible answer.

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