★ Exam Tip: Everything in this chapter (stationary waves,
interference, diffraction grating) comes under the principle of superposition. Conditions,
definitions and word-perfect phrases earn marks (3–4 marks just for “conditions of interference”). Learn
the formulas given here, and read the question first — many answers can be worked out from a quick sketch.
1. Principle of Superposition
Principle: every “principle” starts with a condition (moments → equilibrium,
momentum → closed system). Here too:
When 2 or more waves of the same nature meet at a point, the resultant displacement is the
algebraic sum of the individual displacements at that point.
Wording: “displacement” is the better word, but mark schemes accept
“amplitude” too — the board answers use resultant amplitude is the algebraic sum of the individual
amplitudes.
Same nature = same family: light + light, sound + sound. Red light + yellow light,
or microwave + UV, cannot superpose.
They must meet at a point.
Algebraic sum = ordinary + and − signs (like 1 + 2), done at every point (+ and − displacements
both count).
Drawing the resultant: take as many points as you can, add the two displacements at each
(zero + zero = 0, A + 2A = 3A, −A + −2A = −3A…) and join them.
(red wave 1 = A, blue wave 2 =
2A, green = resultant) Constructive: resultant is always greater
than both (3A = A + 2A). Waves cross only where both are 0.Destructive: 2A + (−A) = A — resultant is never greater than either wave.Neither: in some places the resultant is bigger than both, in others smaller → do not call it
constructive or destructive.
★ Learn the identification rule:Constructive ⇒ resultant is
greater than both individual waves at every point. Destructive ⇒ resultant is never greater
than the individual waves. Anything in between is neither (sir: “rough sketches are fine, take andaza — but
plot many points; if two points look unreasonably close to 2.5A, check the in-between points”).
Examiner style: if they give you two values at a point (e.g. at
0.3 m), just add the two displacements. If they say draw, plot as many sum-points as possible, using
positive+positive = bigger positive, negative+negative = bigger negative.
2. Interference & its Conditions
Interference pattern: a pattern of alternating maximum and minimum
displacement (alternate constructive & destructive regions).
Whenever 2 waves superpose they produce an interference pattern IF:
Same nature and frequency
They should meet at a point
They should have the same polarisation
They should have a constant phase relationship — they must be
coherent(same T, same λ)
Why same T and λ? A phase difference between two waves only makes sense
(is constant) if their period and wavelength are equal — otherwise the examiner would never ask you to find it.
Coherent: constant phase difference. θ can be anything (0°, 90°, 120°…) but it must
be constant.
Same T, same λ, fixed θ = 90° ⇒ coherent.Not coherent: phase difference keeps changing ⇒ no pattern.
Board QExplain
Why do car headlights not produce an interference pattern?
1) They are not coherent (no constant phase relationship)2) They
don’t have the same frequency (e.g. 1×10¹⁴ Hz vs a different value)3) They may not be
polarised similarly (e.g. one vertical, one horizontal; mostly unpolarised)4) The fringe
separation is far too small: λDa — ↓x
as a ↑ (headlights are far apart, so fringes are packed too close to see)
Waves overlap, but the sources are independent ⇒ no fixed phase relation.
3. Producing Light Interference · Path
Difference
How to get light interference? Two separate sources will never be coherent, so ★ split a single source into 2 (a double slit). Then both waves are coherent ✓, same
frequency ✓, same polarisation ✓ and they start with a constant phase difference = 0°.
one source split into two ⇒ A and B are coherent, in phase at the slits
Path difference: the difference in the distances travelled by 2
coherent waves interfering at a point (P.D. = d₁ − d₂).
Constructive interference: when 2 waves, 0° out of phase, superpose at a point,
the resultant displacements are always greater than the individual displacements.
If the waves start in phase (constant phase difference =
0°)Constructive: P.D. = nλ (n = 0, 1, 2, 3… whole numbers)
Phase difference:Phase diff = Δxλ × 360°= nλλ × 360°= 360°
n (whole cycles ⇒ always back in step)
Destructive interference: when 2 waves, 180° out of phase, superpose at a point,
the resultant displacements are never greater than the individual displacements.
If the waves start in phase (constant phase difference =
0°)Destructive: P.D. = (n + ½)λ e.g. 0.5λ, 1.5λ, 2.5λ, 6.5λ
Not examinable: the phase-difference derivation above need not be
remembered (it will not be asked). Exams also do not ask you to explain why the path difference gives
interference — they just give you numbers.
in phase at P ⇒ constructive180° out of phase ⇒ destructive
Proving a point is constructive / destructive: work out P.D.λ. If it comes out as a whole number n ⇒ constructive;
if it is n + ½ ⇒ destructive.
Note — flip rule: everything above assumes the waves start with 0°
phase difference. If the constant phase difference at the sources is 180°, the results swap:
Constructive: P.D. = (n + ½)λ | Destructive: P.D. = nλ. e.g. P.D. = 3.25λ − 2.75λ
= 0.5λ ⇒ constructive here.
Past-paper: two sound waves Y and Z at a fixed point P
Q4 · Fig 4.1structured · coherence &
phase
Two progressive sound waves Y and Z meet at a fixed point P. Graph of displacement x (µm) vs t (ms): Y has
amplitude 4 µm, Z has amplitude 2 µm, both with T = 3 ms; Z is shifted 1 ms from Y.
(a)(i) quantity the same: time period /
frequency(a)(ii) quantity different: amplitude(b) coherent? Yes — they have a
constant phase difference
a = slit separation (for light ≈ 0.5 – 1 mm works best — if asked to “suggest
a value”, write one value in this range)
Fringe separation (width)x = λDa
Where: x = fringe width | λ = wavelength | D = distance double slit →
screen | a = slit separation
Use for: finding an unknown wavelength (e.g. of green light) and for
proportionality questions: x ∝ λ, x ∝ D, x ∝ 1/a. Single slit in front of the double slit = to make sure
the light is incident on both slits (coherent source).
Learn the variables: know exactly what x, λ, D and a mean — questions
often give a setup and ask you to “write an expression relating λ, a, D and x”.
Q28 · MCQdouble slit · ratio
Red light, single slit, then double slit, then screen. Bright fringes are x = 4 mm apart. The double
slit is replaced by one with half the spacing, and the screen is moved so that its distance is twice
as great. New fringe spacing?
2. λ → 1.5λv = f λ (v const): λ↑ ⇒ f↓ ⇒ I ∝ f²
↓f ∝ 1/λ ⇒ I ∝ 1/λ²I₁λ₁² = I₂λ₂² =
I₂(2.25λ₁²)I₂ = I₁/2.25
Tip: brightness is qualitative — “less bright” is enough; the factor
(÷4, ÷2.25) only needed if intensity is asked.
Q6 (a)(b)structured · diffraction & two
slits
(a)(i) Part played by diffraction in producing the fringes: As the light
passes through the slit it spreads into its geometric shadow, so the 2 waves superpose at many spots and
create an interference pattern.
(ii) Why a double slit instead of two separate sources? So that the 2 waves have a
constant phase relationship (coherent).
(b) Slits S₁, S₂ separated by 0.48 mm = a, screen at D = 2.4 m; 16 fringes span 36 mm.
x = 36 mm16 =
2.25×10⁻³ mx = λDa ⇒ 2.25×10⁻³ = λ(2.4)0.48×10⁻³λ = 4.5×10⁻⁷ m = 450×10⁻⁹ m = 450 nm (blue light)
Board follow-up question (asked, not answered on board): to
which group of the EM spectrum does this belong? → 450 nm lies in the visible region (blue).
Q6 (c)water dippers · phase
Two dippers D₁, D₂ make identical waves on water; phase difference at the sources = 0, λ = 1.6 cm. P is 7.2 cm
from D₁ and 11.2 cm from D₂.
(i) What is observed at P?P.D. = 11.2 − 7.2 = 4.0
cmP.D.λ = 4.01.6 = 2.5= (n + ½) ⇒ destructive: water at P is flat (no vibration)
(ii)1. sources 180° out of phase: there will be vibration
at P — waves now interfere constructively(ii)2. intensity from D₁ is less than
from D₂: there will be vibration at P — the amplitudes differ, so complete
cancellation does not take place
Microwaves, same λ and amplitude, emitted in phase from X and Y. A detector moved parallel to the line
XY finds an interference pattern: central maximum at A, adjacent minimum at B. The microwaves have λ = 0.040
m.
(b) at B (first minimum)P.D. = 0.5 × 0.040 = 0.020 mphase diff = 0.5λλ × 360° = 180°
(c) amplitudes from X and Y both doubled — intensity at A: increases by a factor of 4(I ∝ a²; resultant amplitude 2a → 4a, so I₂ =
4I)(d)(i) separation of the sources increased: fringes (max &
adjacent min) move closer together (x = λD/a, ↑a ⇒ ↓x)(d)(ii) phase
difference becomes 180°: positions of maximums and minimums flip
Q5 (a)(b)(c)superposition · intensity · linear
law
(a)(i) Principle of superposition: When 2 waves of the same nature meet at a point,
the resultant amplitude is the algebraic sum of the individual amplitudes.
(ii) Coherence: the 2 waves have a constant phase difference. ⚠ Board wrote “0° phase difference” with the note “0° ⇒ constructive” — 0° is only the special
case; the definition is constant phase difference.
(b) Coherent P and Q in phase (0° ⇒ constructive, so amplitudes add: resultant = 1.5 +
aQ). aP = 1.5 cm, intensity I; resultant intensity 3I. Find aQ.
I ∝ a² ⇒ I₁a₁² = I₂a₂²I(1.5)² = 3I(1.5 + aQ)²(1.5 + aQ)² = 3 ×
2.25 = 6.751.5 + aQ = 2.60aQ = 1.1 cm
Board discrepancy: the board working wrote I1.5 = 3I1.5+aQ (forgot to square the amplitudes) and
got 1.5 + aQ = 4.5 ⇒ aQ = 3.0 cm. Because I ∝ a², the squares must stay, giving 1.1
cm. Don’t copy the 3.0.
(c) Laser, λ = 680 nm, double slit (separation a), screen at D. Distance D varied 2.0 → 3.5 m; Fig 5.2
shows x (mm) vs D (m).
Linear-law: x = λa
× D + 0 ⇒ y = m x + 0 ⇒ gradient m = λ/a.
(i) slit separation agradient m = λa(7 − 4)×10⁻³3.5 − 2.0 = 680×10⁻⁹aa = 3.4×10⁻⁴
m
(ii) A laser of shorter wavelength (λ↓) ⇒ gradient λ/a smaller ⇒ sketch a straight line, still
through the origin (when extended), less steep and below the original line.
x = (λ/a)D: smaller λ ⇒ smaller gradient (green line below blue)
Challenge Q · maglev traininterference ·
speed
A 15 MHz signal sent coherently from two antennas 4 km apart (a) to a railway track 20 km away. A train with an
antenna + CRO crosses the interference pattern at constant speed v and receives a frequency of 0.05 Hz. Find v.
Idea: one full cycle on the CRO (T = 20 s) = train travelling from one bright (max) fringe to the
next = one fringe width x.
5. Stationary (Standing) Waves
Stationary wave: can be longitudinal or transverse (a stationary sound wave is
still a sound wave — particles still move along the direction, e.g. up and down in a vertical tube). These waves
don’t transfer energy (progressive waves do).
★ Progressive vs stationary (energy):Progressive waves transfer
energy while stationary waves do not. (Asked in past papers — 1 mark.)
Conditions: the two waves must have the same frequency, same wavelength, same
velocity (and nature), travelling in opposite directions.
Describe the formation of a stationary wave• one wave
travels forwards• another wave travels backwards (mostly achieved
by reflection)• both waves superpose
Full-marks answer (“describe conditions / formation”): One wave must
travel in one direction while the other travels in the opposite direction. They must be of the same nature,
wavelength, velocity and frequency. As the waves pass through each other they superpose and form a stationary
wave. (Sir: you do not need to draw the t = 0, T/4… diagram in the exam — just know the
points below.)
blue wave →, red wave ←, green = resultant. t = 0, T: in phase (0°) ⇒ biggest resultant · t = ¼T, ¾T: 180°
out of phase ⇒ resultant is flat · red dots = nodes never move.
The stationary wave has the same λ as the individual waves.
It also has the same T (frequency) as the individual waves.
At t = 0 and t = T/2, T: waves are 0° out of phase → big antinodes; at ¼T, ¾T: 180° out of phase
→ complete cancellation, everything momentarily flat.
Nodes & antinodes — things to remember
Node (N): point of minimum (or zero) displacement | Antinode
(A): point of maximum displacement
Wording: “minimum” or “zero” displacement are both correct, but mark
schemes usually say minimum — write “a point of minimum displacement”.
Distance between a node and the adjacent antinode = ¼ λ
Phase difference within one loop = 0° (P, Q and X are in phase)
Phase difference between adjacent loops = 180° (X and R are 180° out of phase)
At the point of reflection you always have a node; at the source (speaker, tuning
fork) you always have an antinode.
Stationary waves form only at specific frequencies — harmonics (not at every frequency).
Q6 (a)(b)stationary waves on a string
(a) Conditions for 2 waves to form a stationary wave — see full-marks answer above. (b) Distance between a node
and an adjacent antinode = 6.0 cm.
(i) node: a point of minimum
displacement(iii) phase difference between adjacent antinodes: 180°
(ii) wavelengthN → A = ¼λ6
cm = ¼λλ = 24 cm
Q4 · Fig 4.1 & 4.2progressive →
stationary
A progressive wave on a string, P to R = 0.48 m; period 0.020 s. Fig 4.1 shows the wave at t = 0 (and
later at t = 0.005 s).
(i) λP→R = 6 × λ4 = 0.48λ = 0.32 m
(ii) speedv = fλv =
10.02 × 0.32v = 16 m s⁻¹
(iii) phase difference between Q and Rphase diff =
Δxλ × 360° = 1.25 ×
0.320.32 × 360°= 450° (≡ 90° —
either form accepted on the board)
(iv) Point Q from t = 0 to 0.010 s (half a period): Q goes to a minimum displacement
at t = 0.005 s, then returns to its original displacement at 0.01 s.(0.02 ÷ 4 = 0.005 s)
(c) Stationary wave on a different string fixed at X and Y (Fig 4.2, max displacement): nodes N at X, Y
and one between; antinodes A at W and the other loop. (i) node: point of minimum
displacement · (ii) number of antinodes: 2 · (iii) phase diff W
and Z: 180° (adjacent loops)
(iv) New wave with half the frequency, speed unchanged: λ = v/f ⇒ λ doubles ⇒ only one
loop (one antinode, N at both ends) between X and Y — the fundamental.
f halved ⇒ λ doubled ⇒ one antinode only
6. Stationary Waves between Two Nodes
(At the point of reflection you always have a node — e.g. a stretched rope fixed at both ends.) A
stationary wave forms only at certain frequencies called harmonics; the 1st harmonic is the smallest
f = fundamental frequency. Question first, then sketch — you do not need to memorise the formulas.
Note: same formulae as node–node, with antinodes and nodes
swapped in the picture. Don’t forget to mark the nodes (N) in the middle — students often miss this.
Q5 · vertical tube, open both endsstationary sound
wave
Vertical tube 0.60 m long, open at both ends: antinodes A at both ends, node N at the midpoint (0.30 m). Speed
of sound 340 m s⁻¹.
(a) Explain formation: The wave travels upwards, reflects and enters the tube again;
the 2 waves have the same frequency, wavelength and velocity, so they superpose and form a stationary
wave.
(b) Sketch amplitude vs height h: amplitude is max at h = 0 and 0.60 (antinodes) and zero at h
= 0.30 (node) — a “V-shaped valley” curve touching 0 at 0.30 m.
(c)(i) Direction of oscillation of air at 0.15 m: up and down (it is a sound
wave in a vertical tube) · (ii) phase difference between particles at 0.10 m and 0.20 m: 0° (same loop).
amplitude is a maximum at both open ends (A), zero at the midpoint (N)
8. Stationary Waves between an Antinode and a Node
(Source = antinode, closed end/reflector = node — e.g. a tube closed at one end.)
1st: λ/4 = L ⇒ λ = 4L2nd: 3λ/4 = L ⇒ λ = 4L/33rd: 5λ/4 = L ⇒ λ = 4L/5
1stλ/4 = L ⇒ λ = 4Lf₁ =
v4L
2nd3λ/4 = L ⇒ λ = 4L3f₂ = 3v4L
3rd5λ/4 = L ⇒ λ = 4L5f₃ = 5v4L
λn = 4L2n − 1
fn = (2n − 1)v4L
Remember: 2n − 1 always gives an odd number — only odd harmonics
exist in a closed tube.
Speed of sound using a water column
Lower the tube / water level until a loud sound is heard (antinode–node case, resonance). Play a
constant known frequency (a tuning fork also works).
1st: λ4 = L₁ +
ee = λ4 − L₁
2nd: 3λ4 = L₂ +
e3λ4 = L₂ + λ4 − L₁λ2 = L₂ − L₁
v = fλ = f [ 2 (L₂ − L₁) ]
This is always an antinode-to-node case; the step between two loud positions is L₂ − L₁ = ½λ
(always) — write this for the 2 marks.
Q6 · tuning fork 512 Hzwater column
Long tube with a tap, filled with water; tuning fork (512 Hz) sounded above the top as water runs out. Loud
sound first heard at Fig 6.1 (air column 15.7 cm) and again at Fig 6.2; the water levels differ by 32.4 cm.
(a) On Fig 6.2: (i) sketch the wave — an antinode at the top and a node at the water surface, with 1½
loops (3λ/4) in the air column; (ii) mark N at the water surface and at the other node (as drawn on the
board).
(b) speed of sound32.4×10⁻² = ½λλ = 64.8×10⁻² mv = fλ = (512)(64.8×10⁻²)v = 331.776 ≈ 332 m
s⁻¹
(c) where is the antinode (air column = 15.7 cm)?λ =
64.8 cm ⇒ λ/4 = 16.2 cm16.2 − 15.7 = 0.5 cm⇒ The antinode is most likely 0.5 cm above the end of the tube (end correction
e)
Q6 · hollow tube closed at one endloudspeaker +
signal generator
Tube of length L, closed at one end, open at the other; loudspeaker near the open end, f = 440 Hz; 5
quarter-wavelengths fit in L (nodes N marked at the closed end and in the tube). v = 330 m s⁻¹.
(a)(i) energy: progressive waves transfer energy while
stationary waves don’t(a)(ii) formation: the wave travels
forward, gets reflected and travels in the opposite direction; incident and reflected waves superpose and
create a stationary wave since they have the same wavelength, frequency and
velocity(a)(iii) direction of oscillation of air particle at P: left ↔ right (along the tube — longitudinal)(b) label N at each node
(incl. the closed end — it is a point of reflection)(c) phase difference P and Q: 0° (same loop)
(d)(i) λv = fλ330 = 440
λλ = 0.75 m
(ii) length LL = 5 × λ4 = 54
× 0.75= 0.9375 ≈ 0.94 m
Q4 · microwaves & metal platestationary
microwaves
Transmitter → receiver (at X) → metal plate; λ = 0.040 m; stationary wave between the transmitter and the
plate. A node is at X (minimum).
(a) principle of superposition: same wording as section
1.(b)(i) function of the metal plate: it acts as a reflector
(reflected wave + incident wave ⇒ stationary wave)
(ii) frequencyv = fλ3×10⁸ = f (0.040)f
= 7.5×10⁹ = 7.5 GHz
(iii)1 next minimum from XN → N = ½λ= ½ × 0.040 = 0.020 m (2 cm)
(iii)2 Number of intensity maxima (= antinodes) detected moving X → 9.1 cm: the board marked N at
0, 2, 4… cm and A at 1, 3, 5, 7, 9 cm (every 2 cm from a node, ¼λ = 1 cm to the first antinode). Antinodes at 1,
3, 5, 7, 9 cm ⇒ 5 (9.1 cm just includes the 9 cm one).
9. Diffraction & the Diffraction Grating
Diffraction:spreading of a wave into its geometric shadow as it
passes through an edge or a slit.
Exam phrase (2 marks): “As the light passes through the slit,
it spreads into its geometric shadow.” Both parts are needed. Diffraction is most significant when slit
size ≈ wavelength.
Diffraction grating: an instrument containing many slits that
helps to find the wavelength of light through diffraction. d = slit spacing.
orders on both sides of the zero order; θ measured from the zero order
d sin θ = nλ
Only works if the light is incident normally (wording “incident normally” is what makes this
formula valid).
Where: λ = incident wavelength | n = order number | θ = angle between
the nᵗʰ order and the zero order | d = slit spacing
Maximum order and total number of orders
d sin θ = nλmax when θ = 90° (sin 90° = 1): d(1) =
nmaxλnmax = dλ(round DOWN)Total orders = 2 nmax + 1 (both sides + the zero order)
Why +1? nmax = 9 ⇒ 9 orders above, 9 below = 18, but the
zero order isn’t included in either ⇒ 2(9) + 1 = 19.
If given lines per mm: convert to lines per metre first (“0.001
m has 750 lines ⇒ lines per m = 750 ÷ 10⁻³”), then d = 1 ÷ (lines per metre). Convert every unit (mm, cm,
nm) to metres first.
Q (b) · 590 nm, 750 lines/mmgrating · orders
Light of wavelength 590 nm is incident normally on a grating with 750 lines per millimetre.
(ii) nmaxd sin 90° =
nmaxλnmax = dλ = 1/750 000590×10⁻⁹= 2.26 ⇒ round down ⇒ 2
(iii) Total orders = 2(2) + 1 = 5
Q (b) · second order at 16°grating · wavelength
& highest order
Beam incident normally on a grating with line spacing d = 3.4×10⁻⁶ m; the angle between the 2nd-order beam and
the zero order is 16°.
(i) λd sin θ = nλ(3.4×10⁻⁶) sin 16° = 2λλ = 4.68×10⁻⁷ ≈ 4.7×10⁻⁷ m
(ii) highest order(3.4×10⁻⁶) sin 90° = nmax(4.7×10⁻⁷)nmax = 3.4×10⁻⁶4.7×10⁻⁷ ≈
7.23highest order = 7 (round down)
Q4 (a)–(c)waves, intensity, grating
(a)(i) displacement: distance moved by a particle from its mean
position in a specific linear direction (+ve & −ve)(a)(ii) amplitude: maximum displacement moved by a particle from its mean position (+ve only)
(b) X: 1.2 cm, intensity I; Y: 3.6 cm; waves 180° out of phase ⇒
destructiveresultant amplitude = 3.6 − 1.2 = 2.4 cmI(1.2)² = I₂(3.6 − 1.2)²I₂ = I
(2.4)²(1.2)² = 4.0 I
(c)(i) Diffraction of light at the grating: As the light passes through the slit, it
spreads into its geometric shadow. (They love this question — both parts needed for marks.)
(ii) Light of 540 nm and 630 nm; 3rd-order maxima for each, no higher orders for either. Smallest
possible d?
d sin 90° = 3λ ⇒ d = 3λ (smaller d ⇔ smaller
λ)3(540×10⁻⁹) = 1.62×10⁻⁶ m ✗ (too small for 630
nm)3(630×10⁻⁹) = 1.89×10⁻⁶ ⇒ d ≈ 1.9×10⁻⁶ m✓
(iii) Blue light (λ ≈ 400 – 450 nm) on the same grating: Yes, a third-order maximum
is formed as the wavelength of blue light (≈ 450 nm) is shorter — nmax = d/λ is larger than
for the longer wavelengths, so third order is still produced.
Q4 (a)–(d)laser, grating & sin θ vs λ
graph
(a) wavelength: the distance between 2 successive points in
phase(c) diffraction at the grating: As the light passes
through the slit, it spreads into its geometric shadow.
(b) period of a 460 nm light wavev = (1/T)λ ⇒ T =
λvT = 460×10⁻⁹3×10⁸ = 1.53×10⁻¹⁵T =
1.5×10⁻¹⁵ s
(d) Fourth-order maximum measured for several wavelengths; Fig 4.1 plots sin θ against λ (nm).
(i) d in terms of gradient Gd sin θ =
nλsin θ = nd × λ (y = m
x)gradient G = 4dd = 4 / G
(ii) 2nd-order graphgradient = nd = 2d⇒ half the gradient of the
4th-order linesketch a straight line through the origin (when extended), lower
than the 4th-order line
higher order ⇒ bigger gradient; for a fixed λ a higher order is at a bigger angle (θ₁ < θ₂ < θ₃)
✓ Full Chapter Cheat-Sheet — Superposition
Principle: waves of the same nature meet at a point ⇒ resultant displacement = algebraic sum of
individual displacements.
Constructive: resultant always > individual waves; 0° out of phase; P.D. = nλ (sources in phase).
Destructive: resultant never > individual waves; 180° out of phase; P.D. = (n+½)λ. If sources are
180° out of phase ⇒ swap.
Interference conditions: same nature & frequency · meet at a point · same polarisation · constant
phase relationship (coherent).