Thin-film interference

A film can turn the same pair of reflections from reinforcement to opposition when only its substrate changes. Thickness is one part of the story; the interfaces supply the other.

We follow that phase comparison from one uniform layer to thickness measurements using fringes.

What you will learn to do

Determine reflection phase from the actual order of refractive indices.
Distinguish a phase minimum from complete amplitude cancellation.
Use thickness spacing and order differences to remove unknown offsets.
Connect film geometry to straight fringes and curved rings.

Useful before this chapterInterference: phase, brightness and contrast →Refraction, internal reflection and dispersion →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Why does a reflected internal route contain a factor of two in its normal-incidence optical path?

Compare your reasoning

It crosses the film on the inward and outward journey. Each pass contributes n_f t, so propagation adds 2n_f t before any reflection phase is included.

Revisit the explanation →

01 / Thin films: reflections, colours and precision

A surface can return two versions of the same wave

A thin transparent layer has two interfaces. Some light returns from the first interface; some enters the layer, reflects from the second interface and returns through the first. The detector compares these returned waves. Their fields remember both the propagation through the layer and what happened at reflection.

Begin by naming the incident medium, the film and the substrate. Their indices are n₀, n_f and n_s. Then identify the two waves that reach the same observation direction. Asking about reflected light is different from asking about transmitted light; switching the observation without changing the phase analysis is a common source of reversed answers.

We first use a coherent two-beam model with a uniform, transparent film and ordinary nonmagnetic dielectric interfaces near normal incidence. Weak later reflections are neglected when discussing two-beam intensity. At high reflectivity, repeated internal reflections change the detailed intensity response and require a multiple-beam treatment. The phase ledger remains the right starting point.

Normal-incidence path bookkeeping. The routes are separated sideways so both reflections can be labelled; this is a route diagram, not an oblique-ray construction.

WORKED EXAMPLE

The substrate can change a bright condition without changing the film

A film of index 1.50 is illuminated from air. Compare the reflection phase changes when the substrate index is 1.00 and when it is 2.00.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: A surface can return two versions of the same wave
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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02 / Thin films: reflections, colours and precision

Count propagation and reflection separately

At normal incidence the returning internal wave travels an extra physical distance 2t in the film. Its extra optical path is 2n_f t. Use the vacuum wavelength with optical path, or use the wavelength inside the film with physical path. Mixing those two conventions counts the refractive index twice.

At an ordinary lossless dielectric boundary near normal incidence, reflection toward higher index reverses the electric field, adding π; reflection toward lower index does not. Apply this at the top and bottom interfaces in the directions in which the relevant waves actually arrive. Total internal reflection, metals and strongly oblique polarized light need a more general reflection-phase treatment.

A propagation lag may be written with either sign if the full phase convention is consistent. For the zero-or-π reflection offsets considered here, reversing the overall phase sign does not change cosine or the constructive/destructive conditions. What matters is to combine the propagation difference with the relative reflection offset before choosing the integer order.

\delta=\frac{4\pi n_f t}{\lambda_0}+\Delta\phi_r\pmod{2\pi}

At normal incidence, Δφ_r is the relative reflection phase, equal to zero or π under the stated interface assumptions. Constructive interference has δ = 2mπ; destructive interference has δ = (2m+1)π.

Explain this equation

Read it in words

The relative phase of the two principal reflected rays is their round-trip propagation phase inside the film plus their difference in reflection phase.

Symbols and units

$\delta$: Relative phase at observation.Radians.
$n_f$: Refractive index of the film at the wavelength used.Dimensionless.
$t$: Film thickness here; this symbol is not time.Metres (m).
$\lambda_0$: Vacuum wavelength of the incident light.Metres (m).
$\Delta\phi_r$: Reflection phase of the internal-return ray minus that of the front-surface ray.Radians.
$\pmod{2\pi}$: Phases differing by a whole number of full cycles are equivalent.An operation; a full cycle is 2π radians.

Signs and operations

The 4π comes from 2π per wavelength and a two-way trip through the film. For ordinary lossless dielectric interfaces near normal incidence, reflection toward higher index adds π while reflection toward lower index does not.

When it applies

This displayed propagation term assumes normal incidence and the two principal reflected beams. At oblique incidence it includes the internal-angle cosine. Opposite phase gives a minimum, but zero intensity also needs matching returned amplitudes.

Make it concrete

For film index 1.5, thickness 100 nm and vacuum wavelength 600 nm, propagation contributes π. With one reflection reversal the total is equivalent to 0 modulo 2π, giving constructive phase alignment in reflection.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \delta=\frac{4\pi n_f t}{\lambda_0}+\Delta\phi_r\pmod{2\pi}
Read it as: The relative phase of the two principal reflected rays is their round-trip propagation phase inside the film plus their difference in reflection phase.
Symbols:
\delta: Relative phase at observation.; Radians.
n_f: Refractive index of the film at the wavelength used.; Dimensionless.
t: Film thickness here; this symbol is not time.; Metres (m).
\lambda_0: Vacuum wavelength of the incident light.; Metres (m).
\Delta\phi_r: Reflection phase of the internal-return ray minus that of the front-surface ray.; Radians.
\pmod{2\pi}: Phases differing by a whole number of full cycles are equivalent.; An operation; a full cycle is 2π radians.
Signs and operations: The 4π comes from 2π per wavelength and a two-way trip through the film. For ordinary lossless dielectric interfaces near normal incidence, reflection toward higher index adds π while reflection toward lower index does not.
Conditions: This displayed propagation term assumes normal incidence and the two principal reflected beams. At oblique incidence it includes the internal-angle cosine. Opposite phase gives a minimum, but zero intensity also needs matching returned amplitudes.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

One quarter-wave optical thickness can reinforce reflection

A transparent film in air has n_f = 1.40 and thickness 100 nm. Normally incident light has vacuum wavelength 560 nm. Are the two reflected beams in phase or opposite in phase?

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STOP & THINK

Keep that film and wavelength, but place the film on an index-1.80 substrate. What changes?

It remains a reflected maximum.
The two principal reflected beams become opposite in phase.
The propagation phase becomes zero.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: Count propagation and reflection separately
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
\delta=\frac{4\pi n_f t}{\lambda_0}+\Delta\phi_r\pmod{2\pi}
Meaning: At normal incidence, Δφ_r is the relative reflection phase, equal to zero or π under the stated interface assumptions. Constructive interference has δ = 2mπ; destructive interference has δ = (2m+1)π.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

03 / Thin films: reflections, colours and precision

A measured colour does not determine one unique thickness

For a film with one relative reflection reversal, reflected maxima require a half-integer round-trip optical path. Reflected minima require an integer round-trip optical path. These rules are consequences of the phase sum, not separate laws to memorize.

A single maximum therefore permits a family of thicknesses. The phrase “smallest positive thickness” chooses the first member, while two adjacent maxima in a thickness scan identify a spacing. Without an order or additional measurement, there is usually not enough information to infer a unique thickness.

For a fixed film seen in white light, several wavelengths can satisfy different orders. Actual colour also depends on the source spectrum, refractive-index dispersion, viewing angle and detector response. A single monochromatic maximum predicts one condition; it does not justify assigning a unique everyday colour to every film thickness.

2n_ft=(m+\tfrac12)\lambda_0\quad(\text{one reversal, reflected maximum}),\qquad\Delta t=\frac{\lambda_0}{2n_f}

Successive maxima at fixed wavelength and index are separated by Δt. Two or zero relative reversals interchange the maximum/minimum conditions.

WORKED EXAMPLE

Use the spacing to remove the unknown order

A thickness gauge reports adjacent reflected maxima at 0.12 μm and 0.32 μm for normally incident 600 nm light. Its zero has not been checked. Infer the film index from the spacing, then test whether the absolute readings fit a zero-or-π reflection offset.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: A measured colour does not determine one unique thickness
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
2n_ft=(m+\tfrac12)\lambda_0\quad(\text{one reversal, reflected maximum}),\qquad\Delta t=\frac{\lambda_0}{2n_f}
Meaning: Successive maxima at fixed wavelength and index are separated by Δt. Two or zero relative reversals interchange the maximum/minimum conditions.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

04 / Thin films: reflections, colours and precision

A phase minimum is only part of an antireflection design

A coating between air and a higher-index substrate can produce a phase minimum in reflected light if both principal reflections reverse and the round-trip propagation supplies an extra π. The smallest positive thickness then has n_f t = λ₀/4. The same quarter-wave optical thickness gave a maximum for the air-backed film because the reflection ledger was different.

Opposite phase is not enough for zero reflected intensity: the returning amplitudes must also match. Their amplitudes depend on both interfaces and the transmissions along the internal route. An exact single-layer coating calculation includes repeated internal reflections. In the present two-beam treatment, call the result a reflected phase minimum unless amplitude cancellation has separately been established.

The selected thickness is wavelength-specific. It also depends on incidence angle. A design that suppresses one colour at normal incidence need not suppress every colour or every viewing angle. This explains why a coating can reduce glare while still showing a faint coloured reflection.

WORKED EXAMPLE

Choose the correct quarter-wave role

Light of vacuum wavelength 640 nm arrives normally from air onto an index-1.60 film on an index-2.00 substrate. Find the smallest positive thickness for opposite phase of the two principal reflected beams.

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STOP & THINK

A film in air has index 1.25. What is the smallest positive thickness giving a reflected maximum at normal incidence for 500 nm light?

Answer in nm.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: A phase minimum is only part of an antireflection design
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Thin films: reflections, colours and precision

An angled path needs comparison at a common outgoing wavefront

Inside a tilted film, a ray travels farther than 2t before returning. It is tempting to replace the normal path by 2t/cos r and stop. But the two emerging rays leave at different positions. To compare their phase at the same outgoing wavefront, subtract the extra outside propagation of the earlier reflection.

Using Snell’s law in that common-wavefront comparison reduces the propagation optical path difference to 2n_f t cos r. The cosine is therefore in the numerator, even though the actual internal route length contains its reciprocal. This is a geometrical comparison of two complete paths, not just a measurement of one route.

For a fixed interference order, the selected vacuum wavelength decreases as cos r decreases, if index dispersion can be neglected. Use the propagation result with the appropriate polarization-dependent reflection phases; the simple zero-or-π reflection rule has a more limited angular domain than the path geometry itself.

\Delta\mathrm{OPL}=2n_ft\cos r,\qquad n_0\sin i=n_f\sin r

r is the angle inside the film. Compare waves on the same outgoing wavefront; do not substitute the external incidence angle for r.

WORKED EXAMPLE

Track one maximum while changing the view

A fixed-order reflected maximum of a nondispersive film occurs at 600 nm at normal incidence. The internal angle is changed so that cos r = 0.80, without changing the relative reflection phase. Find the new maximum wavelength.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: An angled path needs comparison at a common outgoing wavefront
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
\Delta\mathrm{OPL}=2n_ft\cos r,\qquad n_0\sin i=n_f\sin r
Meaning: r is the angle inside the film. Compare waves on the same outgoing wavefront; do not substitute the external incidence angle for r.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Thin films: reflections, colours and precision

A slowly changing thickness turns phase into a ruler

Two nearly parallel plates can enclose a thin air wedge. Its thickness increases approximately as t = αx when the angle α is small and measured in radians. Reflections at the two boundaries of the air layer have one relative reversal, so contact is a reflected minimum in the ideal model.

At normal incidence, successive dark fringes occur each time the round-trip path grows by one wavelength. The resulting position spacing is constant for a straight wedge. A curved or irregular gap changes the spacing, so the pattern can diagnose shape as well as thickness.

A fringe-spacing measurement estimates the wedge angle without locating the exact contact. However, converting that angle into a wire diameter or edge separation also requires the relevant length. If the gap is filled with a liquid while its geometry stays fixed, optical path grows faster with x and the fringes move closer together.

t\simeq\alpha x,\qquad2n\alpha x_m=m\lambda_0,\qquad\Delta x=\frac{\lambda_0}{2n\alpha}

Near-normal reflected dark fringes for a small wedge with one relative reflection reversal. n is the index of the gap material; x is measured from ideal contact.

WORKED EXAMPLE

Infer an angle from fringe spacing

An air wedge gives reflected dark fringes 1.50 mm apart under 600 nm light near normal incidence. Find its small angle.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: A slowly changing thickness turns phase into a ruler
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
t\simeq\alpha x,\qquad2n\alpha x_m=m\lambda_0,\qquad\Delta x=\frac{\lambda_0}{2n\alpha}
Meaning: Near-normal reflected dark fringes for a small wedge with one relative reflection reversal. n is the index of the gap material; x is measured from ideal contact.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

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07 / Thin films: reflections, colours and precision

Curvature makes equal phase steps unequal in radius

A spherical surface placed near a plane encloses a curved thin gap. At radius ρ from a point of contact, sphere geometry gives ρ² = 2Rt−t². For t much smaller than the curvature radius R, this becomes t ≈ ρ²/(2R). The gap thickens quadratically with radius rather than linearly as in a wedge.

For near-normal reflected light in an air gap between glass surfaces, one relative reflection reversal gives dark rings at 2t = mλ₀. Combining that with the geometry makes squared radius proportional to order. The rings become closer together as radius increases because the same added thickness needs a smaller radial step.

An unknown uniform separation at the centre shifts all squared radii by a constant. Taking the difference of squared diameters at known order separation removes that offset. This is often a more robust experimental comparison than insisting that the apparent centre is perfect contact.

\rho_m^2=m\lambda_0R,\qquad D_{m+p}^2-D_m^2=4p\lambda_0R

These dark-ring expressions assume a thin air gap, near-normal incidence, spherical curvature and one relative reflection reversal. D is a ring diameter, not its radius.

WORKED EXAMPLE

Use squared diameters to avoid the contact offset

For an air-gap ring pattern at 500 nm, the difference of squared dark-ring diameters five orders apart is 12.0 mm². Find the curvature radius.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thin films: reflections, colours and precision
Focus: Curvature makes equal phase steps unequal in radius
Learning goals:
- Determine reflection phase from the actual order of refractive indices.
- Distinguish a phase minimum from complete amplitude cancellation.
- Use thickness spacing and order differences to remove unknown offsets.
- Connect film geometry to straight fringes and curved rings.
Equations in this section (LaTeX):
\rho_m^2=m\lambda_0R,\qquad D_{m+p}^2-D_m^2=4p\lambda_0R
Meaning: These dark-ring expressions assume a thin air gap, near-normal incidence, spherical curvature and one relative reflection reversal. D is a ring diameter, not its radius.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Name the incident medium, film and substrate before choosing a bright or dark rule.
Use vacuum wavelength with optical path, or in-medium wavelength with physical path.
Opposite phase does not guarantee equal reflected amplitudes.
At oblique incidence, compare paths at a common outgoing wavefront.
Wedge fringes are linear in thickness; squared ring radii are linear in thickness.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

The same fixed air wedge is filled with index-1.5 liquid, retaining one relative reflection reversal. Its original fringe spacing was 1.2 mm. What is the new spacing?

Answer in mm.

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SELF CHECK 2

A two-beam coating has exactly opposite reflected phases but unequal returned amplitudes. What follows?

Reflected intensity is exactly zero.
Reflected intensity is a nonzero minimum for those amplitudes.
There is no interference.

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SELF CHECK 3

Dark-ring squared diameters four orders apart differ by 8 mm² for λ₀ = 500 nm in an air gap. What is the curvature radius?

Answer in m.

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

The substrate changes which thickness reinforces reflectionIntermediate · Hints and guided reasoning Measure index without knowing the fringe orderIntermediate · Hints and guided reasoning Turn fringe spacing into a physical gapIntermediate · Hints and guided reasoning Remove an unknown central gap from a ring measurementAdvanced · Hints and guided reasoning Use the internal angle for the film phaseAdvanced · Hints and guided reasoning Turn a film phase ledger into an actual detector readingAdvanced · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

A coating makes its two principal returned beams opposite in phase. Is zero reflected intensity guaranteed?

Compare your reasoning

No. Their returned field amplitudes must also match, and a precise high-reflectivity treatment includes repeated reflections. A phase minimum alone does not establish zero reflectance.

Revisit the explanation →

A few days later: change the situation

The centre of a ring experiment is not perfect contact. Which measurement can still determine the spherical curvature?

Compare your reasoning

The difference of squared dark-ring diameters at a known order separation removes a constant central gap. With an air gap, ΔD² = 4pλ₀R under the stated thin-gap approximation.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

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