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SCIENCE · INVESTIGATE AND EXPLAIN

Sum interface phases and the complete reflected series

At a lossless normal-incidence interface the electric-field reflection coefficient is r_ij=(n_i−n_j)/(n_i+n_j). A negative coefficient is a π phase reversal; a positive coefficient is not. This interface sign is separate from the propagation phase accumulated while a wave traverses a film.

Learning outcome: Track each interface coefficient and round-trip phase separately, then verify the full power sum and both zero-thickness and matched-index limits.

Starting recap

Retain complex amplitude before intensity, definite-integral boundaries, normalization, calibration covariance and unused evidence.

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Try

For d=100 nm and substrate index 1, r12=+0.2 and φ=π. The complete reflectance is about 0.147929 and transmittance 0.852071. At d=0, the same surrounding media give R=0, not the 0.04 reflectance of an isolated first interface.

Open the experiment

Predict

Can a reflection-phase reversal be replaced by adding an arbitrary geometric path length in every surrounding medium?

Experiment

Track each interface coefficient and round-trip phase separately, then verify the full power sum and both zero-thickness and matched-index limits.

Separate interface sign and propagation phase

Worked teaching model · synthetic values · Lossless normal-incidence stack n0=1, film n1=1.5, substrate n2=b, vacuum wavelength 600 nm. Real Fresnel field coefficients and all internally reflected beams are summed coherently. No absorption, dispersion, roughness or operating instructions.

Reset keeps your prediction, captures and practice history.

Reflection signs and propagation phase enter different factors; the complete lossless series closes R+T=1.

Lossless normal-incidence stack n0=1, film n1=1.5, substrate n2=b, vacuum wavelength 600 nm. Real Fresnel field coefficients and all internally reflected beams are summed coherently. No absorption, dispersion, roughness or operating instructions.

04000.1479290film thickness (nm)reflectance

The vertical axis is rescaled to this case’s actual range; equal display height does not imply equal optical power. The finite curve is illustrative; the ledger preserves the exact case values and stated numerical quadrature.

Original conditional optical ledger
QuantityValueUnit
First interface field reflection-0.200000amplitude ratio
Second interface field reflection0.200000amplitude ratio
Round-trip propagation phase3.14159radians
Total coherent reflectance0.147929incident power fraction
Total coherent transmittance0.852071incident power fraction
Outgoing energy sum1.00000ratio

Virtual original records only. No laser, intense-light source, instrument alignment, physical apparatus operation or completed practical competence is supplied.

Reflection signs and propagation phase enter different factors; the complete lossless series closes R+T=1.

Lossless normal-incidence stack n0=1, film n1=1.5, substrate n2=b, vacuum wavelength 600 nm. Real Fresnel field coefficients and all internally reflected beams are summed coherently. No absorption, dispersion, roughness or operating instructions.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Establish the optical mechanism

At a lossless normal-incidence interface the electric-field reflection coefficient is r_ij=(n_i−n_j)/(n_i+n_j). A negative coefficient is a π phase reversal; a positive coefficient is not. This interface sign is separate from the propagation phase accumulated while a wave traverses a film.

The original stack has incident index 1, film index 1.5, substrate index n2 and vacuum wavelength 600 nm. Therefore r01=−0.2 and r12=(1.5−n2)/(1.5+n2). A round trip through thickness d contributes φ=4π(1.5)d/600. Two interface signs can change relative phase without changing the optical path.

Transmitted light can reflect inside the film repeatedly. Summing that convergent geometric series gives total reflected field r=(r01+r12 exp(iφ))/(1+r01 r12 exp(iφ)). The transmitted field is t01 t12 exp(iφ/2)/(1+r01 r12 exp(iφ)), where t_ij=2n_i/(n_i+n_j). Power transmission must include the impedance factor n2/n0: T=(n2/n0)|t|², not merely |t|².

For real lossless indices the complete result obeys R+T=1. A two-reflection approximation may explain phase qualitatively but does not replace this exact series when a quantitative power balance is required. At zero thickness the stack reduces to a single incident/substrate interface; when n2 equals the film index, the lower interface disappears. Absorption, oblique polarization effects and roughness are outside this model.

2. Work the original evidence

For d=100 nm and substrate index 1, r12=+0.2 and φ=π. The complete reflectance is about 0.147929 and transmittance 0.852071. At d=0, the same surrounding media give R=0, not the 0.04 reflectance of an isolated first interface.

3. Predict and investigate

Can a reflection-phase reversal be replaced by adding an arbitrary geometric path length in every surrounding medium?

Track each interface coefficient and round-trip phase separately, then verify the full power sum and both zero-thickness and matched-index limits.

Transfer the derivation to a changed case

Keep d=100 nm but set substrate index 1.5. What changes?

Retain your prediction, intermediate reasoning and assumptions before opening public feedback.

Compare a worked response

The lower reflection coefficient becomes zero. Reflectance is 0.04 and power transmission 0.96 independent of film thickness; the former cavity interference no longer exists.

Primary mechanism context. Checked 31 October 2026. Original prose, records and diagrams; no copied source exercises or physical light-source operation. Retain first reasoning and public-feedback exposure in the course evidence workspace when registered.

Multiple-reflection derivation context: the lesson independently derives the original unequal-index stack and checks its lossless power balance.

Apply

Why must the transmitted field intensity include n2/n0 here?

Public practice checks a specific claim. It does not establish independent mastery.

Keep d=100 nm but set substrate index 1.5. What changes?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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