Materials Science · Light interactions and optical materials

Wavelength, refractive index and dispersion

Useful background: Atoms, bonding, structure and defects; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Interpret refractive index and dispersion within their optical regime.
Relate phase speed, frequency and in-medium wavelength without changing the source frequency at a stationary linear boundary.
Use angles measured from the surface normal to interpret refraction and a two-wavelength record.
Distinguish a parallel plate’s lateral displacement from a prism’s angular separation.
Assess what wavelength-dependent index estimates and their stated bounds can establish.

A wave changes propagation in matter

Refractive index relates phase propagation to that in vacuum. In a transparent nonmagnetic medium, the familiar real index helps predict refraction. Absorbing media generally need a complex optical description; a single real index omits attenuation.

Consider a fictional transparent material with phase index n = 1.50 for light whose vacuum wavelength is 600 nm. The wave crests move more slowly and are closer together inside this model material. That does not mean the source begins oscillating at a different rate. We will separate the spacing of crests, their timing and the direction of propagation before using a ray diagram.

All numerical records below are original teaching inputs, not measurements of a named glass or evidence for a particular composition. Use c = 3.00 × 10⁸ m/s throughout the new calculations. The first records assume linear, stationary, homogeneous, isotropic, nonmagnetic media with negligible absorption at the stated wavelengths. Ray geometry follows the directly transmitted beam; surface reflection can still remove power, and internal reflected beams are not traced.

One cycle has a spacing and a time

Light is an electromagnetic disturbance. At one location, an electric-field component oscillates; frequency f counts cycles per second in hertz (Hz), and period T = 1/f is the time for one cycle. At a fixed instant, wavelength λ is the distance between successive equal-phase crests along propagation. Phase describes position within a cycle, rather than the position of a material particle.

A simple travelling component can be written E(z,t) = E_a cos[2π(z/λ − ft)]. Here E_a is its amplitude, z is distance along propagation, t is time, and a full cycle of the cosine is 2π radians. Holding a crest’s phase constant gives Δz/Δt = fλ. This is the phase speed v_p. The plotted wavy line represents field value versus position; it is not the curved trajectory of a photon or a moving atom.

Define vacuum wavelength λ₀ = c/f and phase index n = c/v_p. Combining v_p = fλ with these definitions gives λ = λ₀/n. Index n is dimensionless. Units, scales and orders of magnitude supplies the length conversions: 1 nm = 10⁻⁹ m and 1 fs = 10⁻¹⁵ s.

For the 600 nm record, f = c/λ₀ = 5.00 × 10¹⁴ Hz, T = 2.00 fs, v_p = 2.00 × 10⁸ m/s and λ = 400 nm. A quarter-period later, a crest advances 150 nm in vacuum and 100 nm in the material. This compares phase progression, not a measured pulse arrival or information speed.

Original fictional 600-nanometre vacuum-wavelength record with material phase index 1.50. Two stacked field-versus-distance panels use the same horizontal scale, 0.4 pixel per nanometre from 0 to 1200 nanometres. Each field is normalised by its own amplitude; these curves do not specify interface power or absolute field continuity. The upper vacuum wavelength is 600 nanometres; the lower material wavelength is 400 nanometres. In both panels solid blue is time zero and dashed orange is one quarter-period later. Circles identify a time-zero crest and squares the corresponding later crest. The frequency is 5.00 times ten to the fourteenth hertz, period 2.00 femtoseconds, in both media. The marked crest advances 150 nanometres in vacuum and 100 nanometres in the material during 0.500 femtosecond. It illustrates phase progression of two separately normalised travelling components, not a photon trajectory, a pulse arrival or measured experimental data.

Match wave timing and direction at a boundary

A stationary linear interface, with no frequency conversion, preserves f for each transmitted component. The field at the boundary must keep the same temporal oscillation on the two sides. Different phase speeds therefore change λ rather than f. A moving boundary or frequency-converting process is a different problem.

A wavefront joins points of equal phase at one instant. In the isotropic model, the ray points perpendicular to these wavefronts. Let θ₁ and θ₂ be the incident and transmitted angles from the surface normal, the line perpendicular to the boundary. The angles are not measured from the surface itself. In a right triangle, sine is opposite side divided by hypotenuse; inverse sine returns the corresponding angle.

The crests must also match along a planar boundary. At nonzero incidence, successive crests meet that surface a distance λ/sin θ apart. Equating this distance on the two sides and using λ = λ₀/n gives Snell’s law: n₁ sin θ₁ = n₂ sin θ₂. Thus θ₂ = arcsin[(n₁/n₂) sin θ₁] for a propagating transmitted ray with a real angle in this regime.

For a supplied exterior n₁ = 1.00 and material n₂ = 1.50, incidence θ₁ = 45.0° gives θ₂ = 28.1255°. The ray moves closer to the normal while its wavefront spacing decreases. At normal incidence both angles are zero: direction can stay unchanged even though wavelength and phase speed change. If the inverse-sine argument exceeds one, this simple propagating transmitted-ray construction is unavailable; it is not a reason to clip the argument or invent an angle.

The exterior value 1.00 is a teaching approximation for air, not a precision refractivity measurement. The planar, uniform ray picture requires dimensions large compared with wavelength and excludes a graded index, diffraction-scale structure and anisotropic ray walk-off.

Original calculated planar-interface diagram for 600-nanometre vacuum wavelength, incidence 45.0 degrees from the normal, exterior index 1.00 and transparent isotropic material index 1.50. The directly transmitted ray is at 28.1255 degrees from the same normal. Diagram coordinates use equal horizontal and vertical scales: the exterior crest spacing measured along propagation is 96 pixels, and the material spacing is 64 pixels. Solid blue upper lines and dashed orange lower lines are equal-phase wavefronts at one instant, clipped to their respective media. Corresponding wavefronts meet continuously at the surface. The green ray directions are perpendicular to their wavefronts; dashed grey is the surface normal. Frequency is unchanged and no reflected amplitude or transmitted power is drawn. Angle arcs refer to the normal rather than the horizontal surface. This is a calculated geometric phase picture, not an experimental image, a photon track or the general anisotropic-ray construction.

Index can vary with wavelength

Dispersion means the optical response changes with frequency or wavelength. A prism separates colours partly because different wavelengths refract differently. State temperature, polarisation and direction when relevant; anisotropic crystals may require more than one index.

The optical field drives charges in matter. Their frequency-dependent response changes the phase of the total field, so the phase index can vary with frequency. This is a collective field response, not a sequence of slower photon flight between stopping collisions. Bonding and interactions between atoms supplies the material background; Polarisation, dielectric response and breakdown discusses response to an applied field in its own stated regime.

Now supply n = 1.520 at vacuum wavelength 450 nm and n = 1.500 at 650 nm for one fictional material at fixed temperature. Each component retains its own source frequency. At the same 45.0° incidence from n₁ = 1.00, the shorter-wavelength component has θ₂ = 27.7233°, while the longer-wavelength component has θ₂ = 28.1255°. Their internal angular separation is 0.402221°.

The table uses λ₀ to label the material’s index data; its λ column is the different wavelength inside that material. This supplied trend, higher n at shorter λ₀, is common in transparent bands but is not a universal rule across resonances, absorbing bands or all materials. Two points do not determine a continuous dispersion curve or an extrapolated value.

A prism can convert wavelength-dependent refraction into different outgoing directions because its two faces have different normals. A plane parallel plate gives a different result, even when the same material is dispersive.

Fictional two-wavelength record at fixed state and 45.0° incidence from n₁ = 1.00; calculated values rounded. No measured uncertainty is implied.
Vacuum λ₀ (nm)	Phase n	f (10¹⁴ Hz)	Material λ (nm)	Material θ₂ (degrees)
450	1.520	6.66667	296.053	27.7233
650	1.500	4.61538	433.333	28.1255

A parallel plate restores direction, while shifting position

Use the same exterior n₁ = 1.00 on both sides of an ideal plane parallel plate. At the second surface, Snell’s law gives n sin θ₂ = n₁ sin θ_out. Combining it with the first surface gives sin θ_out = sin θ₁; for the forward rays here, θ_out = θ₁ = 45.0° at both wavelengths. The outgoing rays are parallel, even though their paths inside the plate differ.

Let thickness t = 20.0 mm be measured along the common normal, and let the beam enter at transverse coordinate x = 0. At the exit plane, right-triangle geometry gives x_exit = t tan θ₂; tangent is opposite side divided by adjacent side. The 450 nm component exits at 10.5106 mm, and the 650 nm component at 10.6904 mm. The separation along that exit plane is 0.179844 mm. These coordinates use the same entry point and normal direction.

After exit, both rays have the same slope tan 45.0° = 1 in this plane. At any common plane farther along the normal, their transverse separation therefore remains 0.179844 mm. This is not the growing separation of rays with different outgoing angles. The diagram’s enlarged exit-position panel resolves the small difference without enlarging the actual ray angles.

A monochromatic 650 nm beam, for example, would have reached x = t tan θ₁ = 20.0 mm at the exit plane if the plate region had instead contained the exterior medium. Its transverse coordinate changes by 9.30955 mm in the plate model. That difference is measured along the exit plane; it is not the perpendicular shortest distance between the two parallel rays. A window can change beam position without changing its final direction.

This is useful when interpreting a beam-position or alignment comparison. It predicts the primary geometric path under the supplied assumptions, not throughput, an interference pattern, image quality or a real window tolerance. Reflection, absorption and transmission treats the separate power account.

Original plane-parallel-window calculation. The upper panel traces one 650-nanometre vacuum-wavelength primary beam with material index 1.50 and exterior index 1.00 on both sides. The thickness along the normal is 20.0 millimetres; horizontal and vertical geometry both use 12 pixels per millimetre. Solid blue enters at 45.0 degrees, travels internally at 28.1255 degrees and emerges at 45.0 degrees, parallel to its incident direction. A dotted grey counterfactual continues at 45.0 degrees if the plate region had instead contained the exterior medium. At the exit plane, the actual transverse coordinate is 10.6904 millimetres versus 20.0 millimetres for that counterfactual. Their coordinate difference 9.30955 millimetres lies along the exit plane, not perpendicular to the rays. The lower panel enlarges only the exit-position scale: from 10.4 to 10.8 millimetres at 1200 pixels per millimetre. A blue circle marks 450 nanometres with supplied index 1.520 at 10.5106 millimetres; an orange square marks 650 nanometres with index 1.500 at 10.6904 millimetres. Their transverse separation is 0.179844 millimetres. Both actual components leave at 45.0 degrees; the position inset does not exaggerate or imply an outgoing angular split. All values are fictional calculated records, not measured windows, image-quality predictions or throughput.

Infer index without losing angle bounds

For known n₁ and incidence, an angle estimate gives n₂ = n₁ sin θ₁/sin θ₂. It is evidence about the phase response at the specified wavelength and state. One such number does not identify composition, total transmission or dispersion over unmeasured wavelengths.

Consider a separate fictional angular record with n₁ = 1.00 and θ₁ = 45.0° stipulated exact: transmitted angles are 27.70° at 450 nm and 28.10° at 650 nm. The central index estimates are 1.52118 and 1.50125. Now supply a ±0.30° bound on each transmitted angle, with every combination within the two bounds permitted. These are stated bounds, not standard uncertainties or confidence intervals.

Over these acute angles, a larger θ₂ gives a smaller inferred n₂. The first interval therefore gives 1.50618 ≤ n_450 ≤ 1.53652; the second gives 1.48669 ≤ n_650 ≤ 1.51614. They overlap. The central difference alone cannot establish that n_450 exceeds n_650 for every allowed observation. This record is distinct from the exact supplied-index calculations in the table.

Better angular resolution can make the comparison discriminating, but it does not automatically remove normal alignment, source bandwidth, sample gradients or shared calibration errors. Calibration, uncertainty and repeatability explains why the measurement model and correlated effects matter. An index comparison also needs matched temperature, polarisation, direction and specimen state.

Keep the optical regime attached to the claim

A single real n describes phase progression in the transparent scalar model. With appreciable absorption, a complex optical response is needed to describe attenuation as well. Index alone does not determine how much incident power is transmitted through a specimen. Neither an internal wavelength change nor a refracted direction measures absorptance.

In anisotropic materials, index depends on polarisation and propagation direction, and energy flow need not be perpendicular to the phase fronts in the simple way used here. The direction and structure questions continue in Optical anisotropy, scattering and disorder. The two-wavelength plate calculation should not be transferred unchanged to an arbitrary crystal.

Phase speed follows the motion of a fixed phase of one sinusoidal component. A pulse or newly transmitted information combines a range of frequencies and needs their response and boundary history. Dividing c by one index does not establish that speed. Nor is n ≥ 1 a universal statement for every optical regime; a phase speed greater than c would not by itself establish faster-than-light information.

Report the vacuum wavelength or frequency, the kind of index, material state and measurement geometry. Then ask whether the needed result is crest spacing, an internal angle, an outgoing position, transmitted power or a pulse response. They are different observables, even when the same specimen is involved.

Worked examples

Work through the case

A material has n = 1.5 at one wavelength. Estimate phase speed using c = 3 × 10⁸ m/s.

Use v = c/n for the stated phase index.
3 × 10⁸/1.5 = 2 × 10⁸.

2 × 10⁸ m/s phase speed; this does not by itself specify signal propagation.

Keep the clock while changing crest spacing

For λ₀ = 600 nm and n = 1.50, find f, T, the internal λ and the crest advance during T/4. Why is f not v_p/λ₀?

Convert λ₀ to 6.00 × 10⁻⁷ m. The source frequency is c/λ₀ = 5.00 × 10¹⁴ Hz, hence T = 2.00 fs.
The internal phase speed is c/n = 2.00 × 10⁸ m/s and λ = λ₀/n = 400 nm. Check fλ = 2.00 × 10⁸ m/s.
T/4 is 0.500 fs. Crest advance is v_pT/4 = 100 nm in the material, compared with 150 nm in vacuum.
Using v_p/λ₀ incorrectly pairs an internal speed with a vacuum spacing. It would give 3.33333 × 10¹⁴ Hz, which does not describe this boundary record.

f = 5.00 × 10¹⁴ Hz, T = 2.00 fs, λ = 400 nm and internal advance = 100 nm. Frequency is preserved; the wavelength changes.

Compare two wavelengths at one boundary

Use n_450 = 1.520 and n_650 = 1.500 with 45.0° incidence from n₁ = 1.00. Find both transmitted angles and decide which component is closer to the normal.

Compute the common incident factor n₁ sin 45.0° = 0.7071068.
Divide by each index: sin θ_450 = 0.7071068/1.520 and sin θ_650 = 0.7071068/1.500.
Inverse sine gives 27.7233° and 28.1255°. Their difference is 0.402221°.
The shorter-wavelength component has the higher supplied index and the smaller angle from the normal. This is internal angular separation, not yet a prediction after a second surface.

θ_450 = 27.7233° and θ_650 = 28.1255°; the 450 nm component is closer to the normal.

Separate exit angle from exit position

A 20.0 mm plane parallel plate has the supplied two-wavelength indices and the same n₁ = 1.00 on both sides. Both components enter at one point at 45.0°. Find the outgoing angles and their transverse separation at the exit plane and 100 mm farther along the normal.

First-interface Snell gives n₁ sin 45.0° = n sin θ₂. At the second interface, n sin θ₂ = n₁ sin θ_out, so θ_out = 45.0° for each component.
At the exit, x_450 = 20.0 tan 27.7233° ≈ 10.5106 mm and x_650 = 20.0 tan 28.1255° ≈ 10.6904 mm, using the unrounded model angles for final arithmetic.
Subtract to get 0.179844 mm along the exit plane.
At another 100 mm in the exterior, both add 100 tan 45.0° = 100 mm to x. Their positions become 110.5106 and 110.6904 mm; the separation remains 0.179844 mm.

Both outgoing angles are 45.0°. The transverse separation is 0.179844 mm at both stated planes; it does not grow as an angular spread.

Check whether a central trend is resolved

Use the separate measured-angle record 27.70° and 28.10°, each bounded by ±0.30°, with exact θ₁ = 45.0° and n₁ = 1.00. Does it establish n_450 > n_650 for every permitted observation?

Use n = sin 45.0°/sin θ₂ and the monotonic decrease of this expression across the acute angle range.
At 450 nm, endpoints 27.40° and 28.00° give the reversed index endpoints 1.53652 and 1.50618.
At 650 nm, endpoints 27.80° and 28.40° give 1.51614 and 1.48669.
For instance, n_450 = 1.510 and n_650 = 1.515 fall inside their respective allowed intervals and reverse the central ordering. The central values do not exclude that possibility.

No. The allowed index intervals overlap, and an allowed reversal exists. These stated bounds carry no automatic confidence probability.

Try the reasoning

Can one index value describe every colour?
Show the answer and reasoning
Not generally.
Dispersion changes the value.
Is phase speed automatically information speed?
Show the answer and reasoning
No.
Wave-packet and causal propagation need additional analysis.
The fictional material has n = 1.50 at λ₀ = 600 nm, but incidence is normal to its planar surface. Does an unchanged transmitted direction imply unchanged wavelength?
Show the answer and reasoning
No. Both angles are zero, while the internal wavelength is 400 nm.
Snell’s law sets direction; λ = λ₀/n sets spacing at the preserved frequency. Normal incidence can show no bending while phase propagation changes.
At λ₀ = 600 nm, change the exterior index to 1.33 and keep the material index 1.50. Incidence is 30.0° from the normal. Find the material angle and wavelengths on each side; does the source frequency change?
Show the answer and reasoning
θ₂ ≈ 26.3168°; exterior λ ≈ 451.128 nm and material λ = 400 nm. The source frequency remains 5.00 × 10¹⁴ Hz.
sin θ₂ = (1.33/1.50) sin 30.0°. Both wavelengths use the same vacuum label: 600/1.33 and 600/1.50. An index ratio governs bending; the frequency is common.
A separate transparent scalar model has n = 1.60 at λ₀ = 750 nm. Find source f, phase speed and material λ. A report uses v_p/750 nm for f: what is mismatched?
Show the answer and reasoning
f = 4.00 × 10¹⁴ Hz, v_p = 1.875 × 10⁸ m/s and λ = 468.75 nm. The report pairs an internal speed with a vacuum wavelength.
Use c/λ₀ for source frequency, c/n for phase speed and λ₀/n for internal spacing. The incorrect mixed ratio is 2.50 × 10¹⁴ Hz; it is not the common boundary frequency.
Keep the exact two-wavelength indices and 45.0° incidence, but double the plane parallel plate thickness to 40.0 mm. What changes in the primary outgoing angle and exit-plane separation?
Show the answer and reasoning
The outgoing angle remains 45.0° for both components. Their transverse separation doubles to 0.359689 mm.
The two Snell relations do not involve thickness. Exit position is t tan θ₂, so its difference is proportional to t. This is a position change, not a new outgoing angular split.
Keep the separate central angle readings 27.70° and 28.10°, but independently justify bounds of only ±0.05° with all other stipulated quantities exact and every bounded combination allowed. Is their ordering now resolved within this model?
Show the answer and reasoning
Yes within those supplied bounds: n_450 is about 1.51865–1.52371, while n_650 is about 1.49880–1.50371.
The smallest allowed n_450 exceeds the largest allowed n_650. That is a bounded model conclusion, not a statistical confidence level, a proof that alignment errors are absent or a unique composition identification.
Two specimens have the same real phase index at 600 nm. One sends much less power to a narrow transmitted-beam detector. Does the shared index fix their total transmittance or explain the difference?
Show the answer and reasoning
No. A phase-index match does not fix absorption, reflection, scattering or collected power.
The real scalar index is insufficient to describe attenuation or every sample boundary. Thickness and collection geometry also matter. Use the power and collection evidence from Reflection, absorption and transmission rather than assigning the missing power from the index alone.

What to carry forward

Specify wavelength and the kind of index.
Distinguish phase propagation from broader wave behaviour.
At a stationary linear nonconverting boundary, frequency stays common while wavelength follows the local phase index.
Measure refraction angles from the normal and keep vacuum versus material wavelength distinct.
Dispersion can change an internal angle or exit position without producing different outgoing directions through a parallel plate.
A bounded index estimate is evidence at stated conditions, not a complete spectrum or composition test.

References

Teaching resources and research supporting this lesson’s explanations and examples.