Anti-reflection coatings on glasses and camera lenses are thin films about a quarter wavelength thick. Light reflected from the top and bottom of the film interferes destructively, cutting reflections from about 4% to under 1%.
How to use the Thin-Film Optical Coating Calculator
- Incident angle θ₁
- Optical film thickness d
- Wavelength λ
- Refractive index of the first medium n₁
- Refractive index of the film n₂
- Refractive index of the substrate n₃
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the incidence angle, film thickness, wavelength and the three refractive indices (incident medium, film, substrate). A quarter-wave film of index √(n₁n₃) gives zero reflection at normal incidence.
Inputs explained
| Field | What to enter |
|---|---|
| Incident angle θ₁ | Units: °, rad, arcmin, arcsec, mrad |
| Optical film thickness d | Units: nm, µm, mm, m, pm, Å |
| Wavelength λ | Units: nm, µm, mm, m, pm, Å |
| Refractive index of the first medium n₁ | A number |
| Refractive index of the film n₂ | A number |
| Refractive index of the substrate n₃ | A number |
What each result means
- Optical path difference = 2×[Refractive index of the film n₂]×[Optical film thickness d]×[c2]
- s-polarised reflectivity = ([rs12]^2 + [rs23]^2 + 2×[rs12]×[rs23]×cos(2×[delta]))÷(1 + [rs12]^2×[rs23]^2 + 2×[rs12]×[rs23]×cos(2×[delta]))×100
- p-polarised reflectivity = ([rp12]^2 + [rp23]^2 + 2×[rp12]×[rp23]×cos(2×[delta]))÷(1 + [rp12]^2×[rp23]^2 + 2×[rp12]×[rp23]×cos(2×[delta]))×100
- Bare substrate reflectivity (normal incidence) = (([Refractive index of the first medium n₁] – [Refractive index of the substrate n₃])÷([Refractive index of the first medium n₁] + [Refractive index of the substrate n₃]))^2×100
- Minimum anti-reflection coating thickness = [Wavelength λ]÷(4×[Refractive index of the film n₂])
- Ideal coating index √(n₁n₃) = sqrt([Refractive index of the first medium n₁]×[Refractive index of the substrate n₃])
Formula
r_ij from the Fresnel equations, δ = 2π·n₂·d·cos θ₂ / λ
R = (r₁₂² + r₂₃² + 2r₁₂r₂₃cos 2δ) / (1 + r₁₂²r₂₃² + 2r₁₂r₂₃cos 2δ), d_AR = λ/(4n₂)
Worked example
Inputs — Incident angle θ₁: 0 °; Optical film thickness d: 99.6 nm; Wavelength λ: 550 nm; Refractive index of the first medium n₁: 1; Refractive index of the film n₂: 1.38; Refractive index of the substrate n₃: 1.52.
Results — Optical path difference = 274.896 nm; s-polarised reflectivity = 1.2601 %; p-polarised reflectivity = 1.2601 %; Bare substrate reflectivity (normal incidence) = 4.258 %; Minimum anti-reflection coating thickness = 99.63768 nm; Ideal coating index √(n₁n₃) = 1.2329.
Key terms
- Quarter-wave layer
- A film with optical thickness λ/4.
Frequently asked questions
Why are AR coatings a quarter wavelength thick?
Reflections from the two surfaces then cancel by destructive interference.
What material is used for single-layer coatings?
Often magnesium fluoride (n ≈ 1.38).
Why do coated lenses look purple or green?
Reflection is lowest in the middle of the spectrum, so the ends reflect slightly.
What about multilayer coatings?
Stacks of layers reduce reflection over a broad range of wavelengths.