Hong Life CalculatorsUnderstanding the anti-reflection coating calculator — concept and principle
A single thin layer on glass can cut reflections sharply, as on coated lenses and eyeglasses. This calculator finds the quarter-wave thickness of such a layer for a chosen wavelength and coating index, and compares the reflectance of the bare and coated surface. It also shows the ideal coating index for your glass.
How it is calculated
t = m × λ ÷ (4 nc), with m = 1, 3, 5 …
Optical thickness = nc × t; wavelength inside the coating = λ ÷ nc
Bare reflectance R₀ = [(n₀ − ns) ÷ (n₀ + ns)]²
Coated reflectance R = [(n₀ns − nc²) ÷ (n₀ns + nc²)]²
Ideal index = √(n₀ ns)
- λ is the vacuum design wavelength in nm; nc, ns and n₀ are the refractive indices of the coating, substrate and outside medium.
The principle
A coated surface gives two reflections: one at the top of the layer and one at the bottom. When the outside index is lower than the coating and the coating lower than the substrate, both reflections get the same phase flip, so only their path difference matters. A layer a quarter wavelength thick (measured inside the coating) makes the second reflection travel half a wavelength further, so the two waves cancel. They cancel completely only if both reflections are equally strong, which happens when nc = √(n₀ns).
Worked example
Default inputs: λ = 550 nm, nc = 1.38, ns = 1.52, n₀ = 1.000293 (air), order 1.
- t = 550 ÷ (4 × 1.38) = 99.64 nm; optical thickness = 1.38 × 99.64 = 137.5 nm (λ/4).
- Wavelength inside the coating = 550 ÷ 1.38 = 398.55 nm.
- R₀ = [(1.000293 − 1.52) ÷ 2.520293]² = 0.2062² = 4.252 %.
- n₀ns = 1.5204 and nc² = 1.9044, so R = [(1.5204 − 1.9044) ÷ 3.4248]² = 1.257 %.
- Ideal index = √(1.000293 × 1.52) = 1.2331; reflection reduced by (1 − 1.257 ÷ 4.252) × 100 = 70.4 %.
| Refractive index of the coating | Physical coating thickness (nm) |
|---|---|
| 0.345 | 398.55 |
| 0.69 | 199.28 |
| 1.03 | 132.85 |
| 1.38 | 99.64 |
| 1.73 | 79.71 |
| 2.07 | 66.43 |
| 2.76 | 49.82 |
How to read the result
Uncoated glass reflects about 4 % per surface, so multi-element lenses lose light and contrast. A single magnesium-fluoride-type layer (n about 1.38) cuts this to roughly 1 % at the design wavelength. Few durable materials have an index as low as the ideal 1.23, which is why high-end coatings use several layers. Away from the design wavelength the cancellation is incomplete, so a coated lens reflects a faint purple or green tint.
Common mistakes
- Using the vacuum wavelength without dividing by the coating index.
- Choosing an even order (2, 4); that doubles the path and makes reflections add instead of cancel.
Frequently asked questions
Why is 550 nm a common design wavelength?
It is in the green, near the peak of human eye sensitivity, so it gives the best visual improvement.
Why are higher orders like 3 sometimes used?
A thicker layer is easier to manufacture or more durable, though it works over a narrower band of wavelengths.
Why do coated lenses look purple?
Green light near the design wavelength is cancelled best, so the remaining reflection is mostly red and blue.
Sources: C. S. Adams & I. G. Hughes, Optics f2f (Oxford University Press), "Chapter 3: Worksheet on thin films", Durham University, 2019 · OpenStax, University Physics Volume 3, Table 1.1 (refractive indices)