Thin films (soap bubbles, oil slicks, optical coatings) can look colorful because light reflected from the top and bottom film surfaces combines. Depending on the film thickness t, refractive index n, and wavelength λ, the two reflected rays can add (constructive interference) or cancel (destructive interference).
This calculator returns the film thickness that produces a chosen interference condition in reflected light for a specified wavelength at normal incidence, using an integer order m (with m = 0 giving the smallest non‑negative thickness for the selected condition).
At normal incidence, the extra optical path inside the film for the ray that reflects from the bottom interface is 2 n t. Whether that corresponds to constructive or destructive reflection depends on phase reversals upon reflection.
The equations below assume exactly one reflection undergoes a π phase shift (a “half‑wave” phase reversal). This occurs when one interface reflects from lower→higher refractive index and the other does not (typical example: air → film → substrate with indices arranged so that only one reflection is from lower to higher).
With one phase reversal, the reflected‑beam conditions are:
Solving for thickness t gives the formulas implemented by the calculator:
t = ((m + 1/2) λ) / (2 n)t = (m λ) / (2 n)The output thickness is the film thickness that makes the chosen wavelength interfere constructively or destructively in reflection under the stated assumptions. Real “color” under white light happens because many wavelengths are present at once; a single thickness can enhance some wavelengths while suppressing others, and the effect changes with viewing angle.
Suppose you want a film with refractive index n = 1.33 (water‑like) to produce constructive reflection at λ = 550 nm (green) for the smallest nonzero thickness (m = 0), assuming one phase reversal.
Use t = ((m + 1/2) λ) / (2 n):
t = ((0 + 1/2) · 550) / (2 · 1.33) = 275 / 2.66 ≈ 103.4 nm.
If instead you wanted the next constructive solution (m = 1), thickness increases by λ/(2n):
t = ((1.5) · 550)/(2 · 1.33) ≈ 310.5 nm.
For λ = 550 nm and n = 1.33 under the calculator’s one‑phase‑reversal assumption:
| Order m | Constructive thickness t (nm) | Destructive thickness t (nm) |
|---|---|---|
| 0 | 103.4 | 0.0 |
| 1 | 310.5 | 206.8 |
| 2 | 517.3 | 413.5 |
2 n t cos(θ_film), shifting results.