Thin-Film Interference

Equivalent forms

\lambda = \frac{2 n t}{m + 1/2}

2 n t = \left(m + \tfrac12\right)\lambda \;(\text{constructive in reflection})

The same half-wave flip that brightens one color darkens its neighbor, painting whole spectra onto a film a few hundred nanometers thick.

Symbol	Name	SI unit	Dimension	Range
$n$	Film refractive index Refractive index of the thin film	dimensionless	1	1 – 2.5
$t$	Film thickness Physical thickness of the film	m	L	50 – 800
$lambda$	Constructive wavelengthoutput Vacuum wavelength most strongly reflected for order m = 0	m	L	380 – 750

Unit systems

Symbol	SI	CGS	Imperial
$n$	dimensionless	dimensionless	dimensionless
$t$	m	cm	in
$lambda$	m	cm	in

Where it holds

Valid for near-normal incidence on a non-absorbing film with index between the surrounding media (single half-wave flip). Oblique incidence replaces t with t*cos(theta_film); absorbing or metallic films require complex indices.

Discovery

Thomas Young · 1802

Young explained the colors of thin plates and soap films as interference between the two reflected beams, extending Newton's earlier observations of 'Newton's rings' into a full wave-interference account.

Try this

Why does a soap bubble or an oil slick on water shimmer with rainbow colors?

A soap film of index 1.33 is 300 nm thick. Which visible wavelength is most strongly reflected (constructive, m = 0)?

Research status: stable

Real-world applications

Anti-reflection coatings on camera lenses and eyeglasses cancel reflected light at chosen wavelengths.
Iridescent colors of butterfly wings, peacock feathers, and beetle shells arise from layered thin films.
Optical filters and dielectric mirrors stack many films for precise wavelength selection.
Soap-film and oil-slick colors reveal local thickness variations to the naked eye.

Common misconceptions

Thicker films always look more colorful — beyond a few microns the orders overlap and the color washes out to white.
The half-wave shift happens at both surfaces — it occurs only where light goes from lower to higher index.
Transmission and reflection brighten the same color — they are complementary; a color bright in reflection is dim in transmission.

Experimental verification

Newton's rings — a curved glass on a flat plate forms an air wedge whose thickness varies radially; the observed bright and dark rings match the half-integer condition, and the ring radii scale as the square root of the order.

Derivation

The beam reflected at the bottom surface travels an extra optical path 2*n*t inside the film.

Reflection at the top (low-to-high index) adds a half-wave (lambda/2) phase shift, while the bottom reflection (high-to-low) does not.

Constructive interference in reflection therefore requires the total to be a half-integer number of wavelengths:

2*n*t = (m + 1/2)*\lambda

.

Limiting cases

t \to 0⟶ Dark filmA vanishingly thin film gives near-zero path difference; with the half-wave flip the reflected beams cancel, so very thin soap films look black just before bursting.

2\,\mathrm{n} t = m

\lambda⟶ Destructive reflectionInteger multiples of the wavelength yield cancellation in reflection (and bright transmission) for that color.

What if…

What if the film index is higher than both surrounding media on one side only?

Only one of the two reflections flips by a half-wave, so the half-integer condition for constructive reflection holds as written. If both or neither flip, the roles of bright and dark swap.

What if you view the bubble at a steep angle?

The in-film path shrinks to 2*n*t*cos(theta_film), shifting the bright color toward the blue — which is why bubble colors change as you tilt your view.

1

Brightest color from a soap film

Given ·

n:: 1.33
t:: 300

Find · lambda

Steps

Write the constructive condition $\lambda = 2*n*t/(m + 1/2)$ .
$Try m = 0$ : $\lambda = 798/0.5 = 1596\,\mathrm{nm}$ (infrared, not visible).
$Try m = 1$ : $\lambda = 798/1.5 = 532\,\mathrm{nm}$ — green light is most strongly reflected.

Result · For

m = 0

:

\lambda = 2*n*t/(m + 1/2) = 2*1.33*300/0.5 = 798/0.5 = 1596\,\mathrm{nm}

. That lies in the infrared, so for visible color

use m = 1

:

\lambda = 2*1.33*300/1.5 = 532\,\mathrm{nm}

(green).

Young's Double Slit Interference→Refractive Index Definition→Snell's Law→