The colors of a soap bubble or an oil slick come from light reflecting off the top and the bottom of a very thin layer. The two reflections interfere, like the two slits did. Two details decide which colors win: the wavelength inside the film is shorter (Index of Refraction), and a reflection can flip the wave (Reflection at Boundaries).
Key idea
At nearly straight-on viewing, the ray from the bottom surface travels an extra $2t$ inside a film of thickness $t$, where the wavelength is $\lambda/n_{\text{film}}$. A reflection off a higher index adds a half-wave flip. One flip: bright when $2t = (m + \frac{1}{2})\frac{\lambda}{n}$. Zero or two flips: bright when $2t = m\frac{\lambda}{n}$. Dark is the other way around.
Why it works
Count the delay in wavelengths. The extra path $2t$ inside the film is $2t\big/\frac{\lambda}{n}$ wavelengths. Each flip adds half a wavelength. If the two rays end up a whole number of wavelengths apart, they reinforce; a half number, they cancel. A soap film in air flips only the top reflection (air to soap is low to high; soap to air is high to low), so the conditions swap.
Common mistake
Two slips. (1) Forgetting that the wavelength inside the film is $\lambda/n$. (2) Missing a half-wave flip, or adding one where there is none. Check each surface separately: does the light go from lower $n$ to higher $n$? Then that reflection flips.
Worked example: Soap film
A soap film ($n = 1.33$) in air reflects $600\ \text{nm}$ light strongly. What is the thinnest it can be?
- Top: air ($1.00$) to soap ($1.33$), flip. Bottom: soap to air, no flip. One flip.
- One flip, bright: $2t = (m + \frac{1}{2})\frac{\lambda}{n}$. Thinnest: $m = 0$, $t = \dfrac{\lambda}{4n}$.
- $t = \dfrac{600}{4(1.33)} = 113\ \text{nm}$.
Answer: $113\ \text{nm}$
Worked example: Anti-reflection coating
A lens ($n = 1.50$) gets a coating of magnesium fluoride ($n = 1.38$) to cancel the reflection of $550\ \text{nm}$ light. What is the thinnest coating that works?
- Top: air to coating, flip. Bottom: coating ($1.38$) to glass ($1.50$), flip again. Two flips cancel each other.
- Zero net flips, dark: $2t = (m + \frac{1}{2})\frac{\lambda}{n}$. Thinnest: $t = \dfrac{\lambda}{4n}$.
- $t = \dfrac{550}{4(1.38)} = 99.6\ \text{nm}$.
Answer: About $100\ \text{nm}$