Count the shifts, then write the condition

Trace the second ray step by step: in through the top surface, down across the film, reflect off the bottom, and back up to rejoin the first. For light near normal incidence the extra path is

$$\Delta L = 2t.$$

Writing $t$ instead is off by a factor of two in the thickness or in the wavelength, and it is easy to miss because the result still carries sensible units and a believable magnitude.

Where the factor matters most is the antireflection coating, whose answer is a quarter wavelength thick. That quarter falls out of the two passes:

$$2t = \frac{\lambda_{\text{film}}}{2} \qquad \Longrightarrow \qquad t = \frac{\lambda_{\text{film}}}{4}.$$

Start from $t$ and you get an eighth, which is a real and wrong coating.

Entering the film, the frequency is unchanged and the speed drops to $c/n$, so the wavelength shrinks:

$$\lambda_{\text{film}} = \frac{\lambda_{\text{air}}}{n}.$$

What decides the interference is how many wavelengths fit inside the material, so that is the wavelength the condition has to use.

Yellow light of $550$ nm in an $n = 1.33$ film has $\lambda_{\text{film}} \approx 414$ nm. Using the air value gives a thickness too large by the factor $n$, roughly a third too large for a typical film.

The convert-first habit is worth making automatic: write $\lambda_{\text{film}}$ on the page before writing the interference condition, so the air value never reaches it.

A reflection off a medium with a higher index flips the wave by $180^\circ$, which counts as an extra half wavelength of path. A reflection off a lower index produces no shift.

So there are three cases, and only the middle one changes anything:

1. Both reflections shift (or neither does): the two half-wavelengths cancel, and the plain path condition applies.
2. Exactly one shifts: a net half-wavelength offset, which swaps the bright and dark conditions.

A soap film in air is the case that exposes this immediately. Air to soap goes low index to high, so that reflection shifts; soap back to air goes high to low, so it does not. One shift, so $2t = \lambda_{\text{film}}$ is destructive here, not constructive. That is why a soap film thin enough looks dark in reflection just before it pops.

Count the shifts first, then write the condition. Doing the path difference alone gets bright and dark exchanged whenever exactly one surface shifts.

Four steps, and the order is what keeps the sign errors out.

1. Convert the wavelength: $\lambda_{\text{film}} = \lambda_{\text{air}}/n$.
2. Write the path difference: $2t$, for near-normal incidence.
3. Count the shifts at the top and bottom surfaces, comparing the indices on each side of each one.
4. Write the condition, swapping bright and dark if exactly one surface shifted.

Two checks are worth running afterwards. An antireflection coating should come out about a quarter of a wavelength thick, and a soap film that looks dark in reflection should come out much thinner than a wavelength. If either one lands off by a factor of two or of $n$, the step that produced it is the one to look at.