Here's a systematic approach to analyzing a typical thin-film problem, which generally involve "normally incident" light (light directed perpendicular to the interface). A key part of this is to figure out the effective path length difference, how far the wave reflecting off the bottom surface of the film is shifted relative to the wave reflecting off the top surface.

Step 1. Write down D_t, the shift for the wave reflecting off the top surface of the film.

D_t = 0 if the wave reflects from a lower-n medium.

Dt

=

lfilm 2

if the wave reflects from a higher-n medium.

Step 2. Write down D_b, the shift for the wave reflecting off the film's bottom surface.

If the film thickness is t this wave goes down and back through the film, traveling an extra distance of 2t. This wave also has an additional shift of either 0 or l_film/2 depending on whether it reflects from a lower-n or higher-n medium.

D_b = 2t if the wave reflects from a lower-n medium.

Db

=

2t +

lfilm 2

if the wave reflects from a higher-n medium.

Step 3. Calculate the relative shift D by subtracting the individual shifts. Think of this as the effective path length difference for the two waves.

D = D_b - D_t

Step 4. Bring in the appropriate interference condition, depending on the situation. If we shine a red light on a thin film and the film reflects the light, constructive interference is going on. If the film looks dark, the light must be interfering destructively.

For constructive interference: D = m l_film

For destructive interference: D = (m + 1/2) l_film

Step 5. Solve. Rearrange the equation (if necessary) to get all factors of l_film on one side. Your equation should give you a relationship between t, the film thickness, and either the wavelength in vacuum or the wavelength in the film. Remember that the wavelength in the equation is the wavelength in the film, l_film. This is related to l, the wavelength in vacuum by:

lfilm

=

l n

where n is the refractive index of the film.