Applying the five-step method

White light in air shines on an oil film of thickness t that floats on water. The oil has an index of refraction of 1.50, while the refractive index of water is 1.33. When looking straight down at the film, the reflected light is orange, with a wavelength of 600 nm. What is the minimum possible thickness of the film?

Step 1. What is Dt, the shift for the top wave?

  1. 0
  2. lfilm/2
  3. t
  4. t + lfilm/2
  5. 2t
  6. 2t + lfilm/2















Because oil has a higher index of refraction than air, the wave reflecting off the top surface of the film is shifted by half a wavelength.
Dt =
lfilm
2

Step 2. What is Db, the shift for the bottom wave?

  1. 0
  2. lfilm/2
  3. t
  4. t + lfilm/2
  5. 2t
  6. 2t + lfilm/2















Because water has a lower index of refraction than oil, the wave reflecting off the bottom surface of the film does not have a half-wavelength shift, but it does travel the extra distance of 2t.

Db = 2t

Step 3. The relative shift is thus:
D = Db - Dt = 2t -
lfilm
2

Step 4. Now, is this constructive interference or destructive interference?

  1. constructive
  2. destructive
  3. we haven't been given enough information to determine that













Because the film looks orange, there is constructive interference taking place for orange light.
For constructive interference: D = 2t -
lfilm
2
= m lfilm

Step 5. Moving all factors of lfilm to the right:

2t = (m + 1/2) lfilm

This looks like an equation for destructive interference! It isn't, because we used the condition for constructive interference in step 4. It looks like a destructive interference equation because only one reflected wave experienced a half-wavelength shift.

The 600 nm specified in the problem is the wavelength of orange light in vacuum. Bringing lfilm = l/n into the equation gives:
2t = (m + 1/2)
l
n

To solve for the minimum film thickness, use the minimum value of m. In this situation that's m = 0.
Therefore tmin =
l
4n
=
600
4 * 1.5
= 100 nm

This is not the only thickness that gives completely constructive interference for this wavelength. Others can be found by using m = 1, m = 2, etc. in the equation in step 6.

If a film thickness of 100 nm gives constructive interference for orange light, what about the other colors? They are not completely cancelled out, because 100 nm is not the right thickness to give completely destructive interference for most of the other wavelengths in the visible spectrum. The other colors do not reflect nearly as intensely as orange light, however, so the film looks orange.