Monochromatic light (vacuum = 661 nm) shines on a soap film (n = 1.33) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?
The reflective path has to be a multiple of whole wavelengths. Because the back side gives a half wave length shift, that means the thickness has to be 1/4 wavelength (or half wavelength both directions).
1/4*661nm=minimum thickness
For constructive interference, however, wouldnt you use 2t = m +1/2 wavelength. For this question m being 0 as atmospheric is equal to vacuum. Which gives 2t = 1/2 wavelength/n, simplifying to t = wavelength/4n
To find the minimum thickness of the film for which constructive interference causes it to look bright in reflected light, we can use the formula for constructive interference in a thin film:
2nt = (m + 0.5)λ
Where:
- n is the refractive index of the film (in this case, 1.33)
- t is the thickness of the film
- m is the order of the interference (in this case, we want the minimum thickness, so we use m = 0)
- λ is the wavelength of the incident light (in this case, 661 nm)
Rearranging the formula, we get:
t = ((m + 0.5)λ) / (2n)
Plugging in the values, we have:
t = ((0 + 0.5) * 661 nm) / (2 * 1.33)
t = (0.5 * 661 nm) / 2.66
t ≈ 123.94 nm
Therefore, the minimum thickness of the film for which constructive interference causes it to look bright in reflected light is approximately 123.94 nm.