Find the minimum thickness of soap film which appears dark (Wavelength [first digit partly illegible in the source scan; read as 5]896 A°) in reflection when light is incident normally on the film. Given refractive index of the film is 1.33.

Assumption: The first digit of the wavelength is obscured in the scan. We take \(\lambda=5896\) Å (the sodium D-line).

Given: \(\lambda=5.896\times10^{-7}\) m, \(\mu=1.33\), normal incidence (\(\cos r=1\)).

Condition for dark in reflection: \(2\mu t\cos r=n\lambda\). The n = 0 case is a film of zero thickness, so the minimum non-zero thickness is at n = 1:

\[ t_{min}=\frac{\lambda}{2\mu}=\frac{5.896\times10^{-7}}{2\times1.33}=2.217\times10^{-7}\ \text{m} \]

\(t_{min}\approx2.22\times10^{-7}\) m ≈ 2217 Å ≈ 0.222 µm