Find the minimum thickness of a soap film (\(\mu=1.35\)) which appears dark in reflected light of wavelength 6000 Å, when the light is incident normally on the film.

Answer: \(t_{min}=2222\) Å \(\approx2.22\times10^{-7}\) m.

For reflected light from a thin film, the condition for a dark band (minimum) is

\[2\mu t\cos r=n\lambda,\qquad n=0,1,2,\dots\]

(The extra \(\lambda/2\) from reflection at the denser surface is already included in this form.)

Normal incidence: \(r=0\), \(\cos r=1\). The case \(n=0\) gives \(t=0\) (an extremely thin film, which also looks dark). The minimum non-zero thickness is for \(n=1\):

\[t=\frac{\lambda}{2\mu}=\frac{6000\times10^{-10}}{2\times1.35}=2.222\times10^{-7}\ \text{m}=2222\ \text{Å}\]