Explain why an excessively thin film appears black in reflected light?

For a thin film of index \(\mu\) and thickness \(t\), the effective path difference between the two reflected rays is

\[ \Delta=2\mu t\cos r-\frac{\lambda}{2} \]

The extra \(\lambda/2\) comes from the phase change of \(\pi\) on reflection at the upper (rarer to denser) surface.

For an excessively thin film, \(t\to0\), so \(2\mu t\cos r\approx0\) and \(\Delta\approx\lambda/2\).

A path difference of \(\lambda/2\) is the condition for destructive interference, and this holds for every wavelength at once, because the condition no longer depends on \(t\) or \(\lambda\). So all colours of white light cancel in reflection.

Hence an excessively thin film (\(t\ll\lambda\)) appears black (dark) in reflected light. In transmitted light it looks bright, since the two patterns are complementary. Example: a soap film just before it bursts looks black.