Monochromatic light of wavelength $\lambda$ falls normally on a glass film of refractive index $\mu$. The waves reflected from the upper and lower faces of the film will interfere constructively if the least thickness of the film is:

When light falls on a surface, it is partly reflected and partly refracted. Thus light from source $\mathrm{S}$ is reflected as wave 1 at A of the top surface of the plate, refracted into the plate, and reflected as wave 2 at B of the bottom surface. On reaching A, wave 2 has traversed an optical path length $-\mu t+\mu t=2\mu t$ (Fig).

Furthermore, on reflection from the denser medium, a wave suffers a phase change of $\pi$ which implies a path change of $\lambda /2$. Hence on reaching A waves 1 and 2 will have a path change of
$\Delta =2\mu t-\frac{\lambda }{2}$
For constructive interference, $\Delta =n\lambda$. Hence
$$\begin{gathered} 2\mu t-\frac{\lambda }{2}=n\lambda \\ \Rightarrow 2\mu t=\left(n+\frac{1}{2}\right)\lambda \\ \text{ or } t=\frac{\left(n+\frac{1}{2}\right)\lambda }{2\mu } \end{gathered}$$
The minimum value of t corresponds to n=0. Hence
$t_{\min }=\frac{\lambda }{4\mu }$.