Answer
The minimum film thickness is $~~\frac{1}{4}$
Work Step by Step
Since the index of refraction of the film has a greater index of refraction than air, there is a phase change of $\frac{\lambda}{2}$ due to reflection with the front surface of the film.
Since the index of refraction of the lens has a greater index of refraction than the film, there is a phase change of $\frac{\lambda}{2}$ due to reflection with the back surface of the film.
We can think of the net phase change as zero.
To eliminate interference, the path length difference should be $(m+0.5)~\lambda$, where $m$ is an integer.
To find the minimum thickness $L$, we can let $m = 0$:
$2L = (m+0.5)~\lambda$
$L = \frac{(m+0.5)~\lambda}{2}$
$L = \frac{\lambda}{4}$
As a multiple of $\lambda$, the minimum film thickness is $~~\frac{1}{4}$
