Answer
The thickness of the oil film is $~~673~nm$
Work Step by Step
There is a phase shift of $\frac{1}{2}~\lambda$ due to reflection at the front surface of the film.
There is a phase shift of $\frac{1}{2}~\lambda$ due to reflection at the back surface of the film.
We can think of the net phase shift due to reflection as zero.
If there is a minimum, the path length difference must be $~~(m+0.5)~\frac{\lambda}{1.30}~~$, where $m$ is an integer.
We can find possible values for the thickness $L$ of the film using $\lambda = 500~nm$:
$2L = \frac{(m+0.5)~\lambda}{1.30},$ where $m$ is an integer
$L = \frac{(m+0.5)~\lambda}{2~(1.30)}$
$L = \frac{(m+0.5)~(500~nm)}{(2)~(1.30)}$
$L = (m+0.5)~(192.3~nm)$
$L = 96.2~nm, 288~nm, 481~nm, 673~nm,...$
We can find possible values for the thickness $L$ of the film using $\lambda = 700~nm$:
$2L = \frac{(m+0.5)~\lambda}{1.30},$ where $m$ is an integer
$L = \frac{(m+0.5)~\lambda}{2~(1.30)}$
$L = \frac{(m+0.5)~(700~nm)}{(2)~(1.30)}$
$L = (m+0.5)~(269.2~nm)$
$L = 135~nm, 404~nm, 673~nm, 942~nm,...$
We can see that the thickness of the oil film must be $~~673~nm$
