A parallel beam of light having a wavelength of 560 nm is incident...

For strong reflection, constructive interference must occur between the light reflected from the top and bottom surfaces of the oil film.

When light is incident from air onto the oil film, the reflection at the top surface (air to oil) undergoes a phase change of $\pi$. The reflection at the bottom surface (oil to air or water) undergoes no phase change.

The condition for constructive interference is given by:
$2nt = \left(m + \dfrac{1}{2}\right)\lambda$
where $n$ is the refractive index, $t$ is the thickness, $\lambda$ is the wavelength, and $m = 0, 1, 2, \dots$

For the minimum thickness, we take $m = 0$:
$2nt = \dfrac{\lambda}{2}$

$t = \dfrac{\lambda}{4n}$

Substituting the given values $\lambda = 560 \text{ nm}$ and $n = 1.4$:
$t = \dfrac{560}{4 \times 1.4}$

$t = \dfrac{560}{5.6} = 100 \text{ nm}$