A glass surface is covered by a uniform oil film with a thickness...

For complete transmission of light through the oil film, the reflected light must undergo destructive interference.

The light reflects from two interfaces: the air-oil interface and the oil-glass interface. Since the refractive index of air ($n = 1.00$) is less than that of oil ($n = 1.25$), the reflection at the air-oil interface introduces a phase shift of $\pi$. Similarly, since the refractive index of oil is less than that of glass ($n = 1.50$), the reflection at the oil-glass interface also introduces a phase shift of $\pi$.

Because both reflected waves undergo the same phase shift of $\pi$, the relative phase difference due to reflection is zero. Therefore, the condition for destructive interference in the reflected light is given by:

$2 \mu t = \left(m + \dfrac{1}{2}\right) \lambda$

where $\mu$ is the refractive index of the oil, $t$ is the thickness of the film, $\lambda$ is the wavelength of light, and $m$ is an integer ($m = 0, 1, 2, \dots$).

Given $\mu = 1.25$ and $t = 1.00 \times 10^{-4} \text{ cm} = 1000 \text{ nm}$, substituting the values gives:

$2(1.25)(1000) = \left(m + \dfrac{1}{2}\right) \lambda$

$2500 = \left(m + \dfrac{1}{2}\right) \lambda$

$\lambda = \dfrac{2500}{m + 0.5} \text{ nm}$

Substituting integer values for $m$ to find wavelengths in the visible region ($400 \text{ nm}$ to $750 \text{ nm}$):

For $m = 3$, $\lambda = \dfrac{2500}{3.5} \approx 714 \text{ nm}$

For $m = 4$, $\lambda = \dfrac{2500}{4.5} \approx 556 \text{ nm}$

For $m = 5$, $\lambda = \dfrac{2500}{5.5} \approx 455 \text{ nm}$

For $m = 6$, $\lambda = \dfrac{2500}{6.5} \approx 385 \text{ nm}$ (outside the visible region)

The wavelengths completely transmitted in the visible region are $714 \text{ nm}$, $556 \text{ nm}$, and $455 \text{ nm}$.

Answer: $714 \text{ nm}$; $556 \text{ nm}$; $455 \text{ nm}$