91Ó°ÊÓ

The formula for the reflective index of oil:

${\mathrm{μ}}_0=1.20$

Reflection index of water,${\mathrm{μ}}_w=1.33$

Here, $\mathrm{λ}$is wavelength of light in air/vacuum.

The reflective index is the factor of speed and wavelength of the radiation with respect to vacuum values:

${\mathrm{λ}}_0=\frac{\mathrm{λ}}{{\mathrm{μ}}_0}$

Phase difference as follows:

$$\begin{gathered} \mathrm{Δ}=\frac{2\mathrm{π}}{{\mathrm{λ}}_0}\left(\mathrm{Δ}x\right) \\ =\frac{2\mathrm{π}}{{\mathrm{λ}}_0}·2t \end{gathered}$$

Here,$t$ -thickness of oil.

For constructive reflection as follows;

$$\begin{gathered} \mathrm{Δ}=\frac{2\mathrm{π}}{{\mathrm{λ}}_0}·2t \\ =2n\mathrm{π} \\ =2\mathrm{μ}t \\ =n\mathrm{λ} \end{gathered}$$

Here, $2t$-potential difference between incident and reflected,$n=0,1,2,3,...$

For destructive reflection as follows:

$$\begin{gathered} \mathrm{Δ}=2\mathrm{μ}t\left(\frac{2\mathrm{π}}{{\mathrm{λ}}_0}\right) \\ =\left(2n+1\right)\mathrm{π} \\ 2\mathrm{μ}t=\left(n+\frac{1}{2}\right)\mathrm{λ} \end{gathered}$$

Near rim of drop,$t<\frac{\mathrm{λ}}{4\mathrm{μ}}$ ie, constructive outer region –bright.

Hence, the film thickness is$t=\frac{3\mathrm{λ}}{2{\mathrm{μ}}_0}$

The third band from rim, $2\mathrm{μ}t=3\mathrm{λ}$film thickness

$$\begin{gathered} t=\frac{3\mathrm{λ}}{2{\mathrm{μ}}_0} \\ =\frac{3×475 \text{nm}}{2×1.20} \\ =594 \text{nm} \end{gathered}$$

As the oil thickness increases, colours gradually fade and disappear.

Hence, the oil thickness is $594\text{nm}$.

The fundamental cause of color fading and eventual disappearance when oil thickness rises is that the colored bands start to blend too much to be easily differentiated. Furthermore, there would be too much space between the two reflecting surfaces for the light that reflected from them to be coherent.