Question

In Fig 35-59, an oil drop ($\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}$) floats on the surface of water ($\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}$) and is viewed from overhead when illuminated by sunlight shinning vertically downward and reflected vertically upward. (a) Are the outer (thinnest) regions of the drop bright or dark? The oil film displays several spectra of colors. (b) Move from the rim inward to the third blue band and using a wavelength of 475 nm for blue light, determine the film thickness there. (c) If the oil thickness increases, why do the colors gradually fade and then disappear?

Short answer

(a) The film thickness is $t=\frac{3\lambda }{2{\mu }_0}$.

(b) The oil thickness is$594\text{nm}$ .

(c) 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.

Step-by-step solution

Given in the question.

The formula for the reflective index of oil:

${\mu }_0=1.20$

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

Here, $\lambda$is wavelength of light in air/vacuum.

Definition of refractive index.

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

${\lambda }_0=\frac{\lambda }{{\mu }_0}$

 Step 3: (a) Outer region of the drop.

Phase difference as follows:

$$\begin{gathered} \Delta =\frac{2\pi }{{\lambda }_0}\left(\Delta x\right) \\ =\frac{2\pi }{{\lambda }_0}·2t \end{gathered}$$

Here,$t$ -thickness of oil.

For constructive reflection as follows;

$$\begin{gathered} \Delta =\frac{2\pi }{{\lambda }_0}·2t \\ =2n\pi \\ =2\mu t \\ =n\lambda \end{gathered}$$

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

For destructive reflection as follows:

$$\begin{gathered} \Delta =2\mu t\left(\frac{2\pi }{{\lambda }_0}\right) \\ =\left(2n+1\right)\pi \\ 2\mu t=\left(n+\frac{1}{2}\right)\lambda \end{gathered}$$

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

Hence, the film thickness is$t=\frac{3\lambda }{2{\mu }_0}$

(b) Calculate the film thickness. 

The third band from rim, $2\mu t=3\lambda$film thickness

$$\begin{gathered} t=\frac{3\lambda }{2{\mu }_0} \\ =\frac{3\times 475 \text{nm}}{2\times 1.20} \\ =594 \text{nm} \end{gathered}$$

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

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

(c) The increase in the thickness. 

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.