Chapter 19.1, Problem 5PP

To determine

The thickness of the thinnest oil film that would create a reflected red band.

Answer to Problem 5PP

  $t=109\text{ nm}$

Explanation of Solution

Given:

Refractive index of the oil is $n_{oil}=1.45$

Wavelength of red light in the oil is $\lambda =635\text{ nm}$

Formula used:

The thinnest film that can create a red band can be calculated by using the equation for destructive interference,

  $\text{2t = }\left(m+\frac{1}{2}\right)\frac{\lambda }{n}$ $\left(1\right)$

Where, $\text{t }$ is the thickness of the film $\lambda$ is the wavelength of the light in the film $n$ is the refractive index of the film, In this problem $n=n_{oil}=1.45$ $m$ is the order of the fringe,

Calculation:

For the thinnest film, $m=0$

Substituting for $m$ and $n$ in equation $\left(1\right)$ ,

  $\text{2t = }\left(\frac{1}{2}\right)\frac{\lambda }{n_{oil}}$

  $\Rightarrow \text{t = }\left(\frac{1}{4}\right)\frac{\lambda }{n_{oil}}$ $\left(2\right)$

Substituting the numerical values in equation $\left(2\right)$ ,

  $\text{t = }\left(\frac{1}{4}\right)\frac{\left(635\text{ nm}\right)}{1.45}=109\text{ nm}$

Conclusion:

The thickness of the thinnest oil film that would create a reflected red band is $109\text{ nm}$ .