Chapter 19, Problem 36A

To determine

The minimum thickness of a place.

Answer to Problem 36A

The minimum thickness of a place where the oil creates constructive interference for light is $94.0\text{ nm}$ .

Explanation of Solution

Given:

The index of refraction of a thin film of oil is $n_{\text{film}}=1.45$ .

The wavelength of light is $\lambda =545\text{ nm}$ .

Formula used:

The expression for the extra distance that the second wave must travel through the thin film to create the needed interference is,

  $2d=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_{\text{film}}}$

Here, $m$ is an integer, $\lambda$ is the wavelength, and $n_{\text{film}}$ is the index of refraction of a thin film.

Calculation:

Consider the integer number as $m=0$ for minimum thickness. The minimum thickness of a place where the oil creates constructive interference for light is,

  $\begin{array}{l}2d=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_{\text{film}}}\\ 2d=\left(0+\frac{1}{2}\right)\frac{545\text{nm}}{1.45}\\ 2d=188.0\\ d=94.0\text{ nm}\end{array}$

Conclusion:

Thus, the minimum thickness of a place where the oil creates constructive interference for light is $94.0\text{ nm}$ .