Chapter 19, Problem 37A

To determine

The thickness of the thinnest film that will reflect yellow-green light and the thickness of the next-thinnest film that will also reflect yellow-green light.

Answer to Problem 37A

The thickness of the thinnest film that will reflect yellow-green light is $75.8\text{ nm}$ .

The thickness of the next-thinnest film that will reflect yellow-green light is $227\text{ nm}$ .

Explanation of Solution

Given:

The index of refraction of the plastic reflecting film is $n_{\text{film}}=1.83$ .

The index of refraction of the glass window is $n_{\text{plastic}}=1.52$ .

The wavelength of yellow-green light is $\lambda =555\text{ nm}$ .

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 $m=0$ for thinnest film and $m=1$ for next-thinnest film.

The thickness of the thinnest film that will reflect yellow-green light is,

  $\begin{array}{l}2d=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_{\text{film}}}\\ 2d=\left(0+\frac{1}{2}\right)\frac{555\text{nm}}{1.83}\\ 2d=151.6\\ d=75.8\text{ nm}\end{array}$

The thickness of the next-thinnest film that will reflect yellow-green light is,

  $\begin{array}{l}2d=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_{\text{film}}}\\ 2d=\left(1+\frac{1}{2}\right)\frac{555\text{nm}}{1.83}\\ 2d=454.9\\ d=227\text{ nm}\end{array}$

Conclusion:

Thus, the thickness of the thinnest film that will reflect yellow-green light is $75.8\text{ nm}$ and the thickness of the next-thinnest film that will reflect yellow-green light is $227\text{ nm}$ .