Chapter 19, Problem 38A

To determine

The three possible thickness of the plastic to produce a blue strip for the given wavelength.

Answer to Problem 38A

The three possible thickness of the plastic are $61.5\text{nm}$ , $184.4\text{nm}$ , and $307.3\text{nm}$ .

Explanation of Solution

Given :

The wavelength of the light is $\lambda =445\text{nm}$ .

The index of refraction of the plastic is $n=1.81$ .

Formula used:

Consider a light ray falls on the plastic, a part of the light passes through the plastic and the remaining part of the light gets refracted by the plastic.

The expression for the thickness $\left(d\right)$ of the plastic is,

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

Here, $\lambda$ is the wavelength of the light, $n$ is the index of refraction of the plastic, and $m$ is the whole number.

Calculation:

Consider light ray falls on the plastic, a part of the light passes through the plastic and other part of the light gets refracted by the plastic.

Find the three possible thickness $\left(d\right)$ of the plastic as follows.

Consider the value of the whole number, $m=0$ .

  $\begin{array}{l}d=\left(m+\frac{1}{2}\right)\frac{\lambda }{2n}\\ d=\left(0+\frac{1}{2}\right)\times \left(\frac{445\text{nm}}{2\times 1.81}\right)\\ d=61.5\text{nm}\end{array}$

Consider the value of the whole number, $m=1$ .

  $\begin{array}{l}d=\left(m+\frac{1}{2}\right)\frac{\lambda }{2n}\\ d=\left(1+\frac{1}{2}\right)\times \left(\frac{445\text{nm}}{2\times 1.81}\right)\\ d=184.4\text{nm}\end{array}$

Consider the value of the whole number, $m=2$ .

  $\begin{array}{l}d=\left(m+\frac{1}{2}\right)\frac{\lambda }{2n}\\ d=\left(2+\frac{1}{2}\right)\times \left(\frac{445\text{nm}}{2\times 1.81}\right)\\ d=307.3\text{nm}\end{array}$

Conclusion:

Thus, the three possible thickness of the plastic are $61.5\text{nm}$ , $184.4\text{nm}$ , and $307.3\text{nm}$ .