Chapter 19.1, Problem 6PP

To determine

The thickness of the nonreflective layer.

Answer to Problem 6PP

  $t=101\text{ nm}$

Explanation of Solution

Given:

The given ray diagram is shown below.

  

Figure 1

Refractive index of air is ${\text{n}}_{air}=1.00$

Refractive index of magnesium fluoride film is $n_{film}=1.38$

Refractive index of glass is $n_{glass}=1.52$

Wavelength of light is $\lambda \text{ = 555 nm}$

Formula used:

There will be a phase inversion in the light rays reflected from the interface between air and magnesium fluoride layer because ${\text{n}}_{air}<n_{film}$ . Also there will be a phase inversion in the light rays reflected from the interface between magnesium fluoride layer and the glass because ${\text{n}}_{film}<n_{glass}$ .

The thickness of the nonreflective layer to keep yellow green light with a wavelength of $\text{555 nm}$ from being reflected 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 $n=n_{film}=1.38$ $m$ 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 $\left(2\right)$ ,

  $\text{t = }\left(\frac{1}{4}\right)\frac{\left(555\text{ nm}\right)}{1.38}=101\text{ nm}$

Conclusion:

The thickness of the nonreflective layer of magnesium fluoride layer on the glass to keep yellow green light with a wavelength of $\text{555 nm}$ from being reflected is $101\text{ nm}$ .