Chapter 24, Problem 85GP

To determine

The minimum thickness of the film.

Answer to Problem 85GP

  $240\text{nm}$

Explanation of Solution

Given info:

Refractive index of film, $n_{film}=1.34$

Refractive index of flat glass, $n_{flatplate}=1.52$

Reflected wavelength, $\lambda =643\text{nm}$

Formula used:

It is known that: $\text{t}=\frac{\text{1}}{\text{2}}\left(\frac{\text{λ}}{{\text{n}}_{\text{film}}}\right)\left(\text{m}\right)$

Where,

  $t$ is thickness of the object.

  $\lambda$ is wavelength of the reflected light.

  $n$ is refractive index.

Calculation:

  

The incident wave reflects form the top surface of the film has a phase change of ${\varphi }_1=\pi$ .

The incident wave reflects form the bottom surface of the flat glass plate has a phase change of ${\varphi }_2=\left(\frac{2t}{{\lambda }_{film}}\right)2\pi +\pi$ .

For constructive interference, the net phase change:

  $\begin{array}{l}\varphi =\left(\frac{2t}{{\lambda }_{film}}\right)2\pi +\pi -\pi =m\left(2\pi \right)\left(m=1,2,\mathrm{...},\right)\\ t=\frac{1}{2}\left(\frac{\lambda }{n_{film}}\right)\left(m\right)\end{array}$

Since $m=1$

Therefore, the thickness of the film:

  $\text{t}=\frac{\text{1}}{\text{2}}\left(\frac{\text{λ}}{{\text{n}}_{\text{film}}}\right)\left(\text{m}\right)=\frac{1}{2}\left\{\frac{643\text{nm}}{1.34}\right\}\left(1\right)=240\text{nm}$

Conclusion: The minimum thickness of the film is $240\text{nm}$ .