Chapter 24, Problem 47P

To determine

The minimum thickness of the film.

Answer to Problem 47P

  $471\text{nm}$ .

Explanation of Solution

Given:

Refractive index of film, $n_{film}=1.36$

Refractive index of glass plate, $n_{glassplate}=1.51$

Maximum reflected wavelength, ${\lambda }_1=640\text{nm}$

Minimum reflected wavelength, ${\lambda }_2=512\text{nm}$

Formula used:

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

Where,

  $t$ is thickness of the object.

  $\lambda$ is wavelength of the reflected light.

  $n$ is refractive index.

Calculation:

According to the question,

  

For constructive interference, the net phase change:

  $\varphi =\left(\frac{2t}{{\lambda }_1{}_{film}}\right)2\pi +\pi -\pi =m_{1}2\pi ,$

  $m_1=0,1,2,3,\mathrm{...}$

  $t=\frac{1}{2}{\lambda }_{1film}\left(m_1\right)=\frac{1}{2}\left(\frac{{\lambda }_1}{n_{1film}}\right)\left(m_1\right)\left(m_1=1,2,3,\mathrm{...}\right)$

For destructive interference, the net phase change:

  $\varphi =\left(\frac{2t}{{\lambda }_2{}_{film}}\right)2\pi +\pi -\pi =\left(2m_2+1\right)\pi ,m_2=0,1,2,\mathrm{...},$

  $t=\frac{1}{4}\left(\frac{{\lambda }_2}{n_{film}}\right)\left(2m_2+1\right)m_2=0,1,2,\mathrm{...}$

Combine both equations,

  $\frac{1}{2}\left(\frac{{\lambda }_2}{n_{film}}\right)\left(m_1\right)=\frac{1}{4}\left(\frac{{\lambda }_2}{n_{film}}\right)\left(2m_2+1\right)$

  $or\frac{\left(2m_2+1\right)}{2m_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{640nm}{512nm}=1.25=\frac{5}{4}$

Since ${\text{m}}_{\text{1}}{\text{=m}}_{\text{2}}\text{=2}$ and the thickness of the film:

  $\begin{array}{l}\text{t}=\frac{\text{1}}{\text{2}}\left(\frac{{\text{λ}}_{\text{1}}}{{\text{n}}_{\text{film}}}\right)\left({\text{m}}_{\text{1}}\right)\\ =\frac{1}{2}\left\{\frac{640\text{nm}}{1.36}\right\}\left(2\right)=471\text{nm}\end{array}$

Conclusion:

The minimum thickness of the film is $471\text{nm}$ .