Question

A thin film of acetone $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{25}$coats a thick glass plate$\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{50}$White light is incident normal to the film. In the reflections, fully destructive interference occurs at $\mathbf{600}\mathbf{}\mathbf{nm}$and fully constructive interference at$\mathbf{700}\mathbf{}\mathbf{nm}$. Calculate the thickness of the acetone film.

Short answer

The thickness of the acetone coating is $840\mathrm{nm}$.

Step-by-step solution

Given Data

• The refractive index of first medium, that is, air is$n_1=1.00$.
• The refractive index of the thin film is$n_2=1.25$.
• The refractive index of the third medium $n_3=1.50$.
• The maximum intensity occurs at wavelength${\lambda }_{700}=700nm$.
• The minimum intensity occurs at wavelength ${\lambda }_{600}=600nm$.

Interference in thin films

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,1,2,...$(Maxima—bright film in the air)

where $\lambda$is the wavelength of the light in air, $L$is its thickness, and$n_2$is the film’s refractive index.

Here, in this case, light is travelling in an air medium and incident on the acetone film whose refractive index $n_2=1.25$is higher than air. And then the light gets reflected of the thick glass of refractive index $n_3=1.50$while travelling through acetone and again the refractive index is higher than acetone. As a result, the condition for constructive and destructive interference is

$2L=\frac{m{\lambda }_{700}}{n_2}\left(\mathrm{Constructive}\right)$

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{600}}{n_2}\left(\mathrm{Destructive}\right)$

The constructive interference occurs at${\lambda }_{700}=700nm$ and destructive at ${\lambda }_{600}=600nm$.

Determine the thickness of the thin layer

Equating the above two equations,

$$\begin{gathered} \frac{m{\lambda }_{700}}{n_2}=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{600}}{n_2} \\ m{\lambda }_{700}-m{\lambda }_{600}=\frac{{\lambda }_{600}}{2} \\ m\left(700nm-600nm\right)=\frac{600nm}{2} \\ m=3 \end{gathered}$$

Inserting the value of in the constructive condition

$\begin{array}{rcl}L& =& \frac{m{\lambda }_{700}}{2n_2}\\ & =& \frac{3\left(700\mathrm{nm}\right)}{2\left(1.25\right)}\\ & =& 840n\mathrm{m}\end{array}$

The thickness of the acetone coating is $840\mathrm{nm}$.