Question

What is the thinnest film of a coating with $n=1.42$ on glass $(n=1.52)$ for which destructive interference of the red component $(650nm)$ of an incident white light beam in air can take place by reflection?

Short answer

The thinnest film of a coating is 114nm.

Step-by-step solution

Formulas used to solve the question

Destructive interference

$\mathbf{2}\mathit{t}\mathbf{=}\mathbf{(}\mathit{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathit{\lambda }$

Snell’s law:

${\mathit{n}}_{\mathbf{1}}{\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}{\mathit{\lambda }}_{\mathbf{2}}$

Determine the thickness in terms of λ

Given: ${\mathit{n}}_{\mathbf{a}\mathbf{i}\mathbf{r}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$

$$\begin{gathered} {\mathit{n}}_{\mathit{c}\mathit{o}\mathit{a}\mathit{t}\mathit{i}\mathit{n}\mathit{g}}\mathit{=}\mathbf{1}\mathbf{.}\mathbf{42} \\ {\mathit{n}}_{\mathit{g}\mathit{l}\mathit{a}\mathit{s}\mathit{s}}\mathit{=}\mathbf{1}\mathbf{.}\mathbf{52} \\ {\mathit{\lambda }}_{\mathit{a}\mathit{i}\mathit{r}}\mathit{=}\mathit{650}\mathit{n}\mathit{m}\mathit{=}\mathbf{650}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{m} \end{gathered}$$

Light beam reflects from a surface that has a greater index of refraction than the index of refraction of the first medium, the wave experiences a phase change. This means that the first ray experiences the phase change and the second ray also experiences a phase change.

From equation (1), at m = 0 for the thinnest coating

$\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathit{\lambda }$ (3)

Determine the thickness

From equation (2),

$n_{coating}{\lambda }_{coating}=n_{air}{\lambda }_{air}$

$\Rightarrow {\lambda }_{coating}=\frac{n_{air}{\lambda }_{air}}{n_{coating}}$

Plug the given,

$\Rightarrow {\lambda }_{coating}=\frac{{\lambda }_{air}}{n_{coating}}$

Plug into (3),

$t=\frac{{\lambda }_{air}}{4n_{coating}}=\frac{650\ast 10{}^{-9}}{4\ast 1.42}=114nm$

Thus, the thinnest film of a coating is $114nm$.