Q23E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

The thinnest film of a coating is 114nm.

Step 1: 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:

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

Step 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)

Step 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),

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

Thus, the thinnest film of a coating is \(114nm\).