91Ó°ÊÓ

Destructive interference

$\mathbf{2}\mathit{t}\mathbf{=}\left(m+\frac{1}{2}\right)\mathit{λ}$ (1)

Snell’s law:

${\mathit{n}}_{\mathbf{1}}{\mathit{λ}}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}{\mathit{λ}}_{\mathbf{2}}$ (2)

Light wave experience a phase change when it is reflected from a surface that has a greater index of refraction than the medium of the incident ray. Since the light beam reflected from the first surface, which is the coating material which has a greater index of refraction than air, so the first reflected ray experience a phase change. But the second ray will not since the index of refraction of the coating material is greater than that of the glass

Since there is one phase change of the two reflected rays, so we need the path difference between them to be an integral number of wavelength to have destructive interference.

So,

$\mathbf{2}\mathit{t}\mathbf{=}\mathit{m}{\mathit{λ}}_{\mathbf{c}\mathbf{o}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}$

For m = 1

$\mathit{t}\mathbf{=}\frac{{\mathbf{λ}}_{\mathbf{c}\mathbf{o}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}}{\mathbf{2}}$ (3)

From equation (2), for${\mathit{n}}_{\mathbf{a}\mathbf{i}\mathbf{r}}\mathbf{=}\mathbf{1}$

${\mathit{λ}}_{\mathbf{c}\mathbf{o}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}\mathbf{=}\frac{{\mathbf{λ}}_{\mathbf{air}}}{{\mathbf{λ}}_{\mathbf{coating}}}$

Plug into (3) and plug the given

$\mathit{t}\mathbf{=}\frac{{\mathbf{λ}}_{\mathbf{a}\mathbf{i}\mathbf{r}}}{\mathbf{2}{\mathbf{n}}_{\mathbf{c}\mathbf{o}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}}\mathbf{=}\frac{\mathbf{502}}{\mathbf{2}\mathbf{*}\mathbf{2}\mathbf{.}\mathbf{62}}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{4}\mathit{n}\mathit{m}$

To find the next three thickness, it means for and .

For m = 2,

${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{4}\mathbf{*}\mathbf{2}\mathbf{=}\mathbf{193}\mathit{n}\mathit{m}$

For m = 3,

${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{4}\mathbf{*}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{=}\mathbf{289}\mathit{n}\mathit{m}$

For m = 4,

${\mathit{t}}_{\mathbf{3}}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{4}\mathbf{*}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{=}\mathbf{386}\mathit{n}\mathit{m}$

Thus, the minimum film thickness required is 96.4nm . The three thinnest ones are 193nm,289nm and 386nm.