Question

When viewing a paper of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index 1.62 and you use $\mathit{T}\mathit{i}{\mathit{O}}_{\mathbf{2}}$, which has an index of refraction of 2.62 , as the coating, what is the minimum film thickness that will cancel light of wavelength 505nm? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

Short answer

1. The minimum film thickness required is 96.4 nm.
2. The three thinnest ones are193nm, 289nm and 386nm .

Step-by-step solution

Formulas used to solve the question

Destructive interference

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

Snell’s law:

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

Determine the minimum film thickness

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,

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

For m = 1

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

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

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

Plug into (3) and plug the given

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

Determine the three thinnest ones

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.