91Ó°ÊÓ

$n_{cornea}=1.38,n_{film}=n_{drops}=1.45,{\mathrm{λ}}_{air}=600nm$

The first reflected ray that reflects from the upper surface of the eyedrops experiences a phase change since the index of refraction of the air is less than the index of refraction of the eyedrops (film). The red circle, in the figure above, indicates a phase change.

But the second reflected ray that reflects from the second surface of the eyedrops (or the upper surface of your cornea) experiences NO phase change since the index of refraction of the eyedrops (film) is greater than the index of refraction of your cornea

We also know that the red light is enhanced in your eye. So the thickness of the coating, which gives an enhanced light while there are two reflected rays with one phase change, is given by

$2t=\left(m+\frac{1}{2}\right){\mathrm{λ}}_{film}$

So,

$t=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{film}}{2}$

Now we need to find the light wavelength inside the coating, which is given by Snell's law.

$n_1{\mathrm{λ}}_1=n_2{\mathrm{λ}}_2$

So, for this case,

$n_{air}{\mathrm{λ}}_{air}=n_{film}{\mathrm{λ}}_{film}$

solving for${\mathrm{λ}}_{film}$and noting that$n_{air}$= 1.0;

${\mathrm{λ}}_{film}=\frac{{\mathrm{λ}}_{air}}{n_{film}}$

Put into (1);

$t=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{air}}{2n_{film}}$

The minimum thickness of the airdrops is for m = 0.

Hence,

$$\begin{gathered} t=\left(0+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{air}}{2n_{film}} \\ t=\frac{{\mathrm{λ}}_{air}}{4n_{film}} \end{gathered}$$

Put the given;

$$\begin{gathered} t=\frac{600}{4×1.45} \\ t=103.4nm \end{gathered}$$

For any constructive interference of any wavelengths among the visible light range, we need to solve equation (2) for${\mathrm{λ}}_{air}$

${\mathrm{λ}}_{air}=\frac{t}{\left(m+{\displaystyle \frac{1}{2}}\right)}×2n_{film}$

Plug the known;

${\mathrm{λ}}_{air}=\frac{103.4}{\left(m+{\displaystyle \frac{1}{2}}\right)}×2×1.45$

${\mathrm{λ}}_{air}=\frac{300}{\left(m+{\displaystyle \frac{1}{2}}\right)}$ (3)

We know that when m = 0, the enhanced light is 600 nm, so we need to find other wavelengths by plugging the values of m =1,2,3,...

Hence,

$$\begin{gathered} {\mathrm{λ}}_{air}=\frac{300}{\left(1+{\displaystyle \frac{1}{2}}\right)} \\ {\mathrm{λ}}_{air}=200nm \end{gathered}$$

Which is not in the range of the visible light, (the range of the visible light is from 400 nm to 700 nm)

Since increasing m decreases the wavelength, as you see in (3), so there are no other wavelengths of visible light that could be enhanced except the one of 600 nm.

Now we need to find the wavelengths of the visible light in which destructive interference occurs.

We know that there are two reflected rays with one phase change, so we need to use the equation of the thickness of destructive interferences.

$2t=m{\mathrm{λ}}_{film}$

Hence,

${\mathrm{λ}}_{air}=\frac{2t×n_{film}}{m}$

Put the known;

${\mathrm{λ}}_{air}=\frac{2×1.45×103.4}{m}$

Put the known

${\mathrm{λ}}_{air}=\frac{300}{m}$

For m=1.0;

${\mathrm{λ}}_{air}=300nm$

Which is not in the range of the visible light, (the range of the visible light is from 400 nm to 700 nm).

Since increasing m decreases the wavelength, as you see in (3), so there are no other wavelengths of the visible light could be canceled

Therefore, and from all the above, there are no visible wavelengths for which there is destructive or constructive interference.

Given

$n_{lens}=1.50,n_{film}=n_{drops}=1.45,t=103.4nm$

Solution:

In this case, the two reflected rays will experience a phase change, so they will be reflected in phase, as you see below.

The red circle indicates a phase change.

So, the equation of constructive interference is

$2t=m{\mathrm{λ}}_{film}$

Hence,

$2t=\frac{m{\mathrm{λ}}_{air}}{m{\mathrm{λ}}_{film}}$

Solving for ${\mathrm{λ}}_{air}$;

${\mathrm{λ}}_{air}=\frac{2t×n_{film}}{m}$

Put the known;

$$\begin{gathered} {\mathrm{λ}}_{air}=\frac{2×1.45×103.4}{m} \\ {\mathrm{λ}}_{air}=\frac{300}{m} \end{gathered}$$

Put m=1,2,3,…..,

${\mathrm{λ}}_{air}=\frac{300}{1}$

${\mathrm{λ}}_{air}=300nm$

Which is not in the range of the visible light, (the range of the visible light is from 400 nm to 700 nm).

Since increasing m decreases the wavelength, as you see in (3), so there are no wavelengths of visible light could be enhanced.

Now we need to use the equation of destructive interference when both reflected rays are in phase.

$2t=\left(m+\frac{1}{2}\right){\mathrm{λ}}_{film}$

So,

$2t=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{air}}{{\mathrm{λ}}_{film}}$

Solving for${\mathrm{λ}}_{air}$

Put the known and solve for m=0

$$\begin{gathered} {\mathrm{λ}}_{air}=\frac{300}{\left(0+{\displaystyle \frac{1}{2}}\right)} \\ {\mathrm{λ}}_{air}=600nm \end{gathered}$$

For m=1.0

$$\begin{gathered} {\mathrm{λ}}_{air}=\frac{300}{\left(1+{\displaystyle \frac{1}{2}}\right)} \\ {\mathrm{λ}}_{air}=100nm \end{gathered}$$

Which is not in the range of the visible light, (the range of the visible light is from 400 nm to 700 nm)

Since increasing m decreases the wavelength, as you see in (3), so there are no wavelengths of visible light could be canceled except the 600 nm- wavelength (which is the red light).

Therefore, from all the above, when you use your contact lenses there will be no enhanced wavelengths of the visible light at all, but there is one canceled wavelength of the visible light which is the one that has a 600 nm wavelength.