Q38P - University Physics with Modern Physics

Question

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Step-by-Step Solution

Answer

a) 103.4 nm

b) Non, except enhancing the 600 nm wavelength.

c) Non, except enhancing the 600 nm wavelength.

Step 1: Given

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

Step 2: Ray diagram

Step 3: Solving part (a) of the problem.

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

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

So,

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

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

$n_1{\lambda }_1=n_2{\lambda }_2$                                                           

So, for this case, 

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

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

${\lambda }_{film}=\frac{{\lambda }_{air}}{n_{film}}$                                                         

Put into (1); 

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

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

Hence,

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

Put the given;

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

Step 4: Solving part (b) of the problem.

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

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

Plug the known; 

      ${\lambda }_{air}=\frac{103.4}{\left(m+{\displaystyle \frac{1}{2}}\right)}\times 2\times 1.45$   

${\lambda }_{air}=\frac{300}{\left(m+{\displaystyle \frac{1}{2}}\right)}$                                                         (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,

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

                                                     

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.

$2t=m{\lambda }_{film}$                                                           

Hence,

${\lambda }_{air}=\frac{2t\times n_{film}}{m}$                                                           

Put the known;

${\lambda }_{air}=\frac{2\times 1.45\times 103.4}{m}$                                                       

Put the known

${\lambda }_{air}=\frac{300}{m}$                                                       

For m=1.0;

${\lambda }_{air}=300nm$                                                           

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.

Step 5: Solving part (c) of the problem.

Given

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

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 

$2t=m{\lambda }_{film}$                                                         

Hence,

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

Solving for ${\lambda }_{air}$;

${\lambda }_{air}=\frac{2t\times n_{film}}{m}$                                                     

Put the known;

$$\begin{gathered} {\lambda }_{air}=\frac{2\times 1.45\times 103.4}{m} \\ {\lambda }_{air}=\frac{300}{m} \end{gathered}$$                                                      

       

Put m=1,2,3,…..,

${\lambda }_{air}=\frac{300}{1}$ 

${\lambda }_{air}=300 nm$                                                      

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.

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

So,

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

Solving for ${\lambda }_{air}$

                                                 

Put the known and solve for m=0

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

                                                         

For m=1.0

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

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.