Q37P - University Physics with Modern Physics

Question

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, 

(a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? 

(b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

Step-by-Step Solution

Answer

a) 69.0 nm

b) No

Step 1: Given

$n_{lenses}=1.62, n_{film}=n_{coating}=1.432, {\lambda }_{air}=5.50nm=550\times {10}^{-9} m$

Step 2: Concept.

The two reflected rays will go toward the person's eye directly but we draw it with some angle to make the two reflected rays more distinguish. In the real, the two rays are above each other.

                   

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

The first reflected ray experiences a phase change since the index of refraction of the air is less than that of the coating. And the second reflected ray also experiences a phase change since the index of refraction of the coating is less than that of the lenses.

The red circle, in the figure above, indicates a phase change 

This means that we have two reflected rays with two-phase change, which means that the two reflected rays are in phase. 

So, the thickness of the coating material that makes the two in-phase-reflected rays interfere destructively is given by

      $$\begin{gathered} 2\mathrm{t}=\left(m+\frac{1}{2}\right){\lambda }_{film } \\ \mathrm{t}=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{\mathrm{fflm} }}{2} \end{gathered}$$                  (1)

We know, from Snell's law, that 

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

So, 

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

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

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

Substitute into (1)

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

The thinnest coating material for this case is for m=0. 

Hence,

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

Substitute the given

$t=\frac{550}{4\times 1.432}$                                                       

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

We need to find the wavelengths of the visible light that will interfere destructively when the thickness of the coating is 96.0 nm, in addition to the 550 nm. 

We will use equation (2) above and solve for ${\lambda }_{air}$

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

Substitute the known;

${\lambda }_{air }\frac{96.0}{\left(m+\frac{1}{2}\right)}\times 2\times 1.62$                                                   

Now we know that when m = 0, the destructive light is 550 nm, so we need to plug the other values of m which are 123, 

When m =1.0;

$$\begin{gathered} {\lambda }_{air }\frac{96.0}{\left(1+\frac{1}{2}\right)}\times 2\times 1.62 \\ {\lambda }_{air }=207\mathrm{nm} \end{gathered}$$

                                                     

which is not in the range of the visible light (Noting that the range of the visible light is $(400nm<{\lambda }_{visible}<700nm)$).

Since increasing m decreases the net value of ${\lambda }_{air}$. So all other value is shorter than 207 nm.

Now we need to find the wavelengths of the visible light that would interfere constructively. 

Since the two reflected rays are in phase, as we mentioned above and as you see in the figure above, so

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

And hence,

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

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

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

Substitute the known

${\lambda }_{air }=\frac{2\times 96.0\times 1.62}{m}$                                                 

For constructive interference, when m = 1.0

$$\begin{gathered} {\lambda }_{air }=\frac{2\times 96.0\times 1.62}{1} \\ {\lambda }_{air }=311\mathrm{nm} \end{gathered}$$                                                   

which is not in the range of the visible light (Noting that the range of the visible light is $(400nm<{\lambda }_{visible}<700nm)$. 

Since increasing m decreases the net value of ${\lambda }_{air}$, So all other value is shorter than 207 nm. 

From all the above, all of the other wavelengths of visible light will not be canceled or enhanced.