Q48P - University Physics with Modern Physics

Question

Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating 

(a) for which there is maximum transmission of the light into the rod; 

(b) for which transmission into the rod is minimized?

Step-by-Step Solution

Answer

a) 155 nm

b) 77.3 nm

Step 1: Given

${\lambda }_{air}=510nm=510\times {10}^{-9}m,n_{plastic}=1.3,n_{coating}=1.65$

Step 2: Concept.

The light rays reflected from the upper surface of the coating material will experience phase change since the index of refraction of the coating is greater than the index of refraction of the air. We represented these rays by ray 1, as you see below. Noting that the red circle indicates a phase change. 

But the light rays reflected from the rod will experience no phase change since the index of refraction of the rod is less than the index of refraction of the coating. We represented these rays by ray 2 

From all the above, we have two reflected rays with one phase change.

                                           

Step 3: The thickness of the coating at transmission into the rod is maximized .

The question here is a little tricky, the author asks about the minimum thickness that makes the transmission of the light, into the rod, maximized. This means that the author needs the thickness that makes the reflected light minimized. 

So, we seek the thickness that makes the two reflected rays interact destructively. 

Hence, the thickness that gives a destructive interference for two reflected rays with one phase change is given by 

                                                                $$\begin{gathered} 2t=m\lambda coating \\ t=\frac{m{\lambda }_{coating}}{2} \end{gathered}$$                                                             …(1)

We know, from Snell's law, that 

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

So, in this case,

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

Solving for ${\lambda }_{coating}$ and remembering that $n_{air}$ =1.0

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

Substitute into (1); 

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

Substitute the given and note that the minimum non-zero thickness is for m = 1.0;

                                                                        $$\begin{gathered} t=\frac{1.0\times 510}{2\times 1.65} \\ t=155nm \end{gathered}$$

Step 4: The thickness of the coating at transmission into the rod is minimized.

The author here asks about the minimum thickness of the coating that makes the transmitted light, into the rod, minimized. This means that he needs the minimum thickness of the rod that makes the reflection maximized So, actually, we are seeking the thickness that makes the two reflected rays interfere constructively. 

So, the thickness of the coating that makes two reflected rays, with one phase change, interfere constructively is given by 

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

Plug ${\lambda }_{coating}$ from above and solve for t. 

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

Plug the given and note that the minimum thickness is for m = 0;

                                                                           $$\begin{gathered} t=\frac{\left(0+{\displaystyle \frac{1}{2}}\right)\times 510}{2\times 1.65} \\ t=77.3nm \end{gathered}$$