Radiation perpendicularly incident to a surface

In summary, the reflected wave can be eliminated by covering the plane with a second dielectric layer with index of √n and thickness 1/4 wavelength. This is due to the interference between the reflected waves, which require a phase shift of 1/2 wavelength to cancel out. This means that the film must be 1/4 wavelength thick.
  • #1
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Homework Statement


Consider radiation perpendicularly incident to a plane whose index of refraction is n. Show that the reflected wave can be eliminated by covering the plane with a second dielectric layer with index of [itex]\sqrt{n}[/itex] and thickness 1/4 wavelength.


Homework Equations


Boundary conditions: [itex]E_{1}^{||}=E_{2}^{||}[/itex], and [itex]B_{1}^{||}=B_{2}^{||}[/itex]


The Attempt at a Solution



Electric and Magnetic field at z=0 can be given by:

[itex]{E_{I}(z,t)=E_{I}exp[i(k_{1}z-\omega t)],E_{R}(z,t)=E_{R}exp[i(-k_{1}z-\omega t)]},{B_{I}(z,t)=\frac{1}{v_{1}}E_{I}exp[i(k_{1}z-\omega t)],B_{R}(z,t)=\frac{1}{v_{1}}E_{R}exp[i(-k_{1}z-\omega t)]}[/itex]

For 0<z<[itex]\frac{\lambda}{4}[/itex]:

[itex]{E_{i}(z,t)=E_{i}exp[i(k_{2}z-\omega t)],E_{r}(z,t)=E_{r}exp[i(-k_{2}z-\omega t)]},{B_{i}(z,t)=\frac{1}{v_{1}}E_{i}exp[i(k_{2}z-\omega t)],B_{r}(z,t)=\frac{1}{v_{1}}E_{r}exp[i(-k_{2}z-\omega t)]}[/itex]

For z>[itex]\frac{\lambda}{4}[/itex]:

[itex]{E_{T}(z,t)=E_{T}exp[i(k_{3}z-\omega t)]},{B_{T}(z,t)=\frac{1}{v_{1}}E_{I}exp[i(k_{3}z-\omega t)]}[/itex]

These boundary conditions allow us to find simultaneous equations for z=0 and z=[itex]\frac{\lambda}{4}[/itex]:

z=0 => [itex]{E_{I}+E_{R}=E_{r}+E_{i}, E_{I}-E_{R}=\beta (E_{r}-E_{i})}[/itex]

z=[itex]\frac{\lambda}{4}[/itex]=> [itex]{E_{r}exp[ik_{2}\frac{\lambda}{4}]+E_{i}exp[-ik_{2}\frac{\lambda}{4}]=E_{T}exp[ik_{3}\frac{\lambda}{4}], E_{r}exp[ik_{2}\frac{\lambda}{4}]-E_{i}exp[-ik_{2}\frac{\lambda}{4}]=\alpha E_{T}exp[ik_{3}\frac{\lambda}{4}]}[/itex]

Here, [itex]\alpha =\frac{v_{2}}{v_{3}}[/itex], [itex]\beta =\frac{v_{1}}{v_{2}}[/itex]

I have done plenty of algebra to find [itex]E_{I}[/itex] in terms of [itex]E_{R}[/itex], but this seems to require more information than I have. I am just not seeing the end from here, does anyone have a suggestion on where to go from this point?
 
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  • #2
I can offer only limited help ! I do not know where √n comes into the full mathematical analysis BUT, at my level, it is easy to show that the film needs to be λ/4 thick.
Here goes: Light from the low n to high n interface undergoes a phase shift of ∏ (λ/2)
Light striking the high n to low n interface does not undergo a phase shift.
In this example BOTH reflections undergo a λ/2 phase shift because both are from low n to high n.
To eliminate the reflected wave needs interference between the 2 reflected waves and the phase shift must be λ/2 (or 3λ/2, 5λ/2 ...etc).
This means that the film must be λ/4 thick ...light travels there and back through the film.
That is the level of my understanding !
Hope it helps
 

1. What is meant by "radiation perpendicularly incident to a surface"?

"Radiation perpendicularly incident to a surface" refers to electromagnetic radiation that is directed or hitting a surface at a 90-degree angle. This means that the radiation is traveling in a straight line and is perpendicular to the surface it is hitting.

2. How does radiation perpendicularly incident to a surface affect the material it is hitting?

The effects of radiation perpendicularly incident to a surface depend on the type and intensity of the radiation, as well as the properties of the material. In general, it can cause heating, ionization, and damage to the molecular structure of the material.

3. Can radiation perpendicularly incident to a surface be harmful to humans?

Yes, some types of radiation perpendicularly incident to a surface, such as ultraviolet and X-rays, can be harmful to humans. Exposure to high levels of these types of radiation can cause skin damage, burns, and even increase the risk of developing cancer.

4. How is the intensity of radiation perpendicularly incident to a surface measured?

The intensity of radiation perpendicularly incident to a surface is typically measured in units of watts per square meter (W/m²). This measures the amount of energy that is transferred to a surface over a certain area in a given amount of time.

5. What are some common sources of radiation perpendicularly incident to a surface?

Some common sources of radiation perpendicularly incident to a surface include the sun, X-ray machines, and nuclear reactors. However, there are also many natural and man-made sources of radiation that can cause perpendicularly incident radiation, such as cosmic rays, microwaves, and cell phones.

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