# Radiation perpendicularly incident to a surface

1. Nov 17, 2011

### Demon117

1. The problem statement, all variables and given/known data
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 $\sqrt{n}$ and thickness 1/4 wavelength.

2. Relevant equations
Boundary conditions: $E_{1}^{||}=E_{2}^{||}$, and $B_{1}^{||}=B_{2}^{||}$

3. The attempt at a solution

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

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

For 0<z<$\frac{\lambda}{4}$:

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

For z>$\frac{\lambda}{4}$:

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

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

z=0 => ${E_{I}+E_{R}=E_{r}+E_{i}, E_{I}-E_{R}=\beta (E_{r}-E_{i})}$

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

Here, $\alpha =\frac{v_{2}}{v_{3}}$, $\beta =\frac{v_{1}}{v_{2}}$

I have done plenty of algebra to find $E_{I}$ in terms of $E_{R}$, 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?

2. Nov 18, 2011

### technician

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