I am trying to solve a problem in which light is normally incident on a material of refractive index n which is coated with an anti-reflective coating of refractive index $n^{\frac 1 2}$ and thickness equal to $\frac 1 4 \lambda$ ($\lambda$ being the wavelength). I need to show that under these conditions there is no reflected wave leaving the surface of the anti-reflective coating. I am taking the wave to be travelling along the z-axis, with the coating and material on the xy-plane.

The boundary conditions for interfaces between the materials are: $$\epsilon_1 E^{\perp}_1= \epsilon_2 E^{\perp}_2$$ $$B^{\perp}_1= B^{\perp}_2$$ $$\vec E^{\parallel}_1= \vec E^{\parallel}_2$$ $$\frac 1 {\mu} \vec B^{\parallel}_1=\frac 1 {\mu} \vec B^{\parallel}_2$$

In order to do this I believe I need to show that the wave reflected from the anti-reflective coating interferes destructively with one transmitted through the coating after having been reflected from the inner material.

So I would start with expressions for the electric and magnetic fields incident on the coating:

$\tilde {\vec E_I} (z,t) = \tilde E_{0_I} e^{i(k_1z- \omega t)} \hat x$

$\tilde {\vec B_I} (z,t) = \frac 1 v \tilde E_{0_I} e^{i(k_1z- \omega t)} \hat y$

This would then generate expressions for reflected and transmitted waves in the coating (to keep it shorter I'll only show the electric field related expressions):

$\tilde {\vec E_{R_1}} (z,t) = \tilde E_{0_{R_1}} e^{i(-k_1z- \omega t)} \hat x$

$\tilde {\vec E_{T_1}} (z,t) = \tilde E_{0_{T_1}} e^{i(k_2z- \omega t)} \hat x$

I can then apply the boundary conditions at the interface between air and the coating to relate the expressions: $$ \tilde {\vec E_{R_1}} + \tilde {\vec E_{T_1}} = \tilde {\vec E_I}$$

$\tilde {\vec E_{T_1}}$ is then incident on the inner material of refractive index n, again reflecting and transmitting:

$\tilde {\vec E_{R_2}} (z,t) = \tilde E_{0_{R_2}} e^{i(-k_2z- \omega t)} \hat x$

$\tilde {\vec E_{T_2}} (z,t) = \tilde E_{0_{T_2}} e^{i(k_3z- \omega t)} \hat x$

Boundary conditions can again be applied to relate expressions at the interface between the coating and the inner material: $$\tilde {\vec E_{R_2}} +\tilde {\vec E_{T_2}} =\tilde {\vec E_{T_1}}$$

Now is where I start to struggle a bit. Once the wave reflects from the inner material I suppose I need to add another reflection and transmission between the coating and air media like so: $$\tilde {\vec E_{R_2}} +\tilde {\vec E_{R_3}} =\tilde {\vec E_{T_3}}$$ Where $\tilde {\vec E_{T_3}}$ is the wave leaving the surface of the anti-reflective coating. At this point it seems like I have more variables than I can solve for with my equations, and thats not even including the magnetic field analysis. I think I must be overcomplicating the problem or missing some key simplification so I'd appreciate it if someone could give me some advice. Thank you!


1 Answer 1


You don’t need a magnetic field analysis; it would be redundant. You don’t need to worry about the boundary conditions. That’s how you derive the Fresnel reflection/transmission coefficients for an interface. Simply use those coefficients (unless your instructor intends you to derive them from scratch here, which seems unlikely), and keep track of the phase accumulation within the dielectric slab.

  • $\begingroup$ I think I see what you mean, so do I just need to use the relations $\tilde E_{0_R}=|\frac{n_1-n_2}{n_1+n_2}|\tilde E_{0_I}$ and $\tilde E_{0_T} = (\frac{2n_1}{n_1+n_2}) \tilde E_{0_I}$ to calculate my amplitudes and find the phase changes as the wave travels from $z=0$ to $z=\frac 1 4 \lambda$ and back again? $\endgroup$
    – Allod
    Nov 22, 2020 at 13:15
  • $\begingroup$ @Allod Right! Except in the reflection coefficient, don’t take the absolute value. The sign is important! $\endgroup$
    – Gilbert
    Nov 22, 2020 at 15:32

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