Answer to Question #284482 in Optics for luna

Question

Refraction of a plane wave at the interface between two dielectrics.

Answer 1

We know that

Incidence wave$E=E_0e^{j(k.r-wt)}$

Reflected wave

$E=E_{01}e^{j(k_1.r-w_1t)}$

Refracted wave

$E_2=E_{02}e^{j(k_2.r-w_2t)}$

$k_1$ =Reflected wave propagation constant

$k_2=$ Refracted wave propagation constant

$k$ =Incidence wave Propagation constant

tangential of E

$$(E_0)_te^{(k.r-wt)}+(E_0)_te^{(k_1.r-w_1t)}=(E_0)_te^{(k_2.r-w_2t)}$$

$k.r=k_1.r=k_2.r\\k_x=k_{1x}=k_{2x}$

For refracted

$k_{2x}=k_{1x}$

$k_{1x}sin\theta_1=k_{1x}sin\theta_2$

$\frac{w_1}{v_1}sin\theta_1=\frac{w_2}{v_2}sin\theta_2$

$w_1=w_2$

$\frac{c}{v_1}sin\theta_1=\frac{c}{v_2}sin\theta_2$

$n_1sin\theta_1=n_2sin\theta_2$

Refracted law