Final answer:
The electric fields for the reflected and transmitted waves at the air-salty lake interface are determined using reflection and transmission coefficients, which are calculated based on boundary conditions and the medium properties. These fields are found for normally incident plane waves at the interface between air and a lossless, salty lake with given permittivity and permeability at 1GHz.
Step-by-step explanation:
To find the electric fields of the reflected and transmitted waves at the air-salty lake interface, we use the boundary conditions at the interface and the properties of plane waves in media. Since the lake is assumed lossless (zero conductivity) and the wave is normally incident, there won't be any phase change upon reflection. The reflection and transmission coefficients can be computed using the permittivity (εr) and permeability (μr) of the salty lake.
The given incident electric field in the air can be written as Ei(z,t) = y100sin(2π×109t − βz)V/m. At z=0, using the boundary conditions, we need to find the electric fields of the reflected and transmitted waves, denoted Er(z,t) and Et(z,t) respectively.
The reflection coefficient R, given by R = (1-βt/βi)/(1+βt/βi), where βi and βt are the wave number in air and the salty lake respectively. The wave number is related to the properties of the medium by β = (2π/λ)√(εrμr). Substituting εr = 78.8 and μr = 1 at 1GHz for the lake, and εr = 1 and μr = 1 for air, we can find βi and βt and thus calculate R.
Once R is known, the electric field of the reflected wave is Er(z,t) = y100Rsin(2π×109t + βz)V/m, remembering that the reflected wave is moving in the -z direction, hence the positive sign in front of βz. The transmission coefficient T is given by T = 2/(1 + βt/βi), and the electric field of the transmitted wave is Et(z,t) = y100Tsin(2π×109t − βtz)V/m, since the transmitted wave continues to propagate in the +z direction within the lake.