91Ó°ÊÓ

To find the frequency, use the formula for wave speed, which is \(c=\lambda f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency. Solving for \(f\), we have \(f=c/ \lambda\). Substituting the given values \(c = 3.0 \times 10^8 m/s\), and \(\lambda = 435 nm = 435 \times 10^{-9} m\), we get \(f = (3.0 \times 10^8 m/s) / (435 \times 10^{-9} m) = 6.90 \times 10^{14} Hz\)

The magnitude of the magnetic field is related to the electric field by the formula \(B = E/c\), where \(E\) is the electric field amplitude, and \(c\) is the speed of light. Substituting the given \(E = 2.70 \times 10^{-3} V/m\) and \(c = 3.0 \times 10^8 m/s\), we get \(B = (2.70 \times 10^{-3} V/m) / (3.0 \times 10^{8} m/s) = 9.0 \times 10^{-12} T\) where \(T\) is Tesla, the unit for the magnetic field

The electric field \(\overrightarrow{\boldsymbol{E}}(z, t)\) can be expressed in general form as \(\overrightarrow{\boldsymbol{E}}(z,t) = E_0\cos(kz-wt + \phi)\hat{i}\) where \(\hat{i}\) is the unit vector in the x direction and the wave travels in the -z direction with a phase constant \(\phi\) of 0 here. Substituting \(E_0 = E = 2.70 \times 10^{-3} V/m\), \(k = 2\pi/\lambda\), \(f = 6.90 \times 10^{14} Hz\)(which gives the angular frequency \(w = 2\pi f\)), we get \(\overrightarrow{\boldsymbol{E}}(z, t) = 2.70 \times 10^{-3} \cos\left(\frac{2\pi z}{\lambda} - 2\pi f t\right)\hat{i}\). The magnetic field \(\overrightarrow{\boldsymbol{B}}(z, t)\) can be represented similarly as \(\overrightarrow{\boldsymbol{B}}(z,t) = B_0\cos(kz-wt + \phi)\hat{j}\) where \(B_0 = B = 9.0 \times 10^{-12}T\) and \(\hat{j}\) is the unit vector in the y direction.