91Ó°ÊÓ

To find the frequency \(f\) of the wave, we use the wave equation relating speed \(c\), frequency \(f\), and wavelength \(\lambda\): \[ c = f \cdot \lambda \]Given that the speed of light is \(c = 3 \times 10^8 \text{ m/s}\), and the wavelength \(\lambda = 432 \text{ nm} = 432 \times 10^{-9} \text{ m}\), we can solve for \(f\):\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{432 \times 10^{-9} \text{ m}} \]\( f \approx 6.94 \times 10^{14} \text{ Hz} \).

Use the relationship between the magnitudes of the electric field \(E\) and the magnetic field \(B\) amplitude for electromagnetic waves:\[ E = cB \]The amplitude of the magnetic field is given as \(B = 1.25 \times 10^{-6} \text{ T}\), so:\[ E = (3 \times 10^8) \times (1.25 \times 10^{-6}) \approx 375 \text{ V/m} \]

The equations for the electric and magnetic fields of a sinusoidal wave can be written as:For the electric field:\[ E(x,t) = E_0 \sin(kx - \omega t) \]For the magnetic field:\[ B(x,t) = B_0 \sin(kx - \omega t) \]Where:- \(E_0 = 375 \text{ V/m}\) is the amplitude of the electric field.- \(B_0 = 1.25 \times 10^{-6} \text{ T}\) is the amplitude of the magnetic field.- \(k = \frac{2\pi}{\lambda}\) is the wave number: \[ k = \frac{2\pi}{432 \times 10^{-9}} \approx 1.45 \times 10^7 \text{ m}^{-1} \]- \(\omega = 2\pi f\) is the angular frequency:\[ \omega = 2\pi \times 6.94 \times 10^{14} \approx 4.36 \times 10^{15} \text{ rad/s} \]Putting it all together, the equations are:\[ E(x,t) = 375 \sin\left(1.45 \times 10^7 x - 4.36 \times 10^{15} t\right) \]\[ B(x,t) = 1.25 \times 10^{-6} \sin\left(1.45 \times 10^7 x - 4.36 \times 10^{15} t\right) \]