Question

Green light has wavelength 5×10−7m in a vacuum. The light is travelingin the +ydirection and is polarized so that the electric field is in the direction. The Peak electric field has a value peak= 1 V/m.(a) Write expressions for the electric and magnetic fields as a function of position andtime. Make sure you indicate the direction as well as the magnitude.

Answer 1

Answer:

$E=1cos(3.76991\times 10^{15}t-12566370.61435y)$

$B=3.33\times 10^{-9}sin(3.76991\times 10^{15}t-12566370.61435z)$

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength = $5\times 10^{-7}\ m$

f = Frequency

Angular frequency is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi \dfrac{c}{\lambda}\\\Rightarrow \omega=2\pi \dfrac{3\times 10^8}{5\times 10^{-7}}\\\Rightarrow \omega=3.76991\times 10^{15}\ Hz$

Wave number is given by

$k=\dfrac{2\pi}{\lambda}\\\Rightarrow k=\dfrac{2\pi}{5\times 10^{-7}}\\\Rightarrow k=12566370.61435$

Magnetic field is given by

$B=\dfrac{E}{c}\\\Rightarrow B=\dfrac{1}{3\times 10^8}\\\Rightarrow B=3.33\times 10^{-9}\ T$

The magnetic field amplitude is $3.33\times 10^{-9}\ T$

The electric field is given by

$E=Ecos(\omega t)\\\Rightarrow \mathbf{E=1cos(3.76991\times 10^{15}t-12566370.61435y)}$

Magnetic field is given by

$B=Bsin(\omega t)\\\Rightarrow \mathbf{B=3.33\times 10^{-9}sin(3.76991\times 10^{15}t-12566370.61435z)}$