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

`E=1cos(3.76991* 10^(15)t-12566370.61435y)`

`B=3.33* 10^(-9)sin(3.76991* 10^(15)t-12566370.61435z)`

Step-by-step explanation:

c = Speed of light =
`3* 10^8\ m/s`

`\lambda` = Wavelength =
`5* 10^(-7)\ m`

f = Frequency

Angular frequency is given by

`\omega=2\pi f\\\Rightarrow \omega=2\pi (c)/(\lambda)\\\Rightarrow \omega=2\pi (3* 10^8)/(5* 10^(-7))\\\Rightarrow \omega=3.76991* 10^(15)\ Hz`

Wave number is given by

`k=(2\pi)/(\lambda)\\\Rightarrow k=(2\pi)/(5* 10^(-7))\\\Rightarrow k=12566370.61435`

Magnetic field is given by

`B=(E)/(c)\\\Rightarrow B=(1)/(3* 10^8)\\\Rightarrow B=3.33* 10^(-9)\ T`

The magnetic field amplitude is
`3.33* 10^(-9)\ T`

The electric field is given by

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

Magnetic field is given by

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