Question

An incident electromagnetic wave is polarized and its E vector has a maximum value of 6 Volts/m. It is incident on a polarizer whose transmission axis is rotated at an angle of 45 with respect to the incident E vector. What is the maximum value of the transmitted E vector?

Answer 1

To solve this problem it is necessary to apply the equations related to the law of Maus.

By the law of Maus we know that

`I = I_0 cos^2 \theta`

Where,

`I_0` = Intesity of incident light

I = Intensity of polarized light

With our values we have that

`I_0 =` 6V/m

`\theta = 45\°`

Then

`I = 6*cos^2(45)`

`I = 3V/m`

Therefore the maximum value of the transmitted E vector is 3V/m