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