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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 :

cjmejiab

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

[tex]I = I_0 cos^2 \theta[/tex]

Where,

[tex]I_0[/tex] = Intesity of incident light

I = Intensity of polarized light

With our values we have that

[tex]I_0 =[/tex] 6V/m

[tex]\theta = 45\°[/tex]

Then

[tex]I = 6*cos^2(45)[/tex]

[tex]I = 3V/m[/tex]

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

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