Question

A linearly-polarized electromagnetic (EM) wave has an electric field with amplitude 1000 V/m. You hold a a linear polarizer in front of it and rotate the polarizer so that all of the EM wave is blocked. You then rotate the polarizer 30 degree. What is the amplitude of the electric field of the EM wave that is transmitted through the polarizer? Select one: 250 V/m500 V/m Depends on rotation direction 866 V/m 750 V/m

Answer 1

Answer:

E1_max = 866 V/m...................................... option D

Explanation:

We know that for linearly polarized light, relation between intensity and electric field is given by:

I_avg = (1/2)*c*e0*E_max^2

I_avg = (1/2)*3*10^8*8.854*10^-12*1000^2

I_avg = 1328.1 W/m^2

Now given that light is already polarized, So Using Malus's law, Intensity of light after passing through polarizer will be:

I1 = I_avg*(cosФ  )^2

Ф = 30 deg, So

I1 = 1328.1*(cos 30 deg)^2 = 996.1 W/m^2

Now electric field corresponding to above Intensity will be:

I1 = (1/2)*c*e0*E1_max^2

E1_max = sqrt (2*I1/(c*e0))

E1_max = sqrt (2*996.1/(3*10^8*8.854*10^-12))

E1_max = 866 V/m