# Electromagnetic Waves: What must be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is (a) one-third the maximum intensity of the transmitted...

Electromagnetic Waves: What must be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is (a) one-third the maximum intensity of the transmitted beam or (b) one-third of the incident beam?

Unpolarized light falls on two polarizing sheets placed one on lop of the Other. What must be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is (a) one-third the maximum intensity of the transmitted beam or (b) one-third of the incident beam? Assume that the polarizing sheet is ideal, that is, that it reduces the intensity of unpolarized light by exactly

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The intensity of light after passing through one polarizer is (Malus Law):

`I = I_0*cos^2(theta)`

where `theta` is the angle between the initial polarization of the light and polarizer axis. Since natural light has all polarization directions, after one polarizer the intensity of the light will be

`I = I_0*int_0^(2*pi)cos^2(theta)d(theta) =(1/2)*I_0`

Thus the maximum intensity of transmitted light through 2 polarizers will be when the axis of both polarizers are parallel.

`I_2/I_0 =(1/2)`

For one third of the maximum possible intensity transmitted we have

`1/3 =I_2/I_1 =cos^2(theta_12)`

`theta_12 =arccos(1/sqrt(3)) =54.73 degree`

For one third of the initial intensity after the second polarizer we have

`(1/3) =I_2/I_0 =I_2/I_1*I_1/I_0 =I_2/I_1 *(1/2)`

`2/3 =I_2/I_1 =cos^2(theta_12)`

`theta_12 =arccos(sqrt(2/3)) =35.26 degree`