A linearly polarized beam of light of intensityI0 is propagating in the +z direc

Question

A linearly polarized beam of light of intensityI0 is propagating in the +z direction and its direction of polarization makes an angle of 66? with the x-axis (the angle between the initial direction of polarization and the transmission axis of the first polarizer is 24? as described below). Two linear polarizing sheets are lined up perpendicular to the beam. The first sheet has its transmission axis at an angle q 1 = 90?with respect to thex-axis. The transmission axis of the second is variable. (See lower panels of the figure.)

(a) Calculate the intensity I1 of the light after it has passed through the first polarizer. Express your answer as a fraction of I0.

I1 / I0 = ______________

(b) Calculate the electric field amplitudeE1 of the light after it has passed through the first polarizer. Express your answer as a fraction of the electric field amplitude E0 of the initial beam.

E1 / E0 = ______________

(c) The second polarizer is set at various angles within the rangeq2= 0 to 90?.Calculate the intensity of the light after it has passed through thesecond polarizer for the following values ofq2. Express each answer as a fractionofI1.

Atq2= 0?:I2/I1=

Atq2= 37?:I2/I1=

Atq2= 90?:I2/I1=

Explanation / Answer

Malus law states that intensity I after passing through a polarizer I=I0 [cos(theta)]^2 where I0 is the initial intensity, and theta is the angle between the light's initial polarization direction and the axis of the polarizer. (a)so I1/I0=[cos(24)]^2=0.834 (b)Etransmitted=Eincident*cos(theta) E1/E0=cos(24)=0.913 (c)after passing through 1st polariser light is at an angle 66 degrees with x axis 0 degrees I2/I1=(cos66)^2=0.165 37 degrees I2/I1=(cos(66-37))^2=0.764 90 degrees I2/I1=(cos24)^2=0.834

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