Polarizing Sunglasses physics problem

[stunner5000pt]

Sunlight illuminates a 2mm hole in the blinds and falls on Polarizing Sunglasses. A plastic linear polarizer is placed (between the blinds and the sunglasses) ahead of the sunglasses with the lines aligned horizontally at time = 0. At this time the polarizer is rotated in a clockwise direction at a rate of 360 degrees per second.The sunlight has lambda ranges between 400nm and 700nm and intensity of 1kW/m^2.

a) after 0.5 seconds will the light that passes through the polairzer be half right hand circular polarized? Explain what the polarization will be as a function of time

well since it is rotated at 360 degrees for second after 180 degrees half of the beam has been polarized (right?) Would this be a sinusoidal function??

b) as a function of times plot the intensity and power of the light that falls on the sunglasses
would hte intensity be a constant function??
Wouldnt the power too be a constant function??

c) Plot the intensitry of the light that makes it through the glasses
Wouldnt it be a constant function as well??

d) Write down the expression for the elctric field at time = 0.625s that reaches the sunglasses for the wavelength = 514nm.
after 0.625s the axis has covered 225 degrees so beta - alpha = 45 degrees

it isn't specified whether this is a half wave or quarter wave plate
would i use this formula
$$E = E_{0} [cos(\alpha) cos(\beta - \alpha) + sin(\alpha) sin(\beta-\alpha)]$$

and both alphab ad beta - alpha = 45 degrees or pi/4 radians?

What is the value for omega and k? omega = 2pi / T - 2pi / 1 = 2pi?
k = 2 pi / v but what is v? the speed of light?

im guessing that the polarizer used in a full wave or a half wave plate because a subsequent question starts using a quarter wave plate

 

[thepatientmental]

a) Yes, after 0.5 seconds, the light that passes through the polarizer will be half right-hand circular polarized. This is because after 0.5 seconds, the polarizer would have rotated 180 degrees, which is half of the full rotation of 360 degrees. The polarization of the light passing through the polarizer will be a sinusoidal function, with the maximum polarization occurring at 0.5 seconds and the minimum polarization occurring at 1 second.

b) The intensity and power of the light that falls on the sunglasses will not be constant functions. The intensity will vary sinusoidally as the polarizer rotates, with the maximum intensity occurring at 0.5 seconds and the minimum intensity occurring at 1 second. The power will also vary in a similar manner, as it is directly proportional to the intensity.

c) The intensity of the light that makes it through the sunglasses will also vary sinusoidally, with the maximum intensity occurring at 0.5 seconds and the minimum intensity occurring at 1 second.

d) The expression for the electric field at time = 0.625 seconds that reaches the sunglasses for the wavelength = 514nm would be:

E = E_0 [cos(45 degrees) cos(45 degrees) + sin(45 degrees) sin(45 degrees)]

= E_0 [cos^2(45 degrees) + sin^2(45 degrees)]

= E_0 [1]

= E_0

The value for omega would be 2pi/0.5 = 4pi radians per second, as the polarizer is rotating at a rate of 360 degrees per second. The value for k would be 2pi/lambda, where lambda is the wavelength of the light passing through the polarizer. In this case, lambda = 514nm = 5.14 x 10^-7m, so k = 2pi/(5.14 x 10^-7m) = 1.22 x 10^7 m^-1.

The value for v would be the speed of light, which is approximately 3 x 10^8 m/s.