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Problems on Polarization

Problems on Polarization

Problems on Polarization

Problems on Polarization

Problem : A polaroid sheet and an analyzer are placed such that their transmission axes are co-linear. The analyzer is then rotated by 22.5o. What is the irradiance of the transmitted light as a fraction of its previous value?

When the axes are co-linear, the amplitude or irradiance of the transmitted light is a maximum. After rotation, the irradiance is given by Malus' Law: I = I0cos2θ = I0cos2(22.5o) = 0.85I0. Thus the fraction of transmitted light is 85% of its previous value.

Problem : You have three ideal linear polarizers. Light of irradiance 1000 W/m2 is shone through two of the polarizers, with their transmission axes placed at a relative angle of 40o. What is the intensity of the transmitted light? Now place the third polarizer at an angle of 20o between the other two. What is the irradiance?

In the first case, we just apply Malus' Law: I = 1000 cos2(40o) = 587 W/m2. In the second case, we must first calculate how much light is transmitted through the first two polarizers, at a relative angle of 20o: I = 1000 cos2(20o) = 883 W/m2. This light is then incident on the second polarizer, which is at 20o relative to the middle one: I = 883 cos2(20o) = 780 W/m2.

Problem : What is Brewster's angle for reflection from air off a glass (n = 1.52) surface?

We have tanθp = , thus θ = tan-1(nt/ni) = tan-1(1.52) = 56.7o.

Problem : Consider the following waves:

Ex = Excos(kz - σt)  
Ey = Eysin(kz - σt + 2Π)  

If these waves overlap at a particular point, what is the polarization of the resulting wave (assume Ex = Ey)?

The polarization depends on the phase difference between the two waves. We can turn wave in y into a cosine expression by subtracting Π/2 from the phase, making the overall phase difference either ε = 3Π/2. This is equal to - Π/2 + 2Π so the resulting wave is right circularly polarized.

Problem : Write the equation for the linearly polarized light wave of angular frequency σ and amplitude E0 propagating along the x-axis with its plane of vibration at 30o to the xy-plane (assume that E = 0 when x = 0 and t = 0).

The wave will consist of two waves, one oscillating in the y-direction and one in the z-direction, both propagating in x. The amplitude of the y-wave is given by E0y = E0cos(30o) = E0/2 and the z-wave by E0z = E0sin(30o) = E0/2. Thus we can write the wave as:

E(x, t) = (/2 +1/2)E0sin(kx - σt)    

This satisfies the condition at x = t = 0.