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Optical Phenomena

Problems on Polarization

Polarization

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Problem : A polaroid sheet and an analyzer are placed such that their transmission axes are co-linear. The analyzer is then rotated by 22.5 o . 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 = I 0cos2 θ = I 0cos2(22.5 o ) = 0.85I 0 . Thus the fraction of transmitted light is 85% of its previous value.

Problem : You have three ideal linear polarizers. Light of irradiance 1000 W/m 2 is shone through two of the polarizers, with their transmission axes placed at a relative angle of 40 o . What is the intensity of the transmitted light? Now place the third polarizer at an angle of 20 o between the other two. What is the irradiance?

In the first case, we just apply Malus' Law: I = 1000 cos2(40 o ) = 587 W/m 2 . In the second case, we must first calculate how much light is transmitted through the first two polarizers, at a relative angle of 20 o : I = 1000 cos2(20 o ) = 883 W/m 2 . This light is then incident on the second polarizer, which is at 20 o relative to the middle one: I = 883 cos2(20 o ) = 780 W/m 2 .

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

We have tanθ p = , thus θ = tan-1(n t/n i) = tan-1(1.52) = 56.7 o .

Problem : Consider the following waves:


E x   = E xcos(kz - σt)  
E y   = E ysin(kz - σt + 2Π)  

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

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 E 0 propagating along the x-axis with its plane of vibration at 30 o 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 E 0y = E 0cos(30 o ) = E 0 /2 and the z -wave by E 0z = E 0sin(30 o ) = E 0/2 . Thus we can write the wave as:

E(x, t) = (/2 +1/2)E 0sin(kx - σt)    

This satisfies the condition at x = t = 0 .

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