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 _{0}cos^{2} θ = I _{0}cos^{2}(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 cos^{2}(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 cos^{2}(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 cos^{2}(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 _{x}cos(kz - σt) | ||
E _{y} | = E _{y}sin(kz - σt + 2Π) |
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 _{0}cos(30^{ o }) = E _{0} /2 and the z -wave by E _{0z} = E _{0}sin(30^{ o }) = E _{0}/2 . Thus we can write the wave as:
E(x, t) = (/2 +1/2)E _{0}sin(kx - σt) |