Polarized light of amplitude \"1\" and described by the wave form \"sin x\" (wav
ID: 1997519 • Letter: P
Question
Polarized light of amplitude "1" and described by the wave form "sin x" (wavelength taken as 2PI for convenience and simplicity), directed from the left to right, passes through a polariser, the axis of which is tilted relative to the light's polarisation direction by 45 degrees. What is the amplitude of the light exiting the polariser? Suppose past the polariser is another linearly polarised light source, again directed left to right, with its polarisation direction the same as that of the polariser behind it. This second light source also has an amplitude "1" and wavelength "2PI", but is described by the wave form A sin(x+ x)". In terms of the given numerical values and the parameter " x", provide an expression for the aggregate amplitude due to the two light sources interfering on the right. If the aggregate amplitude is " 1.414", determine x, the phase shift.Explanation / Answer
Intensity is just the square of the amplitude. So the Intenisty of original polarized light was Io = A^2 = 1
From Malus law,
I = Io(cos^2)
= 1*(cos^2)45
= 0.5
Therefore amplitude of exiting light is sqrt(0.5) = 0.707
b) If two waves (with the same amplitude, frequency, and wavelength) are travelling in the same direction, using the principle of superposition, the resulting wave amplitude is given by
2Asin(/x) = 2*1sin(/x)
= 2sin(/x)
c) Given 2sin(/x) = 1.414
=> sin(/x) = 0.707
=> (/x) = 45 degrees = 0.785 radians