Supplement 3.3: Polarisation of Electromagnetic Waves: Stokes Vectors and Müller Matrices
Solutions to the task on page 2

1. Please calculate the Müller Matrix of a λ/4 retarder with the fast axis in the direction of the z coordinate.

The equation of a component which is exposed to a light beam at an angle of α is:

M(α)=R(α)MR(α)

In this case:

Q λ/4 (90°)=R(90°) Q λ/4 R(90°)

It is cos(α)=cosα , sin(α)=sinα . With α=90° for the orientation with the fast axis in direction of the z coordinate follows:

R(90°) Q λ/4 R(90°) =( 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 )( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )( 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 ) =( 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 )( 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 ) =( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )= Q λ/4 (90°)

This is the required result.

2. Calculate the type of polarisation of the transmitted light when the λ/4 retarder has this orientation, with incoming light of the intensity 1 and the following polarisations:

a) linear along y

Q λ/4 (90°) 0 =( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )( 1 0 0 0 )=( 1 0 0 0 )= 0 '

The outcoming light is still polarised linear along y.

b) linear diagonally in the first and third quadrant

Q λ/4 (90°) 45 =( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )( 1/2 1/2 1 0 )=( 1/2 1/2 0 1 )= c l '

The outcoming light is of a left cicular polarisation.

c) linear along z

Q λ/4 (90°) 90 =( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )( 0 1 0 0 )=( 0 1 0 0 )= 90 '

The outcoming light is still polarised linear along z.

d) linear diagonally in the second and fourth quadrant

Q λ/4 (90°) 135 =( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 )( 1/2 1/2 1 0 )=( 1/2 1/2 0 1 )= c r '

The outcoming light is polarised in a right circular manner.

Throughout these results, the λ/4 retarder may be used to convert linear polarised light into circular polarised and the other way round.

Further questions:
- how does the polarisation of the transmitted light change for other kinds of polarisation than the linear we dealt with?
- how does the polarisation change if the retarder is hit by the beams at a different angle α?
This can be determined by a similar calculation.