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PHY227 Homework 1, answers
Question 6
Consider electric field oscillations of the electromagnetic wave propagating towards you in a fixed
plane (i.e. only the temporal evolution will be taken into account). Any polarisation state will be
represented by two linearly polarised components polarised along the fast (y) and slow (x) axes. The
right-circularly polarised light is then given by
E x  E0x sin ωt
E y  E0y sin(ωt   )
Here =/2+2m (m – integer). After passing through the QWP an additional phase shift of /2 will
be acquired by the y-component so that
E x  E0x sin ωt
E y  E0y sin(ωt  2πm  π)
This is linearly polarised light at an angle =-45 degrees with respect Ox. The relation between E0x
and E0y is obtained from
E0x  E0 | cos  |
E0y  E0 | sin  |
E0x / E0y | cos  | / | sin  |
The linear polariser (LP) blocking this light has the transmission axis at =+45 degrees with respect
Ox.
For the elliptically polarised light, again linearly polarised light is produced by the QWP, as the
light can be completely blocked by the LP. Now its axis is rotated by 30 degrees clockwise, i.e. forms
=-75 degrees with respect Ox.
Following the expression above
E0x / E0y | cos(75) | / | sin(75) | 0.268
Linearly polarised light produced by the QWP can be described as before as
E x  E0x sin ωt
E y  E0y sin(ωt    2πm)
In a similar way, an additional phase shift of /2 has been acquired by the y-component so that the
original waves before the QWP are:
E x  E0x sin ωt
E y  E0y sin(ωt  2πm  π/2)
Thus the answer is an elliptically polarised light with the electric field vector rotating clockwise and
the ratio of the horizontal and vertical components of 0.268.