Measuring the Faraday effect with ac
magnetic fields
Figure 15 gives a representation of an experimental setup
designed to measure the Faraday effect with ac magnetic fields.
Fig. 15 Diagram of the
experimental setup.
We start with linearly polarized light:
(Eq.15.1)
Then for a Faraday rotation of
, this becomes:
(Eq.15.2)
For
a Wollaston prism rotated at an angle
we then have:
(Eq.15.3)
where we recognize that:
(Eq.15.4)
1st case:
If
we position the Wollaston prism along the same axis as the polarizer,
= 0 and the two separated beams going to the photodiodes will
correspond to the x and y components of the electric field:
(Eq.15.5)
And
it follows that:
(Eq.15.6)
For
a small angle , we see that:
(Eq.15.7)
2nd case:
If
we position the Wollaston prism along 45°, Eq. 15.4 becomes:
(Eq.15.8)
and therefore:
(Eq.15.9)
and for small value of the angle
, we see that:
(Eq.15.10)
We could then simply measure the signal on one of the
photodiodes, for instance the one that detects IY. We have
then:
(Eq.15.11)
and since we know that
is proportional to the oscillating magnetic field, i.e.
, we can write:
(Eq.15.12)
and we can identify the parts of the equation to obtain:
(Eq.15.13)
It follows immediately that:
(Eq.15.14)
3rd case:
Another way to evaluate the Faraday rotation is to subtract
the x and y signals from the Wollaston prism. From equation 15.10, we obtain then:
(Eq.15.15)
Since , once again we can write:
(Eq.15.16)
Note that in this case I0 = 0 since we are subtracting that part of
the overall signal by measuring IY– IX.
Nevertheless, both inputs of the lock-in amplifier are still measuring their
respective DC components I0,Y and I0,X. Henceforth,
we can substitute in Eq. 15.13 in order to
obtain:
