[Meep-discuss] Right-left polarised light

Discussion:

Kenan Cicek

2015-12-10 16:33:31 UTC

Dear Meep Users,

I wonder if there is a way to measure the intensity distribution of right
or left circular polarised light beams separately in case of a beam that
consists of the superposition of this both beams.

Many thanks

Kenan

Владимир Борисович Новиков

2015-12-10 19:42:53 UTC

Permalink

Dear Kenan Cicek,

Circular polarizations are

left
*El*= ( 1/Sqrt(2) )
( i/Sqrt(2) )
right
*Er*= ( 1/Sqrt(2) )
( i/Sqrt(2) )

You should measure field electric vector in leght beam. You obtain vector
*E*= ( Ex )
( Ey )
Then consider this field in new basis (left and right polarizations):
L *El*+R *Er*=
*E*
Solving this linear system you receive amplitudes left and right polarized
light.
Intensity is proportional to L^2 and R^2 (for left and right polarized
light respectively).

Sincerely yours,
Vladimir Novikov

Post by Kenan Cicek
Dear Meep Users,
I wonder if there is a way to measure the intensity distribution of right
or left circular polarised light beams separately in case of a beam that
consists of the superposition of this both beams.
Many thanks
Kenan
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Kenan Cicek

2015-12-11 10:48:28 UTC

Permalink

Thanks Vladimir, appreciated it.

Cheers

Kenan

On 10 December 2015 at 19:42, ÐÐ»Ð°ÐŽÐžÐŒÐžÑ ÐÐŸÑÐžÑÐŸÐ²ÐžÑ ÐÐŸÐ²ÐžÐºÐŸÐ² <

Post by ÐÐ»Ð°Ð´Ð¸Ð¼Ð¸Ñ ÐÐ¾ÑÐ¸ÑÐ¾Ð²Ð¸Ñ ÐÐ¾Ð²Ð¸ÐºÐ¾Ð²
Dear Kenan Cicek,
Circular polarizations are
left
*El*= ( 1/Sqrt(2) )
( i/Sqrt(2) )
right
*Er*= ( 1/Sqrt(2) )
( i/Sqrt(2) )
You should measure field electric vector in leght beam. You obtain vector
*E*= ( Ex )
( Ey )
L *El*+R *Er*=
*E*
Solving this linear system you receive amplitudes left and right polarized
light.
Intensity is proportional to L^2 and R^2 (for left and right polarized
light respectively).
Sincerely yours,
Vladimir Novikov