Talk:Optical isolator

Latest comment: 11 months ago by El Lobo in topic Issue with Faraday isolator figure

Merge (with Faraday isolator) ? edit

Seems straightforward to me; Yes!

Atlant 15:59, 10 October 2005 (UTC)Reply

Yes. Also beacause the term optical isolator is mostly(?) connected to the Faraday isolator. Maybe Optical isolator should be a fork page to the two.

sadly the merge leaves the article ignoring optical isolators not based on Faraday rotators. As noted in 2013 below they can also use [absorbing] linear polarisers and quarter-wave plates. Currently some of the mentions of optical isolator look as if they only apply to the Faraday isolator (as if that article had been editted and substituted here). Better? if we demerge - eg rename this article back to Faraday isolator and start a more general optical isolator article. Need to say clearly if Polarization independent isolator is a type of Faraday isolator or not ? - Rod57 (talk) 13:41, 11 October 2016 (UTC)Reply

Faraday rotator talk edit

According to the link it is also possible to isolate unpolarized light.

what exactly does it mean when you say "can provide non-reciprocal rotation while maintaining linear polarization". And perhaps provide an explanation of why a 1/4 waveplate can't maintain linear polarization in this manner. (just a thought) Whoopsi 01:06, 14 April 2006 (UTC)Reply

1/4 rotation by sugar-water-solution can maintain linear polarization (optical activity) . --Arnero 21:31, 23 June 2006 (UTC)Reply

This needs diagrams. — Omegatron 01:53, 18 July 2006 (UTC)Reply

The Faraday isolators I have used all have very strong magnets in them. Why is that? Needs explaining!

Polarization Independent Isolator edit

The change by 203.200.35.12 made the description to vague. This is a common problem with explainations of polarization independent isolators. Grahamwild 18:22, 14 August 2007 (UTC)Reply

umm? edit

why the polarized and all, put a source on one end and a detector on the other , light can only travel from the source, can it not? —Preceding unsigned comment added by 220.227.207.194 (talk) 13:09, 19 December 2007 (UTC)Reply


Some optical sources, such as super luminescent diodes, can be extremely sensitive to back reflection from anywhere in the optical circuit, even from connectors and splices. The isolator significantly reduce light travelling in the reverse direction. Regards, Grahamwild (talk) 14:35, 13 October 2008 (UTC)Reply

Kirchhoff's laws edit

Kirchhoff's laws are about absorption and emission - the generation and destruction of photons which is an exchange of energy format. A fellow called Mungans has a PDF linked in the notes where he totally confuses optical deflection (reflection, refraction etc - all variations of the basic process of scattering) in which the wave never changes its optical charater, thus no exchange of energy. --Damorbel (talk) 18:04, 28 July 2012 (UTC)Reply

Wave-plates edit

As I understand it, polarisation-dependent isolators are also commonly realised using a polariser and a quarter-wave plate: The incident wave is polarised (linearly), and as it passes through the wave-plate, it is converted to circular polarisation. On the return path, it is converted again to horizontal polarisation. I guess the article could be updated to mention this. Papa November (talk) 15:46, 9 October 2013 (UTC)Reply

I agree, a description of a 1/4 wave plate isolator should be included in this article, but it should be noted it doesn't maintain linear polarization, and doesn't isolate against feedback in an arbitrary polarization state. — Preceding unsigned comment added by DrErso (talkcontribs) 13:21, 7 September 2019 (UTC)Reply

What loss of energy edit

All three designs (1/4 wave plate , and 2 with Faraday rotators) all start with a linear polariser. Does that mean that they all absorb/loose 50% of the energy from an unpolarised source ? - Rod57 (talk) 13:57, 11 October 2016 (UTC)Reply

A polarization-independent isolator can be implemented either with a birefringent plate, or can split the rejected light from a linear polarizer to a separate Faraday rotator, and recombine it at the output. A polarization-dependent isolator will lose 50% from an unpolarized source. DrErso (talk) 13:04, 7 September 2019 (UTC)Reply

Issue with Faraday isolator figure edit

The figure demonstrating a polarization-dependent Faraday isolator (File:Faraday_isolator.svg) appears to have an error in the polarization rotation direction in the "backwards direction" portion of the figure. Rather than rotating counterclockwise by 45 degrees to arrive at horizontal polarization, it rotates clockwise by 135 degrees. This fails to demonstrate the critical property of Faraday rotators wherein the polarization rotation direction is dependent on the propagation direction (and therefore opposite for forward and backwards propagating light). --Tarheels 100 (talk) 17:06, 13 January 2021 (UTC)Reply

After 2,5 years the figure still needs to be fixed. Also, the phrase in the article “(the direction of rotation is not sensitive to the direction of propagation)” seems to be more ambiguous than the accurate statement in the above comment: “the polarization rotation direction is dependent on the propagation direction”. El Lobo (talk) 16:49, 29 May 2023 (UTC)Reply