Circular polarizer followed by a linear analyser

[Kaguro]

Homework Statement

A beam of unpolarised light of intensity I0 falls on a linear polariser. Then this is incident on a quarter wave plate inclined at 45 degrees to the axis of the polariser. This produces circularly polarized light. Then if it is incident on an analyser inclined perpendicular to the first polariser,then what is the polarization state and intensity of the light leaving the analyser?

Relevant Equations

Nothing.

I think that the intensity of output of circularly polarised light falling on an analyser should not depend upon the axis of analyser. Because it is circularly polarized. It keeps rotating.

So, the output should be linearly polarised. The amplitude of E field in the direction of analyser is a cosine function of time. So the intensity should be average value of this, and hence is 1/2 times the circularly polarised light.

So original light I0, after falling on 1st polariser it becomes I0/2. Circular polarisation doesn't alter intensity. And finally another 1/2. So final intensity is I0/4.

Is all the above correct?

 

[TSny]

Sounds correct to me. When stating that the output of the analyzer is linearly polarized light, it would probably be best to state the orientation of the polarization (even if it seems obvious to you).

 

[Kaguro]

Got it!

Thanks.