# Double refraction in quarter wave plate and nicol prism

I have read in one of the textbook (local author) that when an unpolarized light is passed through nicol prism (parallel to the long sides of principle section),the light ray splits into two namely ordinary ray and extraordinary ray,and if the angle of incidence is greater than critical angle then o-ray will undergo total internal reflection whereas e-ray will emerge out(and thus a plane polarized light is obtained). In another section where production of circularly polarized light is described its said that when a plane polarized light is incident into a quarter wave plate at perpendicular direction to optic axis the light ray will split into o-ray and e-ray(with phase difference of 90degrees). My confusion is in the diagram they have shown for analysing a plane polarized light by nicol prism.....that is in the diagram of analyser its shown that there is no splitting of plane polarized light into o-ray and e-ray by analyser.But its already said that if light passes through quarter wave plate it splits.Now since quarter wave plate is also made from calcite crystal both nicol and qwp should show similar interaction to plane polarized light.ps:i am completely new to this topic so it could be that i an asking totally absurd stuff.

The situation shown in the figure pertains to the situation where the polarizer and the analyzer having their "pass-axis" aligned.

Since any one of them shall eliminate light/electric-field component along the axis perpendicular to the pass-axis, if linearly polarized light is incident onto a polarizer, then the output of the polarizer shall have light with electric field oscillations only along one definite axis, and nothing along the perpendicular axis. Now, if the pass-axis of the analyzer is also aligned with this axis, output light of the analyzer shall be the entire light input onto the analyzer, since no component is eliminated. That is to say, this would be the position of maximum light intensity through the analyzer.

However, if you rotate the pass-axis of the analyzer about the pass-axis of the polarizer through an angle $\theta$, only the $\cos \theta$ component of the light shall be parallel to the pass axis, and shall pass through, while the other $\sin \theta$ component would be blocked by the analyzer, being perpendicular. Thus, the intensity in this position would be $I = I_0 \cos^2 \theta$, where $I_0$ is the maximum light intensity. (Malus Law.)

If you increase $\theta$, moving it completely between $0$ and $2\pi$, i.e. full rotation in a circle, there will be two points $\theta = \pi/2$ and $3\pi/2$, where light oscillations shall be exactly perpendicular to the pass axis, and hence observed intensity would vanish. On the other hand, there are also two positions $\theta = 0$ and $\pi$, where $I = I_0$ (maximum).

This is how you analyze linearly polarized light using these instruments. Also, if the nature of the original light is unknown, this is how you learn that the incident light was originally linearly polarized.

• As you said when plane polarized light is passed through a polarizer it will have vibrations only along one plane.As i mentioned that in my text it says plane polarized light is split into two.... o-ray and e-ray (plane of vibration for these rays are mutually perpendicular for each other) by qwp then this statement in text is wrong.Am i wrong in assuming that qwp and polarizer are somewhat similar in the case of double refraction? Apr 2, 2017 at 15:04
• @AbhishekPallipparagopakumar - See, when rays "split" into two, they experience different refractive indices, and hence travel with different velocities. That means that after travelling some distance through that material, there shall be some path difference and hence some phase difference. In a QWP, you merely choose this distance to be such that the phase lag between the two components is exactly $\pi/2$, that's all. With everything else same, if you doubled the distance, the phase lag would grow to $\pi$, and you will have what is called a half wave plate! Apr 2, 2017 at 15:12
• @AbhishekPallipparagopakumar - The other part of the story which you perhaps might be missing is, a circularly polarized light has two perpendicular components of equal intensity $\pi/2$ radians apart. So, if CP light passes through a QWP, this $\pi/2$ due to QWP adds to the $\pi/2$ originally present, and the phase difference changes to $\pi$. Then, you have linearly polarized light. Thereafter, polarizer and analyzer would produce the action mentioned in the post. That's how you analyze circularly polarized light. :) Apr 2, 2017 at 15:15
• I didn't understand the last sentence in last comment "thereafter.......".I support the fact that there are two mutually perpendicular components for a CP light and it is the phase difference of 90 together with it which makes it CP .My first question is that so does plane polarized light too has two mutually perpendicular components?i just thought about it from what you said like when a CP light is passed through qwp 90 phase difference is introduced between two components again(an absurd question).Also in CP light i guess the two mutually perpendicular components are called o-ray and e-ray. Apr 2, 2017 at 15:48
• ans 1. In general, plane polarized light may have two components perpendicular to the direction of propagation of light (due to the transverse nature of the same). But they don't have any phase difference between them, i.e. they grow and fall together with time. e.g. $E_x = E_1\sin(\omega t)$ and $E_y = E_2\sin(\omega t)$, so that $E_y/E_x = E_2/E_1 = {\rm constt}$ at all times. On the other hand, in CP light, things are different. The two components have a $\pi/2$ phase difference, if one is a sine, other is a cosine. Thus, when one peaks, the other one extinguishes. In this case, (contd.) Apr 2, 2017 at 18:04