Why does the intensity of unpolarized light remain unchanged when it pass through a quarter wave plate? A quarter wave plate produces a phase difference between e- ray and o-ray. But the intensity is changed for plane polarized light.
[Physics] Intensity of unpolarized light acted upon by a quarter wave plate
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Related Solutions
Malus's law is about the effect of a polariser on polarised light. You've clearly read a badly written version of it. What your author likely meant to say was:
One begins with unpolarised light;
The first polariser quells the unaligned component of the unpolarised light and outputs polarised light (with half the input's intensity). This polarised output has intensity $I_0$ in your notation;
Of the polarised output from the first polariser, the second polariser lets through a fraction $(\cos\theta)^2$ where $\theta$ is the angle between the axes of the polarisers. So I say again: $I_0$ is the intensity of the polarised input to the second polariser, not the intensity of the unpolarised input to the system of two polarisers. With this proviso, the output intensity is $I_0\,(\cos\theta)^2$.
In Answer to:
But I don't understand why the intensity is lowered to half the input's intensity after the first polariser?
Depolarised light is actually quite a subtle and tricky concept: I discuss ways of dealing with it in my answers here and here. You can think of it intuitively: without a preference for polarisation, perfectly depolarised light must dump half its energy into a polariser: you can take this as a kind of "definition" of depolarised light if you like. The quantum description is much simpler than the classical, so I reproduce it here. We imagine the source producing photons each in pure polarisation states but with random direction. That is, each photon's polarisation axis makes some random, uniformly distributed angle with the polariser's axis. Its pobability of absorption is therefore $(\cos\theta)^2$. So now we simply average this quantity given a uniformly distributed angle:
$$\left<(\cos\theta)^2\right> = \frac{1}{2\pi}\int_0^{2\pi} (\cos\theta)^2\,\mathrm{d}\theta = \frac{1}{2}$$
to find the overall probability of passage through the polariser, and thus the proportion of photons that make it through. For more info on what polarisation actually means for a single photon, see my answer here.
A waveplate is a birefringent crystal. Birefringence is a particular kind of anisotropy where the refractive index depends on the plane of polarization of the light: there are two, orthogonal linear polarization planes which have a relatively lower ("fast axis") and higher ("slow axis") refractive index. An waveplate of angle $\phi$ is of such a thickness that the phase delay differs for these two polarization states by $\phi$ as light passes through the crystal.
Therefore, the way to analyze the action of such a crystal is to translate the above into equations. We represent a general polarization state by a $2\times2$ vector:
$$\left(\begin{array}{c}\alpha\,\cos(\omega\,t+\delta)\\\beta\,\cos(\omega\,t+\epsilon)\end{array}\right)$$
where $\alpha,\,\beta,\,\delta,\,\epsilon$ encode the magnitude and phase of the component of the electric field vector along the fast and slow axes.
In a circular polarization state, $\beta = \alpha = 1$ and $\epsilon = \delta\pm \pi/2$, which means that the fast polarization state leads/ lags the slow by $\pi/2$. This in turn means that if the fast polarization amplitude oscillates as $\cos\omega\,t$, the slow oscillates by $\sin\omega\,t$. It's not hard to show that the head of the vector:
$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\mp\sin(\omega\,t)\end{array}\right)$$
undergoes uniform circular motion and that this is therefore circular polarization.
If such a state is input to a quarter wave plate, then the plate imparts a further $\pi/2$ phase delay between the states. So now our state becomes
$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2}\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\pm\cos(\omega\,t)\end{array}\right)$$
and the fast and slow components are now either in or exactly out of phase. It's not hard to see that the head of this vector undergoes simple harmonic rectillinear motion, and that therefore we're now dealing with linear polarization.
Best Answer
An ideal quarter wave plate does not absorb any light.
The speed of light through the quarter wave depends on the orientation of the electric field vector relative to the plate. So one orientation is slowed down more than the other and the thickness of the plate is designed so that there is a change of phase of $90^\circ$ for one plane relative to another.
This represents a change of phase equivalent to a quarter of a wavelength.
An ordinary Polaroid actually absorbs light.
The light which emerges has the electric field oscillating in one plane the plane at right angles having been absorbed.