# Why is unpolarized light halved in intensity when polarized?

• Elmer Correa
In summary: The cycle averaged intensity of an unpolarized light is$$I = \langle |E_x(t) + E_y(t)|^2 \rangle$$$$= \langle E_x(t) E_x^*(t) \rangle + \langle E_y(t) E_y^*(t) \rangle + \langle E_x(t) E_y^*(t) \rangle + \langle E_y(t) E_x^*(t) \rangle = 2\langle A(t) \rangle$$
Elmer Correa
I would appreciate if the explanation didn't derive this phenomenon using Malsus' Law exclusively and would go into the horizontal and vertical components of polarized light, and how the blocking of the horizontal components results in a halved intensity...so more of a geometric explanation would be preferable. Also, if this has anything to do with it, as I understand it the amplitude of an polarized wave after passing through a polarizer is equal to cosθ where θ is the angle at which the wave is being polarized, and I don't quite understand why this is. On top of this amplitude squared is proportional to intensity, making the intensity then cosθ squared, the average of which is somehow 1/2. If this all ties into the geometric explanation please let me know. Also, how would I justify that the intesity is halved using a diagram?

It can't be properly answered without considering components. Let ##\vec{p}## be the polarization direction of the polarizer, and assume that is perpendicular to the direction of propagation. For a given ray R let ##\theta_R## be the angle the ray's polarisation vector makes with ##\vec{p}##. Assume ##\theta_R## has a uniform distribution in ##[0,2\pi]##. Then integrate the component ##A_R\cos\theta_R## over the range of ##\theta_R##.

blue_leaf77
Another alternative is to use the identifying relations for an unpolarized light. The perpendicular components of an unpolarized light satisfies
$$\langle E_x(t) E_x^*(t) \rangle = \langle E_y(t) E_y^*(t) \rangle = \langle A(t) \rangle$$
$$\langle E_x(t) E_y^*(t) \rangle = \langle E_y(t) E_x^*(t) \rangle = 0$$
Therefore the cycle averaged intensity of an unpolarized light is
$$I = \langle |E_x(t) + E_y(t)|^2 \rangle$$
$$= \langle E_x(t) E_x^*(t) \rangle + \langle E_y(t) E_y^*(t) \rangle + \langle E_x(t) E_y^*(t) \rangle + \langle E_y(t) E_x^*(t) \rangle = 2\langle A(t) \rangle$$
That's why if you blocked one component, say ##y##, it will be ##I' = \langle |E_x(t)|^2 \rangle = \langle A(t) \rangle = \frac{1}{2} I##.

andrewkirk said:
Then integrate the component ##A_R\cos\theta_R## over the range of ##\theta_R##.
Wouldn't that give zero?

blue_leaf77 said:
Wouldn't that give zero?
Yes, I forgot the bit about squaring it. I was in a bit of a rush.

## 1. Why does polarizing light decrease its intensity?

Unpolarized light contains waves that vibrate in all directions, while polarized light contains waves that vibrate in only one direction. When unpolarized light passes through a polarizing filter, the filter only allows the waves that vibrate in the same direction as the filter to pass through. This results in a decrease in the intensity of the light because only a portion of the waves are able to pass through the filter.

## 2. What causes light to become polarized?

Light can become polarized through various processes such as reflection, scattering, and transmission. When light is reflected off of a non-metallic surface at a specific angle, it becomes partially polarized. This is known as Brewster's angle. Scattering of light by particles in the atmosphere also causes polarization, as does transmission through certain materials such as polarizing filters.

## 3. Can polarized light be converted back to unpolarized light?

Yes, polarized light can be converted back to unpolarized light through a process called depolarization. This can occur naturally when polarized light is reflected or scattered multiple times, causing its polarization to become randomized. Depolarization can also be achieved artificially by using a depolarizing filter or by passing the light through a birefringent material.

## 4. How is the intensity of polarized light measured?

The intensity of polarized light can be measured using a polarimeter. This device works by passing polarized light through a polarizing filter and then measuring the intensity of the light that passes through the filter. Another method is to use a photometer, which measures the amount of light that is absorbed by a polarizing filter placed between the light source and the photometer.

## 5. Why is it important to polarize light in certain applications?

Polarized light has many practical applications, such as in 3D glasses and sunglasses, where it reduces glare and improves visibility. It is also used in photography and cinematography to improve image quality and reduce reflections. In scientific research, polarized light is used in microscopy to enhance contrast and reveal more details in the sample being observed.

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