Question

When unpolarized light is passed through a polarizer its intensity?
A. There is no effect on the unpolarized light.
B. The intensity of the unpolarized light increases.
C. The intensity of the unpolarized light decreases.
D. The intensity of the unpolarized light becomes zero.

Answer 1

Hint:Let us first get some idea about light. Light, often known as visible light, is electromagnetic radiation that falls within the region of the electromagnetic spectrum that the human eye can perceive. Between the infrared (with longer wavelengths) and the ultraviolet (with shorter wavelengths), visible light is characterised as having wavelengths in the range of $$400-700$$ nanometres (nm) (with shorter wavelengths).This wavelength corresponds to a frequency range between $$430$$ and $$750$$ terahertz (THz).

Complete answer:
A light wave that vibrates in more than one plane is known as unpolarized light. Unpolarized light is that which is emitted by the sun, a classroom lamp, or a candle flame. Electric charges vibrate in a number of directions, causing an electromagnetic wave that vibrates in a variety of directions, resulting in such light waves. Unpolarized light is a challenging idea to conceptualise. In general, visualise unpolarized light as a wave with half of its vibrations in the horizontal plane and half of its vibrations in the vertical plane.
Unpolarized light can be transformed into polarised light. Light waves that are polarised have vibrations that occur in a single plane The process of polarisation is the transformation of unpolarized light into polarised light.
When unpolarized light is transmitted through a polarizer, its intensity is lowered by a factor of ${\displaystyle \frac{1}{2}}$.
According to Malus' law, the intensity of plane-polarized light passing through an analyzer varies as the square of the cosine of the angle between the polarizer's plane and the analyzer's transmission axis. The following expression represents this relationship:
$I=I_{\circ }{\cos }^2{\theta }_i$
where $$I_{\circ }$$ is the initial intensity and ${\theta }_i$ is the angle between the light's initial polarisation direction and the polarizer's axis.
At all angles, an unpolarized beam of light is a homogeneous combination of linear polarizations. The transmission coefficient becomes ${\displaystyle \frac{I}{I_{\circ }}}={\displaystyle \frac{1}{2}}$ since the average value of $$co{s}^2\theta$$ is ${\displaystyle \frac{1}{2}}$ .
So, option (C) is correct.

Note: By refracting light off of nonmetallic surfaces, unpolarized light can become polarised. The amount of polarisation that happens is determined by the angle at which light strikes the surface and the material that it is composed of.