This satisfies 1st point mentioned above.įor the 2nd point, let us consider \(\phi = 0\)Īs the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave, the ratio of transmitted amplitude and the incident amplitude is \(\cos^2\phi\). When the unpolarized light passes through the first filter, the intensity is cut in half and comes out polarized at (0o). An ideal polarizing filter allows 100% of the incident unpolarized light to pass through, which is polarized in the direction of the filter’s polarizing axis.įrom the above two points, it can be assumed that, \(I = I_0 \cos^2 \phi\).It represents the strength of an effective linearly polarized eld that would correspond to the same intensity as (6.4). The overall eld strength Eeff is often unimportant in a discussion of polariza- tion. In this case we let Eeff jEyjeiy, B 1, and 0. When unpolarized light is incident on an ideal polarizer, the intensity of the transmitted light is exactly half of the intensity of the incident unpolarized light, regardless of how the polarizing axis is oriented. 146 Chapter 6 Polarization of Light other hand, if Ex happens to be zero, then its phase eix is indeterminant. This law is useful in quantitatively verifying the nature of polarised light.Ĭoming to the expression of Malus law, let us first see two points
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