**1 Answer**

written 7.5 years ago by | • modified 7.5 years ago |

When a beam of light, polarized by reflection at one plane surface is allowed to fall on the second plane surface at the polarizing angle the intensity of the reflected beam varies with the angle between the planes of the two surfaces.

In the

**Biot’s polariscope**it was found that, ihe intensity of the twice reflected beam is maximum when the two planes are parallel and zero when the two planes are at right angles to each other.The same is also true for the twice transmitted beam from the polarizer and analyzer.

**The law of Malus**states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer.In the case of the Biot’s polariscope this angle is between the two reflecting planes. In the figure AP represents the transmission plane of a polariser and AE represents that of analyser.

The angle between two planes of analyser and polariser be 0(theta) . Let the electric vector of the polarised light emerging out from the plane of polariser is resolved into two components as shown.

To derive the law, let us suppose that the angle between the two planes of transmission is 0 (theta) at any instant.

The light vector, $AP = a$, in the plane polarised light emerging from the polariser may be resolved into two components, as shown in the figure, $ AE = (a cos \theta) $ .

Along the plane of transmission ‘$ AO = ( a sin \theta) $ perpendicular to the plane of transmission Perpendicular component is eliminated in the analyser while the parallel component is freely transmitted through it.

Therefore, the intensity I of light that emerges from the analyser is given by $ I = (a^2cos^2\theta)$ .