According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer. |

i.e I ∞ cos^{2}θ |

Suppose the angle between the transmission axes of the analyzer and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyzer. If E_{0} is the amplitude of the electric vector transmitted by the polarizer, then intensity I_{0} of the light incident on the analyzer is I ∞ E _{0}^{2} |

The electric field vector E_{0} can be resolved into two rectangular components i.e E_{0} cosθ and E_{0} sinθ. The analyzer will transmit only the component ( i.e E_{0} cosθ ) which is parallel to its transmission axis. However, the component E_{0}sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is, |

I ∞ ( E_{0} x cosθ )^{2} |

I / I_{0} = ( E_{0} x cosθ )^{2} / E_{0}^{2} = cos^{2}θ |

I = I_{0} x cos^{2}θ |

Therefore, I ∞ cos^{2}θ. This proves law of malus. |

When θ = 0° ( or 180° ), I = I_{0} cos^{2}0° = I_{0} That is the intensity of light transmitted by the analyzer is maximum when the transmission axes of the analyzer and the polarizer are parallel. |

When θ = 90°, I = I_{0} cos^{2}90° = 0 That is the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other. |

## Saturday, 24 April 2010

### Malus's Law

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