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State and obtain Malus' law. |
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Answer» Solution :The proof of Malus's law : (i) Consider the plane of polariser and ANALYSER are inclined to each other at an angle `theta`. Let `I_0`be the intensity and 'a' be the amplitude of the electric vector transmitted by the polariser. (ii) The amplitude 'a' of the incident light has two rectangular components, `(a cos theta)` and `(a sin theta)` which are the PARALLEL and perpendicular components to the axis of transmission of the analyser. (iii) Only the COMPONENT `(a cos theta)` will be transmitted by the analyser. The intensity of light transmitted from the analyser is proportional to the square of the component of the amplitude transmitted by the analyser. `Iprop ( a cos theta)^2` `I = k (a cos theta)^2` Where k is constant of proportionality. ` I = Ka^2 cos^2 theta` ` I = I_0 cos^2 theta` (iv) Where, `I_0 = ka^2` is the maximum intensity of light transmitted from the analyser. The following are few special cases. Case (i) When ` theta = 0^@ , cos 0 = 1 , I = I_0` When the transmission axis of polariser is ALONG that of the analyser, the intensity of lighttransmitted from the analyser is equal to the incident light that falls on it from the polariser. Case (ii) When ` theta= 90^@ , cos 90^@ = 0, I = 0 ` When the transmission axes of polariser and analyser are perpendicular to each other, the intensity of light transmitted from the analyser is ZERO. |
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