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How does an unpolarised light get polarised when passed through a polaroid ? Two polaroids are set in crossed position. A third polaroid is placed between the two making an angle theta with the pass axis of the first polaroid. Write the expression for the intensity of light transmitted from the second polaroid. in what orientations will the transmitted intensity be (i) minimum and (ii) maximum ? |
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Answer» Solution :A polaroid consists of long chain molecules aligned in a particular direction. If an unpolarised light wave is incident on such a polaroid then the polaroid absorbs the electric vectors of propagating light wave along the direction of the aligned molecules. THUS, in emergent light we have electric vector oscillating only along a direction perpendicular to the molecules (this direction is known as the pass axis of the polaroid). as a result the light wave gets linearly POLARISED. Let intensity of unpolarised light incident on the first polaroid be `I_(0)` then intensity of plane polarised light being transmitted by 1st polaroid will be `I_(1)=I_(0)//2`. When TWO polaroids are set in crossed positions and a third polaroid is placed between the two making an angle `theta` with the pass axis of the first polaroid, it makes an angle `90^(@)-theta)` with the pass axis of the second polaroid. therefore, in accordance with Malus law, the intensity of light transmitted from the second polaroid will be `I_(2)=I_(1)cos^(2)thetacos^(2)(90-theta)=Icos^(2)thetasin^(2)theta=(I)/(4)sin^(2)(2theta)=(I_(0))/(8)sin^(2)(2theta)` (i) The transmitted intensity will have a minimum value of zero when `sin2theta=0` i.e, `2theta=0` or `PI` that is `theta=0 or (pi)/(2)`. it MEANS when the third polaroid is held parallel to either first or the second polaroid, no light will be transmitted through the second polaroid. (ii) The transmitted intensity will have a maximum value of `(I_(0))/(8)` when `sintheta=pm1 or 2 theta=pm90^(@) or theta=45^(@)`. |
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