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  • A polariser and an analyser are inclined to each other at 45°c the intensity of polarized light emerging from the analyser is I the intensity of unpolarised light incident on the polariser is</

A polariser and an analyser are inclined to each other at 45°. The intensity of polarized light emerging from the analyzer is I, then the intensity of unpolarized light incident on the polariser is?

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When an unpolarised light is incident on a polariser, its intensity is reduced by half. Then, if this polarized light passes through an analyzer inclined at an angle θ to the polarizer, the transmitted intensity is given by Malus’ law as $$
I= \frac {I_{unpol}}{2} cos^2θ
$$

Given θ= 45°

So, $$
I= \frac {I_{unpol}}{2} cos^245^°
$$ 

$$
I= \frac {I_{unpol}}{2}. \frac {1}{2} = \frac {I_{unpol}}{4}
$$

As the emerging intensity is I, so $$
\frac {I_{unpol}}{4} = I
$$ 

$$
I_{unpol} = 4I
$$

Therefore, the unpolarized light incident on the polarizer must have an intensity four times that of the polarized light emerging from the analyzer.

Posted by

Saniya Khatri

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