A polarizer and an analyzer are oriented so that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the analyzer is rotated through 45⁰?

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1.25

2.25

3.25

0.5

answer is D.

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Detailed Solution

$I_{1} = \frac{I_{0}}{2};$

For maximum amount of transmission, angle between the polarizer and analyzer should be zero degree,

$\Rightarrow I_{2} = \frac{I_{0}}{2} \cos^{2} 0^{o} = \frac{I_{0}}{2} = I_{\max}$

Now, $I'_{2} = \frac{I_{0}}{2} \cos^{2} 45^{o} = I_{2} (\frac{1}{2}) = 0.5 I_{2}$

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