Unpolarized light of intensity I₀ passes through a polarizing filter P1. The transmitted light then passes through a second filter P2 whose transmission axis is oriented at 60° relative to P1. What is the final intensity of the transmitted light according to Malus's Law?
Answer and explanation
Correct answer: B. (1/8) * I₀ = 0.125 I₀
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- Step 1: When unpolarized light of intensity passes through the first ideal polaroid , the transmitted intensity is halved due to averaging over all random polarization angles:
- Step 2: This transmitted light is now completely linearly polarized along the transmission axis of .
- Step 3: According to Malus's law, when polarized light of intensity passes through analyzer oriented at angle relative to , the transmitted intensity is:
- Step 4: Substitute and :
Answer
(1/8) * I₀ = 0.125 I₀
Why not the others: '(1/4) * I₀' calculates forgetting that the first polaroid cuts unpolarized intensity by half (); '(1/2) I₀' ignores the second analyzer altogether; '(3/8) I₀' uses instead of .
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