QuestionJuly 30, 2025

1.2 We pass initially unpolarized light through a polarizing filter that is oriented at 60^circ degrees with respect to the horizontal. If the light had some initial intensity I, what is the intensity after it goes through the filter? (I)/(2) Zero (I)/(4) (3I)/(4) I (5 points)

1.2 We pass initially unpolarized light through a polarizing filter that is oriented at 60^circ degrees with respect to the horizontal. If the light had some initial intensity I, what is the intensity after it goes through the filter? (I)/(2) Zero (I)/(4) (3I)/(4) I (5 points)
1.2 We pass initially unpolarized light through a polarizing filter that is oriented at 60^circ degrees with respect to the horizontal. If the light had some initial intensity I, what is the intensity after it goes through the filter? (I)/(2) Zero (I)/(4) (3I)/(4) I (5 points)

Solution
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Answer

\frac{I}{8} Explanation 1. Apply Malus's Law For initially unpolarized light, the intensity after passing through a polarizer is **I' = \frac{I}{2}**. 2. Calculate Intensity with Angle After passing through a second polarizer at angle \theta, the intensity is **I'' = I' \cos^2(\theta)**. Here, \theta = 60^\circ. 3. Substitute Values I'' = \frac{I}{2} \cdot \cos^2(60^\circ) = \frac{I}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{I}{8}.

Explanation

1. Apply Malus's Law<br /> For initially unpolarized light, the intensity after passing through a polarizer is **$I' = \frac{I}{2}$**.<br />2. Calculate Intensity with Angle<br /> After passing through a second polarizer at angle $\theta$, the intensity is **$I'' = I' \cos^2(\theta)$**. Here, $\theta = 60^\circ$.<br />3. Substitute Values<br /> $I'' = \frac{I}{2} \cdot \cos^2(60^\circ) = \frac{I}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{I}{8}$.
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