Question:medium

A beam of light of intensity \(I_0\) falls on a system of three polaroids such that each pass axis is turned through \(60^{\circ}\) with respect to preceding one. The fraction of intensity that passes through is:

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The first polaroid always cuts intensity by half; subsequent ones follow $\cos^2 \theta$.
Updated On: May 14, 2026
  • \(\frac{1}{8}\)
  • \(\frac{1}{32}\)
  • \(\frac{1}{16}\)
  • \(\frac{1}{2}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
Unpolarized light of intensity \(I_0\) passes through three polarizers. Each polarizer's axis is rotated by \(60^{\circ}\) relative to the one before it.
Step 2: Key Formula or Approach:
1) After the first polarizer, intensity becomes \(I_1 = \frac{I_0}{2}\).
2) Malus' Law: \(I = I' \cos^2 \theta\), where \(\theta\) is the angle between the pass axis of the polarizer and the plane of polarization of incident light.
Step 3: Detailed Explanation:
Let the initial intensity be \(I_0\).
After passing through the first polaroid (\(P_1\)):
\[ I_1 = \frac{I_0}{2} \]
The light is now linearly polarized along the axis of \(P_1\).
It then passes through the second polaroid (\(P_2\)), which is at \(60^{\circ}\) to \(P_1\):
\[ I_2 = I_1 \cos^2 60^{\circ} = \frac{I_0}{2} \times \left( \frac{1}{2} \right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8} \]
The light is now polarized along the axis of \(P_2\).
It then passes through the third polaroid (\(P_3\)), which is at \(60^{\circ}\) to \(P_2\):
\[ I_3 = I_2 \cos^2 60^{\circ} = \frac{I_0}{8} \times \left( \frac{1}{2} \right)^2 = \frac{I_0}{8} \times \frac{1}{4} = \frac{I_0}{32} \]
The fraction of the incident light intensity is:
\[ \frac{I_3}{I_0} = \frac{1}{32} \]
Step 4: Final Answer:
The fraction is \(\frac{1}{32}\).
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