Question:
Three identical polaroids \(P_1\), \(P_2\) and \(P_3\) are placed one after another. The pass axis of \(P_2\) and \(P_3\) are inclined at angle of \(60^{\circ}\) and \(90^{\circ}\) with respect to axis of \(P_1\). The source has an intensity \(I_0\). The intensity of light finally coming out is
Three identical polaroids \(P_1\), \(P_2\) and \(P_3\) are placed one after another. The pass axis of \(P_2\) and \(P_3\) are inclined at angle of \(60^{\circ}\) and \(90^{\circ}\) with respect to axis of \(P_1\). The source has an intensity \(I_0\). The intensity of light finally coming out is
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The first polaroid halves unpolarised light; Malus law applies after that.
Updated On: Oct 1, 2026
- \(\frac{I_0}{2}cos^230^{\circ}cos^290^{\circ}\)
- \(\frac{I_0}{2}cos^260^{\circ}cos^230^{\circ}\)
- \(I_0cos^260^{\circ}cos^230^{\circ}\)
- zero
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The Correct Option is B
Solution and Explanation
Step 1: First polaroid
Unpolarised light of intensity \(I_0\) becomes \(\frac{I_0}{2}\) after \(P_1\).
Step 2: Second polaroid
The axis of \(P_2\) is \(60^{\circ}\) to \(P_1\), so the intensity is \(\frac{I_0}{2}\cos^260^{\circ}\).
Step 3: Third polaroid
\(P_3\) is at \(90^{\circ}\) to \(P_1\), which is \(30^{\circ}\) to \(P_2\). So the final intensity is \(\frac{I_0}{2}\cos^260^{\circ}\cos^230^{\circ}\). Option (B).
Final Answer:
The final intensity is (I0/2) cos^2 60 cos^2 30. \[ \boxed{\text{(B)}\ \frac{I_0}{2}\cos^260^{\circ}\cos^230^{\circ}} \]
Unpolarised light of intensity \(I_0\) becomes \(\frac{I_0}{2}\) after \(P_1\).
Step 2: Second polaroid
The axis of \(P_2\) is \(60^{\circ}\) to \(P_1\), so the intensity is \(\frac{I_0}{2}\cos^260^{\circ}\).
Step 3: Third polaroid
\(P_3\) is at \(90^{\circ}\) to \(P_1\), which is \(30^{\circ}\) to \(P_2\). So the final intensity is \(\frac{I_0}{2}\cos^260^{\circ}\cos^230^{\circ}\). Option (B).
Final Answer:
The final intensity is (I0/2) cos^2 60 cos^2 30. \[ \boxed{\text{(B)}\ \frac{I_0}{2}\cos^260^{\circ}\cos^230^{\circ}} \]
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