Question:
Three polarised sheets are co-axially placed. Pass axis of polaroids 2 and 3 make \(30^{\circ}\) and \(90^{\circ}\) with pass axis of polaroid sheet. If \(\mathrm{I}_0\) is the intensity of unpolarised light entering sheet 1 , the intensity of the emergent light through sheet 3 is
\[ \left(\cos 30^{\circ}=\sqrt{3} / 2, \cos 90^{\circ}=0, \cos 60^{\circ}=1 / 2\right) \]
Three polarised sheets are co-axially placed. Pass axis of polaroids 2 and 3 make \(30^{\circ}\) and \(90^{\circ}\) with pass axis of polaroid sheet. If \(\mathrm{I}_0\) is the intensity of unpolarised light entering sheet 1 , the intensity of the emergent light through sheet 3 is
\[ \left(\cos 30^{\circ}=\sqrt{3} / 2, \cos 90^{\circ}=0, \cos 60^{\circ}=1 / 2\right) \]
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Unpolarised light halves in intensity at the first sheet, then Malus law applies to each following sheet with the angle between successive axes.
Updated On: Nov 13, 2026
- zero
- \(\frac{3\mathrm{I}_{0}}{32}\)
- \(\frac{3\mathrm{I}_{0}}{8}\)
- \(\frac{3\mathrm{I}_{0}}{16}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Unpolarised light of intensity \(I_0\) becomes \(\dfrac{I_0}{2}\) after the first sheet. After that Malus law \(I=I'\cos^2\theta\) applies.
Step 2: Second sheet
Axis of sheet 2 is \(30^{\circ}\) from sheet 1: \(I_2=\dfrac{I_0}{2}\cos^230^{\circ}=\dfrac{I_0}{2}\times\dfrac34=\dfrac{3I_0}{8}\).
Step 3: Third sheet
Sheet 3 is at \(90^{\circ}\) from sheet 1, so \(60^{\circ}\) from sheet 2: \(I_3=I_2\cos^260^{\circ}=\dfrac{3I_0}{8}\times\dfrac14\).
Step 4: Result
\(I_3=\dfrac{3I_0}{32}\).
Final Answer:
The emergent intensity is \(\dfrac{3I_0}{32}\). \[ \boxed{\text{(B) }\dfrac{3\mathrm{I}_{0}}{32}} \]
Unpolarised light of intensity \(I_0\) becomes \(\dfrac{I_0}{2}\) after the first sheet. After that Malus law \(I=I'\cos^2\theta\) applies.
Step 2: Second sheet
Axis of sheet 2 is \(30^{\circ}\) from sheet 1: \(I_2=\dfrac{I_0}{2}\cos^230^{\circ}=\dfrac{I_0}{2}\times\dfrac34=\dfrac{3I_0}{8}\).
Step 3: Third sheet
Sheet 3 is at \(90^{\circ}\) from sheet 1, so \(60^{\circ}\) from sheet 2: \(I_3=I_2\cos^260^{\circ}=\dfrac{3I_0}{8}\times\dfrac14\).
Step 4: Result
\(I_3=\dfrac{3I_0}{32}\).
Final Answer:
The emergent intensity is \(\dfrac{3I_0}{32}\). \[ \boxed{\text{(B) }\dfrac{3\mathrm{I}_{0}}{32}} \]
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