Question

Four polaroids are placed such that the optic axis of each is inclined at an angle of \(30^{\circ}\) the optic axis of the preceding one. If unpolarised light of intensity ' \(\mathrm{I}_0\) ' falls on the first polaroid, the intensity of light transmitted from the fourth polaroid is \(\left[\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\)

• A. \(\frac{9\mathrm{I}_{0}}{32}\)
• B. \(\frac{27\mathrm{I}_{0}}{128}\)
• C. \(\frac{35\mathrm{I}_{0}}{128}\)
• D. \(\frac{27\mathrm{I}_{0}}{32}\)

Hint:

Unpolarised light loses half its intensity at the first polaroid; each next polaroid multiplies the intensity by cos squared of the angle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
First polaroid: \(I_1=\dfrac{I_0}{2}\).
For each next polaroid, Malus' law gives \(I=I_{prev}\cos^2\theta\).

Step 2: Calculation
\(\cos^2 30^{\circ}=\dfrac{3}{4}\).
Three more polaroids give \(I_4=\dfrac{I_0}{2}\left(\dfrac{3}{4}\right)^3\).

Step 3: Simplify
\(I_4=\dfrac{I_0}{2}\times\dfrac{27}{64}=\dfrac{27I_0}{128}\).

Step 4: Compare the options
This matches the second option.

Final Answer:
The transmitted intensity is 27 I0 over 128. \[ \boxed{\text{(B) }\dfrac{27\mathrm{I}_{0}}{128}} \]