Question

An unpolarized light beam of intensity \(I_{0}\) passes through a series of \(n\) Polaroids \(P_{1},P_{2},P_{3},\ldots P_{n}\). The pass axis of each Polaroid makes an angle of \(60^{\circ }\) with the other Polaroid. What is the final intensity of the transmitted beam? \((\cos 60^{\circ }=1/2)\)

• A. \(\frac{I_{0}}{2^{2n-1}}\)
• B. \(\frac{I_{0}}{2^{n-1}}\)
• C. \(\frac{I_{0}}{2^{2n}}\)
• D. \(\frac{I_{0}}{2^{2n+1}}\)

Hint:

Unpolarised light loses half its intensity at the first Polaroid. Every later Polaroid applies Malus's law with a factor of cos squared 60 degrees.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The first Polaroid cuts unpolarised light to half its intensity. After that, each Polaroid follows Malus's law, \(I=I_{1}\cos^{2}\theta\).

Step 2: Intensity after P1
\(I_{1}=\dfrac{I_{0}}{2}\)

Step 3: Effect of every later Polaroid
The angle is \(60^{\circ}\), so \(\cos^{2}60^{\circ}=\dfrac{1}{4}\).
There are \((n-1)\) more Polaroids, so the intensity is multiplied by \(\dfrac{1}{4}\) a total of \((n-1)\) times.

Step 4: Calculation
\(I=\dfrac{I_{0}}{2}\left(\dfrac{1}{4}\right)^{n-1}=\dfrac{I_{0}}{2\cdot 2^{2n-2}}=\dfrac{I_{0}}{2^{2n-1}}\)

Final Answer:
The final transmitted intensity is I0 divided by 2 raised to (2n minus 1). \[ \boxed{\text{(A) }\dfrac{I_{0}}{2^{2n-1}}} \]