Question

Two Polaroids P₁ and P₂ are placed with their axis perpendicular to each other. Unpolarised light l₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its axis makes an angle 45º with that of P₁. The intensity of transmitted light through P₂ is

• A. \(\frac {l_0}{2}\)
• B. \(\frac {l_0}{4}\)
• C. \(\frac {l_0}{8}\)
• D. \(\frac {l_0}{16}\)

Answer 1

The Correct Option is C

To determine the intensity of transmitted light through the second polaroid \(P_2\) when unpolarized light \(I_0\) is incident on the first polaroid \(P_1\), we use Malus's Law and the properties of polaroids. Let's solve the problem step-by-step:

1. When unpolarized light of intensity \(I_0\) passes through the first polaroid \(P_1\), the intensity of the light transmitted by \(P_1\) becomes: \(I_1 = \frac{I_0}{2}\).

   This is because a polaroid only allows the component of light polarized in its axis to pass through, reducing the intensity by half.

2. The light then passes through the second polaroid \(P_3\), which is oriented at an angle of 45º with respect to \(P_1\). According to Malus's Law, the intensity \(I_2\) of light after passing through \(P_3\) is given by: \(I_2 = I_1 \cdot \cos^2(45^\circ)\).

   Since \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\), we have:

   \(I_2 = \frac{I_1}{2} = \frac{I_0}{4}\).
3. Finally, the light passes through the third polaroid \(P_2\), which is perpendicular to \(P_1\). This means the axis of \(P_2\) is also at an angle of 45º with respect to \(P_3\). Using Malus's Law again: \(I_3 = I_2 \cdot \cos^2(45^\circ)\).

   Similarly, because \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\), we find:

   \(I_3 = \frac{I_2}{2} = \frac{I_0}{8}\).

Thus, the intensity of the light transmitted through \(P_2\) is \(\frac{I_0}{8}\), which matches the correct answer.