Question

A beam of unpolarised light of intensity \( I_0 \) is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of emergent light is:

• A. \(\frac{I_0}{2}\)
• B. \(\frac{I_0}{2\sqrt2}\)
• C. \(\frac{I_0}{4}\)
• D. \(\frac{I_0}{8}\)

Answer 1

The Correct Option is C

When unpolarised light passes through a polaroid, the intensity of the transmitted light \( I \) is given by:

\[ I = \frac{I_0}{2}, \]

where \( I_0 \) is the intensity of the incident unpolarised light.

Passing through Polaroid A: After passing through polaroid A, the intensity becomes:

\[ I_A = \frac{I_0}{2}. \]

Passing through Polaroid B: When the light passes through the second polaroid B at an angle \( \theta = 45^\circ \) relative to the first:

\[ I_B = I_A \cos^2(45^\circ) = \left( \frac{I_0}{2} \right) \cos^2(45^\circ). \]

Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):

\[ I_B = \left( \frac{I_0}{2} \right) \left( \frac{1}{\sqrt{2}} \right)^2 = \left( \frac{I_0}{2} \right) \left( \frac{1}{2} \right) = \frac{I_0}{4}. \]

Thus, the intensity of emergent light after passing through both polaroids is:

\[ \frac{I_0}{4}. \]

Answer 2

The problem asks for the final intensity of a beam of light after it passes through a sequence of two polaroids. The initial light is unpolarised, and the two polaroids are oriented at a specific angle relative to each other.

Concept Used:

1. Polarization by a Polaroid: When unpolarised light of intensity \( I_0 \) passes through a polaroid, the intensity of the transmitted light becomes half of the initial intensity, and the light becomes plane-polarized. The intensity after the first polaroid (\( I_1 \)) is given by:

\[ I_1 = \frac{I_0}{2} \]

2. Malus's Law: When a beam of plane-polarized light of intensity \( I_1 \) passes through a second polaroid (analyzer) whose transmission axis makes an angle \( \theta \) with the polarization direction of the incident light, the intensity of the emergent light (\( I_2 \)) is given by:

\[ I_2 = I_1 \cos^2\theta \]

Step-by-Step Solution:

Step 1: Calculate the intensity of light after passing through the first polaroid, A.

The incident light is unpolarised with an initial intensity of \( I_0 \). When this light passes through the first polaroid (A), it becomes plane-polarized, and its intensity is reduced to half of the original intensity.

Let \( I_A \) be the intensity of light emerging from polaroid A. Then:

\[ I_A = \frac{I_0}{2} \]

Step 2: Apply Malus's Law to find the intensity after passing through the second polaroid, B.

The plane-polarized light from polaroid A, with intensity \( I_A \), is now incident on polaroid B. The principal plane of polaroid B makes an angle of \( \theta = 45^\circ \) with respect to the principal plane of polaroid A.

According to Malus's Law, the intensity of the light emerging from polaroid B, let's call it \( I_B \), is given by:

\[ I_B = I_A \cos^2\theta \]

Step 3: Substitute the known values into the equation to find the final intensity.

We substitute \( I_A = \frac{I_0}{2} \) and \( \theta = 45^\circ \) into the formula from Step 2.

\[ I_B = \left( \frac{I_0}{2} \right) \cos^2(45^\circ) \]

Final Computation & Result:

We know that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \). Therefore, \( \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).

Substituting this value back into the expression for \( I_B \):

\[ I_B = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4} \]

The intensity of the emergent light is \(\frac{I_0}{4}\).