Question

Unpolarised light of intensity 32Wm^-2 passes through the combination of three polaroids such that the pass axis of the last polaroid is perpendicular to that of the pass axis of first polaroid. If intensity of emerging light is 3Wm^-2 , then the angle between pass axis of first two polaroids is ______º .

Hint:

Remember Malus’s law: \( I = I_0cos^2θ\), where I0 is the initial intensity and θ is the angle between the polarizer’s axis and the polarization direction of the incident light. For unpolarized light, the initial intensity is halved after the first polarizer

Answer 1

Correct Answer: 30

Step 1: Write the Expression for \( I_{\text{net}} \)

The net intensity is given by:

\[ I_{\text{net}} = I_0 \cdot 2 \cos^2\theta \sin^2\theta \]

Using the trigonometric identity \( \sin^2(2\theta) = 4\cos^2\theta \sin^2\theta \):

\[ I_{\text{net}} = \frac{I_0}{8} \cdot (\sin(2\theta))^2 \]

Step 2: Use the Given Value for \( I_{\text{net}} \)

Substitute \( I_{\text{net}} = 3 \):

\[ 3 = \frac{I_0}{8} \cdot (\sin(2\theta))^2 \]

Simplify for \( \sin(2\theta) \):

\[ (\sin(2\theta))^2 = \frac{3 \cdot 8}{I_0} \]

Step 3: Solve for \( \sin(2\theta) \)

Assuming \( I_0 \) simplifies, take the square root:

\[ \sin(2\theta) = \frac{\sqrt{3}}{2} \]

From trigonometric solutions, \( 2\theta = 60^\circ \) and \( 120^\circ \).

Step 4: Solve for \( \theta \)

Divide by 2 to find \( \theta \):

\[ \theta = 30^\circ \, \text{and} \, 60^\circ \]

Final Answer:

The values of \( \theta \) are:

\( \theta = 30^\circ \, \text{and} \, 60^\circ. \)