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In the study of light polarization, Malus's Law is fundamental. This law describes how the intensity of light changes as it passes through a polarizing filter. The equation for Malus's Law is expressed as follows: \[ I = I_0 \times \text{cos}^2(\theta) \]Here:

• \( I \) represents the transmitted intensity.
• \( I_0 \) is the initial intensity of the light.
• \( \theta \) is the angle between the initial direction of light polarization and the axis of the polarizing filter.

Imagine shining a flashlight through sunglasses. The sunglasses act as a polarizing filter. Malus's Law helps us understand how dark the light appears after passing through. If the angle \( \theta \) between the light's initial polarization direction and the filter's axis is 90 degrees (right angles), almost no light gets through.

Light can be polarized or unpolarized. Unpolarized light means the light waves are oscillating in multiple planes. Common sources of unpolarized light include sunlight and light bulbs. When unpolarized light encounters a polarizer, it becomes polarized. For unpolarized light passing through the first polarizer, the intensity is reduced by half. This is because only the component of the light that aligns with the polarizer's axis gets through. Mathematically, if the initial intensity is \( I_0 \), after the first polarizer, the intensity is: \[ I_1 = \frac{I_0}{2} \] This reduction happens universally, regardless of the initial intensity of the light.

Light intensity refers to the power of light per unit area. When dealing with polarized light, the intensity changes based on the angle between the light's polarization direction and the polarizing filter. In our exercise, the goal is to find the angle between the polarization directions of two sheets such that the intensity of the transmitted light is one-third of the original intensity. By applying Malus's Law to the second polarizer: \[ I_2 = I_1 \times \text{cos}^2(\theta) \]We know from the second step that \( I_1 = \frac{I_0}{2} \). We also want \( I_2 \) to be one-third of \( I_0 \):

• \( I_2 = \frac{I_0}{3} \)

Putting this in the equation: \[ \frac{I_0}{3} = \left( \frac{I_0}{2} \right) \times \text{cos}^2(\theta) \] We solve this to find out how much \( \theta \) should be: \[ \frac{2}{3} = \text{cos}^2(\theta) \] \[ \theta = \text{cos}^{-1}(\sqrt{\frac{2}{3}}) \] Using a calculator, we find \( \theta \), which is about 35.26 degrees. This angle ensures that the final transmitted light intensity is one-third of the original.