Question

A beam of unpolarized light is incident on a polarizer. The intensity of the transmitted light is measured as it passes through the polarizer. If the angle between the light's initial direction and the axis of the polarizer is \( \theta \), what is the intensity of the transmitted light? The intensity of the transmitted light is given by Malus' law: \[ I = I_0 \cos^2 \theta \]
where:
\( I_0 \) is the intensity of the unpolarized light before passing through the polarizer,
\( \theta \) is the angle between the light's initial direction and the axis of the polarizer.

• A. \( I = I_0 \cos \theta \)
• B. \( I = I_0 \cos^2 \theta \)
• C. \( I = I_0 \sin^2 \theta \)
• D. \( I = I_0 \sin \theta \)

Hint:

Remember that for unpolarized light passing through a polarizer, the intensity is reduced by a factor of \( \cos^2 \theta \), where \( \theta \) is the angle between the incident light's polarization and the polarizer's axis.

Answer 1

The Correct Option is B

When unpolarized light passes through a polarizer, its intensity reduces according to Malus' law, which states that the intensity of the transmitted light \( I \) is related to the initial intensity \( I_0 \) and the angle \( \theta \) between the light’s initial direction and the axis of the polarizer: \[ I = I_0 \cos^2 \theta \] This equation shows that the transmitted intensity depends on the cosine square of the angle between the light’s initial polarization direction and the axis of the polarizer. Thus, the correct answer is Option (B): \( I = I_0 \cos^2 \theta \).