An ideal polarizer with its transmission axis rotated $30 ^ { \circ }$ to the vertical is placed in a beam of unpolarized light of intensity $10 \mathrm {~W} / \mathrm { m } ^ { 2 }$. After passing through the polarizer, what is intensity of the beam?

Question

Quizplus · Accepted Answer

The intensity of the beam after passing through an ideal polarizer is given by $I = I_0 \cos^2(\theta)$, where $I_0$ is the initial intensity and $\theta$ is the angle between the light's initial polarization direction and the polarizer's axis. For unpolarized light, $I_0$ is reduced by half upon passing through the polarizer, regardless of the orientation, because unpolarized light has equal intensity in all polarization directions. Therefore, $I = \frac{1}{2} I_0 = \frac{1}{2} \times 10 \mathrm {~W} / \mathrm { m } ^ { 2 } = 5.0 \mathrm {~W} / \mathrm { m } ^ { 2 }$. The angle $\theta$ does not affect the outcome for unpolarized light in this context.

An Ideal Polarizer with Its Transmission Axis Rotated 30∘30 ^ { \circ }30∘

An Ideal Polarizer with Its Transmission Axis Rotated $30 ^ { \circ }$