Polarized light of intensity S₀ passes through an ideal polarizer.If the E vector of the polarized light is horizontal what,in terms of the initial intensity S₀,is the intensity of the light that passes through a polarizer if that polarizer is tilted $0.449 \mathrm { rad }$ from the horizontal? A)0.812 S₀ B)0.188 S₀ C)0.217 S₀ D)0.284 S₀

Question

Quizplus · Accepted Answer

The intensity of the light after passing through a polarizer is given by Malus's law, $I = I_0 \cos^2(\theta)$, where $I_0$ is the initial intensity and $\theta$ is the angle between the light's polarization direction and the polarizer's axis. Here, $\theta = 0.449$ rad. Thus, $I = I_0 \cos^2(0.449) \approx 0.812 I_0$.

Polarized Light of Intensity S0 Passes Through an Ideal Polarizer 0.449rad0.449 \mathrm { rad }0.449rad

Polarized Light of Intensity S₀ Passes Through an Ideal Polarizer $0.449 \mathrm { rad }$