Malus' Law: The Simple Rule for Analyzing Polarized Light Intensity

Step 1: Identify light type - unpolarized

Step 2: For unpolarized light + single polarizer:

$I = \frac{I_0}{2}$

Step 3: The orientation doesn't matter for unpolarized light

Step 1: Identify light type - polarized

Step 2: Apply Malus' Law directly:

$I = I_0\cos^2(60^\circ)$

Step 3: Calculate: $\cos 60^\circ = 0.5$, so $\cos^2 60^\circ = 0.25$

$I = 0.25I_0$

Step 1: After first polarizer (unpolarized light):

$I_1 = \frac{I_0}{2}$

Step 2: Light is now polarized along first polarizer's axis

Step 3: Second polarizer at 90° to first ⇒ θ = 90°

Step 4: Apply Malus' Law: $I_2 = I_1\cos^2(90^\circ) = I_1 \times 0$

$I_2 = 0$

Step 1: After first polarizer:

$I_1 = \frac{I_0}{2}$

Step 2: Light polarized along first polarizer's axis

Step 3: Angle between polarizers = 60° ⇒ θ = 60°

Step 4: Apply Malus' Law: $I_2 = I_1\cos^2(60^\circ) = \frac{I_0}{2} \times (0.5)^2$

$I_2 = \frac{I_0}{2} \times 0.25 = \frac{I_0}{8}$

$I_2 = 0.125I_0$

Step 1: After P₁ (unpolarized light):

$I_1 = \frac{I_0}{2}$

Step 2: Between P₁ (0°) and P₂ (30°): θ = 30°

$I_2 = I_1\cos^2(30^\circ) = \frac{I_0}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{I_0}{2} \times \frac{3}{4}$

$I_2 = \frac{3I_0}{8}$

Step 3: Between P₂ (30°) and P₃ (90°): θ = 60°

$I_3 = I_2\cos^2(60^\circ) = \frac{3I_0}{8} \times (0.5)^2 = \frac{3I_0}{8} \times \frac{1}{4}$

$I_3 = \frac{3I_0}{32} = 0.09375I_0$

Step 1: After P₁: $I_1 = \frac{I_0}{2}$

Step 2: Between P₁ and P₂: θ₁ = θ

$I_2 = \frac{I_0}{2}\cos^2\theta$

Step 3: Between P₂ and P₃: θ₂ = 90° - θ

$I_3 = I_2\cos^2(90^\circ - \theta) = \frac{I_0}{2}\cos^2\theta\cos^2(90^\circ - \theta)$

Step 4: Use identity: $\cos(90^\circ - \theta) = \sin\theta$

$I_3 = \frac{I_0}{2}\cos^2\theta\sin^2\theta = \frac{I_0}{8}\sin^2 2\theta$

Step 5: Maximum when $\sin^2 2\theta = 1$ ⇒ $2\theta = 90^\circ$ ⇒ $\theta = 45^\circ$

Maximum $I_3 = \frac{I_0}{8}$ at θ = 45°