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**Worked Example:** *A student wants to find the Brewster angle when light travels from air (n1 = 1.0) into water (n2 = 1.33).* **Step 1: Write down the Brewster angle formula** • tan(θB) = n2 / n1 **Step 2: Plug in the refractive indices** • tan(θB) = 1.33 / 1.0 = 1.33 **Step 3: Solve for θB** • θB = arctan(1.33) Using a calculator, arctan(1.33) ≈ 53.1°. **Result:** The Brewster angle for air to water is about 53°. If light hits the water surface at this angle, the reflected light will be completely polarized perpendicular to the surface. **Extra check:** Using Snell’s law (n1 sinθi = n2 sinθt) you can find the transmitted angle: sinθt = (n1/n2) sinθi = (1/1.33) sin53.1° ≈ 0.63, so θt ≈ 39.2°. The sum of θB and θt is about 92.3°, close to 90° as expected for Brewster’s condition.