A light ray in air hits a water surface at 50° to the normal. (a) Find the angle of refraction in water (n = 1.33). (b) A coin lies 3.0 m deep in the pool; how deep does it appear from directly above? (c) What is the speed of light in water? Step 1: Snell's law: sinθ₂ = (n₁/n₂)sinθ₁ = (1.00/1.33)sin50° = 0.7519 × 0.7660 = 0.5760 → θ₂ ≈ 35.2°. The ray bends toward the normal (35.2° < 50°) as expected entering denser medium. Step 2: Apparent depth h′ = h/n = 3.0/1.33 ≈ 2.26 m — the coin looks nearly 0.75 m shallower; aim low! Step 3: v = c/n = 3.00×10⁸/1.33 ≈ 2.26×10⁸ m/s. The slowing from 3.00 to 2.26×10⁸ m/s is what bends the ray — refraction is just the wave adjusting to the new speed.