Question
3. Aray of light cross from air into water and then into a diamond submerged in the water. The angle of incidence at the air-water boundary is 30.0^circ Water has an index of refraction of 133 and diamond an index of refraction of 2.42. Determine the angle of refraction inside the diamond. Include a diagram. (5)
Solution
4.1
(266 Votes)
Seymour
Elite · Tutor for 8 years
Answer
Here's the solution and diagram:**Diagram:**``` air (n1 = 1.00) / / θ1 = 30° /------------ / \ / water (n2 = 1.33) θ2/--------------------\ / θ3 \ diamond (n3 = 2.42) / \------------------/```**Solution:**1. **Air to Water:** We'll use Snell's Law to find the angle of refraction (θ2) in the water: n1 * sin(θ1) = n2 * sin(θ2) Where: * n1 = refractive index of air (approximately 1.00) * θ1 = angle of incidence in air (30.0°) * n2 = refractive index of water (1.33) * θ2 = angle of refraction in water 1.00 * sin(30.0°) = 1.33 * sin(θ2) 0.5 = 1.33 * sin(θ2) sin(θ2) = 0.5 / 1.33 sin(θ2) ≈ 0.376 θ2 ≈ arcsin(0.376) θ2 ≈ 22.1°2. **Water to Diamond:** Now we'll use Snell's Law again to find the angle of refraction (θ3) in the diamond: n2 * sin(θ2) = n3 * sin(θ3) Where: * n2 = refractive index of water (1.33) * θ2 = angle of refraction in water (calculated above, ≈ 22.1°) * n3 = refractive index of diamond (2.42) * θ3 = angle of refraction in diamond 1.33 * sin(22.1°) = 2.42 * sin(θ3) 1.33 * 0.376 ≈ 2.42 * sin(θ3) 0.500 ≈ 2.42 * sin(θ3) sin(θ3) ≈ 0.500 / 2.42 sin(θ3) ≈ 0.207 θ3 ≈ arcsin(0.207) θ3 ≈ 11.9°**Answer:** The angle of refraction inside the diamond is approximately 11.9°.