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)

Here's the solution and diagram:<br /><br />**Diagram:**<br /><br />```<br /> air (n1 = 1.00)<br /> /<br /> / θ1 = 30°<br /> /------------<br /> / \<br /> / water (n2 = 1.33) θ2<br />/--------------------<br />\ / θ3<br /> \ diamond (n3 = 2.42) /<br /> \------------------/<br />```<br /><br />**Solution:**<br /><br />1. **Air to Water:** We'll use Snell's Law to find the angle of refraction (θ2) in the water:<br /><br /> n1 * sin(θ1) = n2 * sin(θ2)<br /><br /> Where:<br /> * n1 = refractive index of air (approximately 1.00)<br /> * θ1 = angle of incidence in air (30.0°)<br /> * n2 = refractive index of water (1.33)<br /> * θ2 = angle of refraction in water<br /><br /> 1.00 * sin(30.0°) = 1.33 * sin(θ2)<br /> 0.5 = 1.33 * sin(θ2)<br /> sin(θ2) = 0.5 / 1.33<br /> sin(θ2) ≈ 0.376<br /> θ2 ≈ arcsin(0.376)<br /> θ2 ≈ 22.1°<br /><br />2. **Water to Diamond:** Now we'll use Snell's Law again to find the angle of refraction (θ3) in the diamond:<br /><br /> n2 * sin(θ2) = n3 * sin(θ3)<br /><br /> Where:<br /> * n2 = refractive index of water (1.33)<br /> * θ2 = angle of refraction in water (calculated above, ≈ 22.1°)<br /> * n3 = refractive index of diamond (2.42)<br /> * θ3 = angle of refraction in diamond<br /><br /> 1.33 * sin(22.1°) = 2.42 * sin(θ3)<br /> 1.33 * 0.376 ≈ 2.42 * sin(θ3)<br /> 0.500 ≈ 2.42 * sin(θ3)<br /> sin(θ3) ≈ 0.500 / 2.42<br /> sin(θ3) ≈ 0.207<br /> θ3 ≈ arcsin(0.207)<br /> θ3 ≈ 11.9°<br /><br />**Answer:** The angle of refraction inside the diamond is approximately 11.9°.<br />