21. Understanding Refraction Wheels on axle rolls along a smooth sidewalk and onto grass. Which picture path is followed? What happens if the motion is reversed?
23. Illustrating Cart Analogy Right front wheel slows down before left front Left front wheel slows down before right front
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27. Law of Refraction: Snell's Law Right front wheel slows down first. Snell's Law: n2 sin Q2 = n1 sin Q1
28. Snell's Law Example n 1 = 1.0 (air) n 2 = 1.52 (glass) Q 1 = 30 degrees ------------------------ n 2 sin Q 2 = n 1 sin Q 1 1.52 sin Q 2 = 1.0 sin 30 sin Q 2 = 0.33 Put calculator in Mode Degree Q 2 = sin -1 (0.33) = 19.3 degrees
31. Apparent Depth in Water Light exits into medium (air) of lower index of refraction, and turns left.
32. More Apparent Depth Spear-fishing is made more difficult by the bending of light. To spear the fish in the figure, one must aim at a spot in front of the fish
33. Refraction at Sunset Why does the sun appear to be flattened at sunset? --------------------------------------------------- The sun actually falls below below the horizon, i.e., it "sets", a few seconds before we see it set.
37. Displacement through a Slab of Glass Entering and exiting rays are displaced from each other, but parallel.
38. Internal Reflection All rays reflect internally, but the top three rays reflect only a small percentage internally; most energy leaves the prism. The fourth and fifth rays are reflected 100 % internally.
44. Critical Angle Calculation What must be Q 1 to get Q 2 = 90 deg ? Snell's Law: n 1 sin Q 1 = n 2 sin Q 2 = n 2 sin 90 sin Q 1 = n 2 / n 1 ------------------------------ Assume water to air: n 1 = 1.33 n 2 = 1.00 q 1 = sin -1 (0.752) = 48.8 degrees Q c = critical angle = 48.8 degrees
45. Cone of Light Crictical angle for water = 48.8 degrees Light within the 48.8 degree cone is detected by fish, while nothing in the air outside that cone can be seen. The only light reaching the fish outside the cone is that light (not shown) which is reflected off the bottom of the pool.
46. Critical Angle of Diamond n = 2.419 Q c = sin -1 (1.00/2.419) = 24.42 degrees 90.00 - 24.42 = 65.58 degrees Light outside of 65.58 degree cone is reflected back inside. Virtually all light entering the top face of the diamond is reflected internally.
When a light beam travels from a medium of a higher index of refraction to a medium of lower index of refraction, the refracted beam bends outwards, away from the normal. You can then imagine that at some point, you are going to hit where the refracted beam exits at 90 degrees. It is at this point where you hit the critical angle.
![Law of Reflection ,[object Object]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)