1. Chapter 14

3.  The brain judges the object location to be the location from which the image light rays originate.

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Wave Model of Refraction Section 1 Refraction

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Refraction Chapter 14 The Law of Refraction The index of refraction for a substance is the ratio of the speed of light in a vacuum to the speed of light in that substance.

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Indices of Refraction for Various Substances Section 1 Refraction

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Refraction Chapter 14 The Law of Refraction, continued When light passes from a medium with a smaller index of refraction to one with a larger index of refraction (like from air to glass), the ray bends toward the normal. When light passes from a medium with a larger index of refraction to one with a smaller index of refraction (like from glass to air), the ray bends away from the normal.

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Refraction Section 1 Refraction

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Refraction Chapter 14 The Law of Refraction Objects appear to be in different positions due to refraction. Snell’s Law determines the angle of refraction.

16. 14.1 (Holt)

17. n = cvcv v = cncn 3.00 x 10 8 m/s 1.33 = 2.25 x 10 8 m/s

19. Snell’s Law: n i sin  i = n r sin  r 1.00 x sin20 o = 1.52 x sin  sin   = = 0.2250 sin20 o 1.52 sin -1   =  = 13.0 o a.

20. Snell’s Law: n i sin  i = n r sin  r  =  = 13.0 o  = 13.0 o 1.00 x sin20 o = 1.52 x sin  1.52 x sin13 o = 1.00 x sin   = 20.0 o

21. Snell’s Law: n i sin  i = n r sin  r 1.00 x sin20 = 1.33 x sin  1.00 x sin 40 o = 1.52 x sin < r < r = 25 o <‘ i = 25 o <‘ r = 40 o 1.00 x sin 60 o = 1.52 x sin < r < r = 35 o <‘ i = 35 o <‘ r = 60 o

22. Snell’s Law: n i sin  i = n r sin  r 1.00 x sin20 = 1.33 x sin  1.00 x sin 80 o = 1.52 x sin < r < r = 40 o <‘ i = 40 o <‘ r = 80 o

24.  A lens is a transparent object that refracts light rays such that they converge or diverge to create an image.  A lens that is thicker in the middle than it is at the rim is an example of a converging lens.  A lens that is thinner in the middle than at the rim is an example of a diverging lens.  The focal point is the location where the image of an object at an infinite distance from a converging lens if focused.  Lenses have a focal point on each side of the lens.  The distance from the focal point to the center of the lens is called the focal length, f.

25. Converging Lenses Diverging Lenses

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Section 2 Thin Lenses Lenses and Focal Length CONVERGING LENSDIVERGING LENS

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Converging and Diverging Lenses Section 2 Thin Lenses

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Focal Length for Converging and Diverging Lenses Section 2 Thin Lenses

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Types of Lenses Ray diagrams of thin-lens systems help identify image height and location. Rules for drawing reference rays

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Characteristics of Lenses Converging lenses can produce real or virtual images of real objects. The image produced by a converging lens is real and inverted when the object is outside the focal point. The image produced by a converging lens is virtual and upright when the object is inside the focal point.

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Ray Tracing for a Converging Lens Section 2 Thin Lenses

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Characteristics of Lenses, continued Diverging lenses produce virtual images from real objects. The image created by a diverging lens is always a virtual, smaller image.

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 14 Ray Tracing for a Diverging Lens Section 2 Thin Lenses

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 The Thin-Lens Equation and Magnification The equation that relates object and image distances for a lens is call the thin-lens equation. It is derived using the assumption that the lens is very thin.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 The Thin-Lens Equation and Magnification, continued Magnification of a lens depends on object and image distances.

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 The Thin-Lens Equation and Magnification, continued If close attention is given to the sign conventions defined in the table, then the magnification will describe the image’s size and orientation.

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem Lenses An object is placed 30.0 cm in front of a converging lens and then 12.5 cm in front of a diverging lens. Both lenses have a focal length of 10.0 cm. For both cases, find the image distance and the magnification. Describe the images.

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 1. Define Given:f converging = 10.0 cmf diverging = –10.0 cm p converging = 30.0 cmp diverging = 12.5 cm Unknown:q converging = ? q diverging = ? M converging = ?M diverging = ?

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 1. Define, continued Diagrams:

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 2. Plan Choose an equation or situation: The thin-lens equation can be used to find the image distance, and the equation for magnification will serve to describe the size and orientation of the image.

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 2. Plan, continued Rearrange the equation to isolate the unknown:

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 3. Calculate For the converging lens:

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 3. Calculate, continued For the diverging lens:

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Thin Lenses Chapter 14 Sample Problem, continued Lenses 4. Evaluate These values and signs for the converging lens indicate a real, inverted, smaller image. This is expected because the object distance is longer than twice the focal length of the converging lens. The values and signs for the diverging lens indicate a virtual, upright, smaller image formed inside the focal point. This is the only kind of image diverging lenses form.

46. Section 2 Thin Lenses (clearly sees objects near) (clearly sees objects far away)

47.  Total internal reflection can occur when light moves along a path from a medium with a higher index of refraction to one with a lower index of refraction.  At the critical angle, refracted light makes an angle of 90º with the normal.  Above the critical angle, total internal reflection occurs and light is completely reflected within a substance.

48.  Snell’s law can be used to find the critical angle. Total internal reflection occurs only if the index of refraction of the first medium is greater than the index of refraction of the second medium.