2. Light Ray, Light Ray A Cruise Through the Wonderful World of Reflection and Refraction (aka Geometric optics) No Light, No Sight

3. What is Light? Electromagnetic Radiation with wavelength (visible) 400 – 700 nm (nanometers) Travels in straight lines Bounces (reflects) off certain materials Refracts (bends) in transparent materials Travels at c = 3 x 10 8 m/sec in vacuum

4. Index of Refraction n = c/v Light slows down in transparent materials other than vacuum(like car slows in sand) v = c/n velocity of light in medium n is called index of refraction n = about 1.5 for glass n = 1.33 for water

5. Reflection Light bounces off objects Consider “rays” – light moving in straight line Law of reflection: angle of incidence = angle of reflection

6. Law of Reflection

7. Law of Reflection  i =  r I - incident ray R - reflected ray N – normal Theta-I is angle of incidence Theta –R is angle of reflection

8. Types of Reflection

9. How Plane Mirror Forms Image

10. Real vs. Virtual Image Real image – light is present at the image position Virtual image – no light is present at the image position

11. Image From Plane Mirror Virtual Behind mirror Right side up(upright) Left-right reversed Located equal Distance behind mirror Same size as object

12. Distance Relationship Image is as far behind mirror as object is in front

13. How Curved Mirror Forms Real Images

14. How Law of Reflection Leads to Images from Spherical Mirror C is center of curvature F is focal point Real image will be located on same side of mirror as object C = 2f

15. Rules of Reflection – Two “Special” Rays An incident ray parallel to principal axis will pass through the focal point after reflection. An incident ray passing through the focal point will leave mirror parallel to principal axis

16. What’s So “Special?” About Special Rays? Its easy to predict where they will go Use Law of Reflection

17. Ray Diagram – Object Beyond C = 2f Real image Inverted Smaller than object Note use of two “special” rays What is another?

18. Third Special Ray – to P P

19. Ray Diagram – Object at C = 2f or between f and 2f Image inverted Located at 2f Same size as object Image inverted Further than 2f Larger than object

20. Object at 2f

21. Object Between f and 2f

22. Object Closer to Mirror Than f “Shaving Mirror” Case Image virtual Behind mirror Upright Larger than object

23. Object Closer to Mirror Than F

24. Object at f Must use different ray(goes to intersection of principal axis and mirror) No image formed

25. Mirror Equation f is positive for concave mirror Question: Where will image formed by a lens with 20 cm focal length be if object is placed 60 cm from mirror? d i = 30 cm Derivation diagram

26. How Big is The Image? M is magnification - sign means image is inverted Question: If the object in the question on previous slide is 2 cm high, how high is image? 1 cm

27. Convex (Diverging) Mirrors f is negative Light rays spread, not focused; images virtual 7-11 mirror foils thieves

28. Ray Diagram for Convex Mirror One ray parallel to p.a. Second ray heads for focus behind mirror Diverging rays must be extended behind mirror (dotted lines) Image virtual, upright, smaller

29. Ray Diagram for Convex Mirror One ray parallel to p.a. Second ray heads for focus behind mirror Diverging rays must be extended behind mirror (dotted lines) Image virtual, upright, smaller

30. Why Are Some Rearview Mirrors Convex?

31. Problem A 1.6 m tall thief stands 10m away from a 7-11 mirror with 2m (negative) focal length. Where is the image and how tall is it? Use Note: negative d i means image behind mirror d i = -1.67 m h i = 0.267m

32. solution 1/d i = -5/10 – 1/10 = -6/10 d i = -10/6 = -5/3 = -1.67 m h i /h o = -d i /d 0 = 1.67/10 h i =.167 x 1.6 = 0.267m

33. Index of Refraction Ratio of speed of light in vacuum to speed of light in material n = c/v = 3.0 x 10 8 m/s /v n always greater than one

34. Refraction: How Much Does It Bend Angle of incidence  i Angle of refraction  r Snell’s Law: n i sin  i = n r sin  r

35. Helpful Analogy: The Band of Sand What happens when a car drives into the sand? highway sand Which way does the car turn?

36. Toward and Away from Normal When light enters a more dense (greater n) medium, it bends toward the normal When light enters a less dense (smaller n) medium, it bends away from the normal

37. Mysteries? Why do a person’s legs appear shorter when they are standing in water? Why Does A Glass Rod Disappear in Mineral Oil?

38. Mysteries? Why do a person’s legs appear shorter when they are standing in water? Why Does A Glass Rod Disappear in Mineral Oil?

39. Mysteries? Why do a person’s legs appear shorter when they are standing in water? Why Does A Glass Rod Disappear in Mineral Oil? Both have same index of refraction

40. Practical Application Fluorocarbon (semi invisible under water) fishing line

41. Problem Light strikes a flat piece of glass(n = 1.5) at 60 degrees to the normal. What is the angle of the light in the glass?  i = 60 o rr

42. Solution n i sin  i = n r sin  r n i = 1 n r = 1.5 sin  r = 1.00x sin  i /1.5= 0.577  r = 35.2 o

43. Total Internal Reflection

44. Total Internal Reflection Light ray leaves more dense medium Angle of refraction approaches 90 0 Past critical angle there is no refracted ray

45. Critical Angle The angle of incidence past which there is no refracted ray n i sin  i = n r sin  r sin  c = n r /n i sin90 0 = n r /n i If ray emerges into air sin  c = 1/n i cc

46. Example What is the critical angle for light rays leaving a swimming pool? What does the world look like to a swimmer at the bottom of the pool? Sin  c = 1.00/1.33 = 0.750  c = 49 0 Swimmer sees outside world compressed into a circle whose edge makes a 49 degree angle to the vertical

47. Applications of TIR Prisms in Binoculars

48. Applications of TIR Prisms in Binoculars

49. Fiber Optics Fiber Optic Amplifier Module Spy under door optics endoscope Photos courtesy JDS Uniphase Inc.

50. Lenses Lenses can focus or diverge light Act like tiny prisms

51. Thin Lens Convex lenses have two focal points

52. Focal Length of Lens That distance, f, from lens at which parallel rays are brought to a point Power of a lens: P = 1/f Unit is “diopter” (m -1) Used by optometrists Focal point

53. Three Special, Predictable Rays Avoid application of Snell’s Law Parallel to principal axis, exiting rays goes through far focus Through near side focus; exiting ray parallel Through center of lens(no refraction)

54. Object Distance Greater than 2f Image real Inverted Smaller Opposite side from object Like camera

55. Object at 2f or between f and 2f Image same size At 2f Image larger Greater than 2f Like projector

56. Object Inside of f – Magnifying Glass Image virtual Upright Same side as object, behind object larger

57. Diverging Lens – Does NOT focus Image virtual Upright Smaller Same side as object Closer than f

58. Thin Lens Equation F positive for converging lens, negative for diverging lens d i positive for real images, negative for virtual images(same side)

59. Magnification Relates height of images to image and object distances

60. Solving Lens Problems Read and re-read problem Draw ray diagram with 2 or 3 special rays Solve lens and magnification equations for unknowns Follow sign conventions including –f positive for converging lenses –f negative for diverging lenses –d i negative on same side of lens as object

61. Lens Problem A one cm high fly is 30 cm from a lens with a focal length of 20cm. Where is the (real) image of the fly located and how tall is it? 1/d i = 1/f – 1/d 0 = 1/20 – 1/30 = 3/60 – 2/60 = 1/60 d i = 60 cm on other side of lens from fly h i = -h 0 d i /d 0 = -2cm - sign means image inverted

62. Object Inside of f – Magnifying Glass Image virtual Upright Same side as object, behind object larger

63. Problem An object is placed 10 cm from a 15cm focal length converging lens. Where is the image? 1/d i = 1/f – 1/d 0 = 1/15 – 1/10 = -1/30 d i = -30 cm - sign means image is virtual and on same side of lens as object

64. Object at f No image formed

65. Diverging Lens Problems Remember, f is negative d i will also be negative

66. Problem Where must a small insect be placed if a 25 cm focal length diverging lens is to form a virtual image 20 cm in front of the lens? 1/d 0 = 1/f – 1/d i = -1/25 + 1/20 = (-4+5)/100 = 1/100 d 0 = 100 cm in front of lens

67. Combinations of Lenses and Lenses with Mirrors The image produced by the first element becomes the object of the second Two converging lenses with focal lengths 20.0 and 25.0 cm respectively are placed 80.0 cm apart. An object is placed 60 cm in front of the first lens. Where is the image? (text example 23-10)

68. Lenses and Mirrors Sketch a ray diagram of a set up in which a mirror and a lens work together Find out how an astronomical telescope works –Refracting –Reflecting Find out how a simple compound (two lens) microscope works

69. Acknowledgements Graphics and animation courtesy of Tom Henderson, Glenbrook South High School, Illinois Photos of fiber optic components courtesy of JDS Uniphase, Inc