2. Optics The Study of Light

3. Areas of Optics Geometric Optics Light as a ray. Physical Optics Light as a wave. Quantum Optics Light as a particle.

4. Reflection Reflection occurs when light bounces off a surface. There are two types of reflection Specular reflection Off a shiny surface Diffuse reflection Off a rough surface

5. Mirrors are great reflectors Plane Mirror shiny + dark - shiny + + dark - - Spherical Mirrors convex concave

6. Light Rays Mathematical rays never bend But light rays can, if they interact with materials!

7. Let’s take a closer look at a plane mirror Plane Mirror +- Incident ray Reflected ray normal A normal is a line that is perpendicular to the mirror.

8. Ray tracing Ray tracing is a method of constructing an image using the model of light as a ray. We use ray tracing to construct optical images produced by mirrors and lenses. Ray tracing lets us describe what happens to the light as it interacts with a medium.

9. Law of Reflection Lab a) Hold a plane mirror upright on a sheet of graph paper that is on top of cardboard. Stick three pins through the graph paper on the same side of the mirror such that they appear to your eye to be in a straight line. (See the whiteboard for details) b) Draw a normal to the surface of the mirror, as well as the incident and reflected rays. c) Measure the angles of incidence and reflection. d) Repeat two or three times with different angles. e) Tabulate your angles of incidence and reflection. What can you say about the angles of incidence and reflection? TURN IN ONE DRAWING PER GROUP. Include each person’s name and period number.

10. Law of Reflection The angle of incidence of reflected light equals the angle of reflection.  r =  I Note that angles are measured relative to a normal to the mirror surface. shiny (+)dark (-) plane mirrorlight source incident ray normal reflected ray rr ii

11. Sample Problem A ray of light reflects from a plane mirror with an angle of incidence of 37 o. If the mirror is rotated by an angle of 5 o, through what angle is the reflected ray rotated?

12. Solution  i = 37 0  r = 37 0 5050 5050  i = 42 0  r = 42 0 42 o + 5 o = 47 o relative to horizontal 47 o - 37 o = 10 o rotation of reflected ray

13. Sample Problem Standing 2.0 m in front of a small vertical mirror, you see the reflection of your belt buckle, which is 0.70 m below your eyes What is the vertical location of the mirror relative to the level of your eyes? If you move backward until you are 6.0 m from the mirror, will you still see the buckle, or will you see a point on your body that is above or below the buckle?

14. Solution buckle mirror eye The mirror must be 0.35 m below the level of your eye. If you move backward, you’ll still see the belt buckle, since the triangles will still be similar. 0.70 m 0.35 m 2.0 m rr ii

15. Optical images Nature real (converging rays) virtual (diverging rays) Orientation upright inverted Size true enlarged reduced

16. Ray tracing: plane mirror Construct the image using two rays. +- object 5 cm Image -5 cm Reflected rays are diverging. Extend reflected rays behind mirror. Name the image: Virtual, upright, true size

17. Experiment Put two plane mirrors at right angles to each other as shown. If a pin is placed on the shiny side of each, how many images are formed?

18. Problem Two plane mirrors meet at right angles as shown. What are the locations of all the images formed by the two mirrors? HINT: Images can form images too!

19. Solution The first two images are obvious. The third image is formed as a reflection of an image rather than an object.

20. Spherical mirrors There are two types of spherical mirrors shiny concaveconvex + +-- (where reflected rays go) (dark side) Focal length, f, is positive Focal length, f, is negative

21. Parts of a Spherical Concave Mirror Principle axis + - These are the main parts of a spherical concave mirror. The focal length is half of the radius of curvature. The focal length is positive for this type of mirror. R = 2f Focus f Center R

22. Demonstrations Blackboard Optics with Plane and Spherical Mirrors. Law of Reflection Identification of Focus What do you notice about Spherical Concave as opposed to Spherical Convex mirrors?

23. Identification of the focus of a spherical concave mirror + - Rays parallel to the principle axis all pass through the focus for a spherical concave mirror.

24. Ray tracing: spherical concave mirror The three “principle rays” to construct an image for a spherical concave mirror are the p-ray, which travels parallel to the principle axis, then reflects through focus. the f-ray, which travels through focus, then reflects back parallel to the principle axis. the c-ray, which travels through center, then reflects back through center. You must draw two of the three principle rays to construct an image.

25. Ray tracing: spherical concave mirror Construct the image for an object located outside the center of curvature. It is only necessary to draw 2 of the three principle rays! CF Real, Inverted, Reduced Image p f c

26. C F Real, Inverted, True Image Ray tracing: spherical concave mirror Construct the image for an object located at the center of curvature. Name the image.

27. CF Real, Inverted, Enlarged Image Ray tracing: spherical concave mirror Construct the image for an object located between the center of curvature and the focus. Name the image.

28. C F No image is formed. Construct the image for an object located at the focus. Ray tracing: spherical concave mirror

29. CF Virtual, Upright, Enlarged Image Construct the image for an object located inside the focus. Name the image. Ray tracing: spherical concave mirror

30. Problem a) Construct 2 ray diagrams to illustrate what happens to the size of the image as an object is brought nearer to a spherical concave mirror when the object outside the focus. b) Repeat part a) for an object which is brought nearer to the mirror but is inside the focus.

31. Solution a) The image becomes larger when you move the object closer.

32. Solution b) The image becomes smaller when you move the object closer.

33. Mirror equation #1 1/s i + 1/s o = 1/f s i : image distance s o : object distance f: focal length

34. Mirror equation # 2 M = h i /h o = -s i /s o s i : image distance s o : object distancem h i : image height h o : object height M: magnification

35. Sample Problem A spherical concave mirror, focal length 20 cm, has a 5-cm high object placed 30 cm from it. a)Draw a ray diagram and construct the image. c)Name the image

36. Sample Problem A spherical concave mirror, focal length 20 cm, has a 5-cm high object placed 30 cm from it. b)Use the mirror equations to calculate i.the position of image ii.the magnification iii.the size of image

37. Demonstrations Real images and the concave mirror. Virtual images and the convex mirror.

38. Parts of a Spherical Convex Mirror These are the main parts of a spherical convex mirror. The focal length is half of the radius of curvature, and both are on the dark side of the mirror. The focal length is negative for this type of mirror. Principle axis CenterFocus + -

39. Ray tracing: spherical convex mirror Construct the image for an object located outside a spherical convex mirror. Name the image. FC Virtual, Upright, Reduced Image

40. Problem Construct 2 ray diagrams to illustrate what happens to the size of the image as an object is brought nearer to a spherical convex mirror.

41. Problem A spherical concave mirror, focal length 10 cm, has a 2-cm high object placed 5 cm from it. a)Draw a ray diagram and construct the image.

42. Problem A spherical concave mirror, focal length 10 cm, has a 2-cm high object placed 5 cm from it. b)Use the mirror equations to calculate i.the position of image ii.the magnification iii.the size of image c)Name the image

43. Problem A spherical convex mirror, focal length 15 cm, has a 4-cm high object placed 10 cm from it. a)Draw a ray diagram and construct the image.

44. Problem A spherical convex mirror, focal length 15 cm, has a 4-cm high object placed 10 cm from it. b)Use the mirror equations to calculate i.the position of image ii.the magnification iii.the size of image c)Name the image

45. Summary Concave vs convex mirrors Concave Image is real when object is outside focus Image is virtual when object is inside focus Focal length f is positive Convex Image is always virtual Focal length f is negative

46. Definition: Refraction Refraction is the movement of light from one medium into another medium. Refraction cause a change in speed of light as it moves from one medium to another. Refraction can cause bending of the light at the interface between media.

47. Index of Refraction speed of light in vacuum speed of light in medium n = c/v n =

48. When light slows down… …it bends. Let’s take a look at a simulation. simulation. URL: http://www.walter- fendt.de/ph14e/huygenspr.htm

49. Snell’s Law n 1 sin  1 = n 2 sin  2 n1n1 n2n2 11 angle of incidence 22 angle of refraction

50. n1 < n2 When the index of refraction increases, light bends toward the normal. n1n1 n2n2 11 22 When n 1 < n 2  1 >  2

51. n1 > n2 n1n1 n2n2 11 22 When the index of refraction decreases, light bends away from the normal. When n 1 > n 2  1 <  2

52. Problem Light enters an oil from the air at an angle of 50 o with the normal, and the refracted beam makes an angle of 33 o with the normal. a) Draw this situation. b) Calculate the index of refraction of the oil. c) Calculate the speed of light in the oil

53. Problem Light enters water from a layer of oil at an angle of 50 o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33. a) Draw this situation. b) Calculate the angle of refraction. c) Calculate the speed of light in the oil, and in the water

54. Refraction Lab Using pins and ray tracing, determine the index of refraction for the glass block or for the water. You must graph the data you collect such that the index of refraction appears as the slope in a graph. Take a look at Snell’s Law and see if you can figure out how to do this. You must turn in your full lab report next week, which must include your ray tracing, data table, calculations, and a graph for which n 2 is the slope.

55. θ2θ2 θ1θ1 θ1θ1 θ1θ1 θ2θ2 θ2θ2 n 1 sinθ 1 = n 2 sinθ 2 n 1 is a known value. (1.00 for air) θ 1 and θ 2 are measured values from your ray tracing. Do at least 3 trials at 3 different incident angles for the substance you select. Figure out how to graph your data so that n 2 is obtained by measuring the slope of the graph. (HINT: Remember the slope- intercept form of a linear equation?) Example glass block ray tracing. pin

56. Problem Light enters a prism as shown, and passes through the prism. a) Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b) If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. c) How would the answer to b) change if the prism were immersed in water? 30 o 60 o glass air

57. Problem Light enters a prism made of air from glass. a)Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b)If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. 30 o 60 o glass air

58. Dispersion The separation of white light into colors due to different refractive indices for different wavelengths is called dispersion. Dispersion is often called the prism effect.

59. Dispersion Which color of light has the greatest refractive index?

60. Critical Angle of Incidence The smallest angle of incidence for which light cannot leave a medium is called the critical angle of incidence. If light passes into a medium with a greater refractive index than the original medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence. If the angle of refraction is > 90 o, the light cannot leave the medium.

61. Critical Angle of Incidence This drawing reminds us that when light refracts from a medium with a larger n into one with a smaller n, it bends away from the normal. n1n1 n2n2 n 1 > n 2

62. Critical Angle of Incidence This shows light hitting a boundary at the critical angle of incidence, where the angle of refraction is 90 o. No refraction occurs! n1n1 n2n2 cc  r = 90 o n 1 > n 2 Ray reflects instead of refracting.

63. Critical Angle of Incidence Instead of refraction, total internal reflection occurs when the angle of incidence exceeds the critical angle. n1n1 n2n2 cc  r = 90 o n 1 > n 2 Ray reflects instead of refracting.

64. Calculating Critical Angle n 1 sin(  1 ) = n 2 sin(90 o ) n 1 sin(  c ) = n 2 sin(90 o ) sin(  c ) = n 2 / n 1  c = sin -1 (n 2 /n 1 )

65. Problem (separate sheet) A.What is the critical angle of incidence for a gemstone with refractive index 2.45 if it is in air? B.If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence?

66. Lenses There are two types of lenses. converging + +- - (where refracted rays go) Focal length, f, is positive Focal length, f, is negative (where refracted rays go) diverging Thicker in middle Thinner in middle

67. Lens ray tracing Ray tracing is also used for lenses. We use the same principle rays we used for mirrors. the p-ray, which travels parallel to the principle axis, then refracts through focus. the f-ray, which travels through focus, then refracts parallel to the principle axis. the c-ray, which travels through center and continues without bending. You must draw 2 of the 3 principle rays.

68. Lens Equations We use the same equations we used for mirrors 1/s i + 1/s o = 1/f M = h i /h o = -s i /s o

69. Identification of the focus All rays parallel to the principle axis refract through the focus of a converging lens. F

70. Ray tracing: converging lens Construct the image for an object located outside 2F. It is only necessary to draw 2 of the three principle rays! CF Real, Inverted, Reduced Image F2F + - p c f

71. Construct the image for an object located at 2F. CF Real, Inverted, True Image F2F + - Ray tracing: converging lens

72. Construct the image for an object located between F and 2F. CF Real, Inverted, Enlarged Image F2F + - Ray tracing: converging lens

73. Construct the image for an object located at the focus. CF No image F2F + - Ray tracing: converging lens

74. Construct the image for an object located inside the focus. CF Virtual, Upright, Enlarged Image F + - Ray tracing: converging lens

75. For converging lenses f is positive s o is positive s i is positive for real images and negative for virtual images M is negative for real images and positive for virtual images h i is negative for real images and positive for virtual images

76. Diverging lens Construct the image for an object located in front of a diverging lens. CF Virtual, Upright, Reduced Image F + -

77. For diverging lenses f is negative s o is positive s i is negative M is positive and < 1 h i is positive and < h o

78. Problem A converging lens, focal length 20 cm, has a 5-cm high object placed 30 cm from it. a)Draw a ray diagram and construct the image. b)Use the lens equations to calculate i.the position of image ii.the magnification iii.the size of image c)Name the image

79. Solution a) and c) CFF + - a) c) Real, Inverted, Enlarged

80. Solution b) i. 1/s i + 1/s o = 1/f 1/s i + 1/30 = 1/20 1/s i = 1/20 - 1/30 = 3/60 –2/60 = 1/60 s i = 60 cm ii. M = - s i /s o = -60/30 = -2 iii. M = h i / h o h i = M h o = (-2)5 = -10 cm

81. Problem A converging lens, focal length 10 cm, has a 2-cm high object placed 5 cm from it. a)Draw a ray diagram and construct the image. b)Use the lens equations to calculate i.the position of image ii.the magnification iii.the size of image c)Name the image

82. Solution a) and c) a) c) Virtual, Upright, Enlarged CFF + -

83. Solution b) i. 1/s i + 1/s o = 1/f 1/s i + 1/5 = 1/10 1/s i = 1/10 - 1/5 = 1/10 –2/10 = -1/10 s i = -10 cm ii. M = - s i /s o = -(-10)/5 = 2 iii. M = h i / h o h i = M h o = (2)2 = 4 cm

84. Problem A diverging lens, focal length -15 cm, has a 4-cm high object placed 10 cm from it. a) Draw a ray diagram and construct the image. b) Use the lens equations to calculate i.the position of image ii.the magnification iii.the size of image c) Name the image

85. Solution a) and c) a) c) Virtual, Upright, Reduced CFF + -

86. Solution b) i. 1/s i + 1/s o = 1/f 1/s i + 1/10 = 1/(-15) 1/s i = -1/10 - 1/15 = -3/30 – 2/30 = -5/30 = -1/6 s i = -6 cm ii. M = - s i /s o = -(-6)/10 = 0.6 iii. M = h i / h o h i = M h o = (0.6)(4) = 2.4 cm

87. Lens and Mirror lab Purpose: Determine the focal lengths of a converging lens and a converging mirror from the positions of an object and an image. Rules and hints: Your image must be that of a clear arrow. You must have at least 2 trials for the mirror and 2 trials for the lens. Images produced must be real and not virtual. Full lab report, which must include procedure, ray diagrams, tabulated data, calculations, etc., must be turned in Wednesday next week.