1. Reflection of Light

2. When light rays hit an object, they change direction. The type of surface the light encounters determines the type of reflection. Reflection can be: 1.Specular (Ex. Mirror) 2.Diffuse (ex. Rough surfaces like concrete)

3. Specular reflection on a plane(flat) mirror The light that hits the mirror is the incident ray. The light that is reflected off of the mirror is the reflected ray. Both the incident ray and the reflected ray are at an equal angle from the normal to the mirror. θ i = θ r

4. Reflection on a spherical mirror There are two main types of curved mirrors: 1.Concave mirrors. The reflective surface is on the inner part of the sphere. 2.Convex mirrors. The reflective surface is on the outer surface of the sphere. The characteristics of the curved mirrors are determined by their principal points.

5. The center of curvature, C, is the center of the original sphere from which the curved mirror originates. The focal point, F, is the point at which all the light rays parallel to the principal axis, P, converge. The vertex, V, is the center of the curved mirror.

6. We use the three principal points to determine the focal length, f, and the radius of curvature, R, for the spherical mirror. The principal points are also used to define the three principal rays.

7. CONCAVE Mirror: First principal ray: Incident ray is parallel to the principal axis and is reflected through the focal point. Second principal ray: Incident ray passes through the focal point and is reflected parallel to the principal axis. Third principal ray: Incident ray passes through the center of curvature and is reflected back on itself.

8. CONVEX Mirror: First principal ray: Incident ray is parallel to the principal axis and is reflected in a direction originating from the focal point. Second principal ray: Incident ray directed towards the focal point and is reflected parallel to the principal axis. Third principal ray: Incident ray directed towards the center of curvature and is reflected back on itself.

9. Light rays reflect off of objects and produce images. The image can either be real or virtual. Real images are created when light rays cross each other after bouncing off of the object. Virtual images are created when light rays only appear to cross (like in a mirror), but in reality diverge from one another.

10. Images formed on plane mirrors The images formed by plane mirrors are virtual. These images have the following properties: 1.The image is as far behind the mirror as the object is in front. 2.The image is unmagnified and upright. 3.The image has a left-right reversal. object image hoho hihi

11. The magnification of a mirror is found by comparing the height of the image to the height of the object. M = image height = h i object height h o For plane mirrors, M= 1

12. Images formed by spherical mirrors 1.Concave mirrors. We can use the properties of the principal rays to find the image. The image produced can be described as: a.Type : Real or virtual b.Orientation: Inverted or upright c.Magnification : enlarged or reduced or same d.Position within the three points relative to the principal axis.

13. To find the image of an object reflected in a concave mirror, we must use the properties of the three principal rays. The image will form where the rays cross.

14. Activity On your stencil, determine the characteristics of the image formed for each of the concave mirrors. Create a table in your notes summarizing the main points.

15. Image characteristics for Concave mirrors Object position Image type Image orientation Image magnification Image position Object at C Object at F Object b/w C and F Object b/w F and V

16. Label the distance between the object and the mirror as d o. Measure d o. Label the distance between the image and the mirror as d i. Measure d i. Label the focal length, f. Measure f.

17. The focal length, f, distance from the object to the mirror, d o, and the distance from the mirror to the object, d i are related to on another. Mirror equation: 1/f = 1/ d o + 1/ d i

18. Also, the magnification equation for spherical mirrors is: M = h i / h o = - d i / d o * The negative sign is included since virtual distances are negative by convention as is the height if it is inverted.

19. Example 1 An object with a height of 5 cm is located 20 cm from a converging mirror whose radius of curvature, R, is 20 cm. Find: a.The image position, d i b.The magnification, M c.The image height

20. a. 1/f = (1/d i + 1/d o ) 1/ 10 cm = (1/d i + 1/20 cm) 0.1 = 1/d i + 0.05 0.1 -0.05 = 1/d i + 0.05 – 0.05 0.05 = 1/d i d i = 20 cm

21. b.M = h i / h o = - d i / d o M = - 20 cm/20 cm = -1 Since this results in a negative number, we understand that the image produced is inverted. c. M = h i / h o -1 = h i / 5 cm h i = - 5 cm, the negative height value also tells us that the image is inverted.

22. Example 2 An object with a height of 2 cm is located 5cm from a converging mirror whose radius of curvature, R, is 8 cm. Find: a.The image position, d i b.The magnification, M c.The image height

24. 2. Images formed by Convex Mirrors All images formed by convex mirrors will have the following characteristics: 1.Type: virtual 2.Orientation: upright 3.Size: smaller than object 4.Position: between F and V, d i < d o

25. Draw the image produced by the convex mirror:

26. Both the mirror equation and the magnification ratio apply to convex mirrors. Example 3: An object with a height of 5 cm is located 32 cm from a diverging mirror with a radius of curvature, R, of 40 cm. Determine: a.The image position, d i b.The magnification, M c.The image height, h i

27. a.1/f = (1/ d i + 1/ d o ) 1/(-20 cm) = (1/d i + 1/(32 cm)) -0.05 = (1/d i + 0.03125) -0.08125 = 1/d i d i = -12.3 cm * The negative value tells us that the image is located behind the mirror and is virtual.

28. b. M = - di / do M = - (-12.3 cm/ 32 cm) = + 0.384 *The positive value for magnification shows us that the image produced is upright. c.M = h i /h o 0.384 = h i / 5 cm, h i = 1.92 cm

29. Example 4 An object with a height of 8 cm is located 20 cm from a diverging mirror with a radius of curvature, R, of 40 cm. Determine: a.The image position, d i b.The magnification, M c.The image height, h i

31. HW p. 70 # 8 -12