1. Image Formed by a Flat Mirror Consider rays from an object, O, a distance p in front of the mirror. They reflect, with angle of reflection = angle of incidence. To an observer, the look like they came from a distance q behind the mirror. I is called the image of O. Note that q = p.

2. s Proof that q = p  =  ’ (Law of reflection))  ’ =  ’’ (interior angles) Therefore tan  = tan  ’’. p/s = q/s Therefore p = q. Note that this is true for all rays from the object (point source) that hit the mirror, i.e. all values of s and .  All reflected rays seem to come from the same point: unique image at q. ’’  ’’  s Because rays do not actually pass through the image (i.e. cannot put a screen there to observe the image), it is called a “virtual image”.

3. This is true even when the point object is not directly in front of the mirror Point source image The image is the same distance behind the plane of the mirror as the source is in front of it, directly behind the source (i.e. the line between the source and image is perpendicular to the plane of the mirror). Plane of mirror p q

4. Magnification  image height/object height: M = h’ / h = +1 + sign: image is upright

5. Extensive object Where is the image and what does it look like?

6. Extensive object A B C D Choose some “extreme points” outlining the object.

7. Extensive object A B C D A B C D Locate the point image of these extreme points.

8. Extensive object A B C D A B C D Image (note that it has the same size as the object).

9. Extensive object A B C D A B C D Note that the mirror image switches the front and back (z) (not top/bottom or sides (x,y)) z x y

10. Problem: Determine the minimum height of a vertical flat mirror in which a person 178 cm (5’10”) can see his whole image.

11. Light from top of head travels path 5-4-3 to eyes. Light from feet travels path 1-2-3 to eyes. Therefore, the minimum length of the mirror = L = a+b. But the height of the man H = 2b + 2a. Therefore, L = H/2 = 178 cm/2 = 89 cm. Note the answer does not depend on how far the man is standing from the mirror.

13. Convex Spherical Mirror

18. Mirrors Object distance = p, image distance = q, radius of curvature = R, focal length = f 1/p + 1/q = 2/R = 1/f If p,q,R,f in “front” of mirror, they are positive (e.g. q > 0  real image). If p,q,R,f in “back” of mirror, they are negative (e.g. q < 0  virtual image) R, f > 0 for concave mirror R,f < 0 for convex mirror R,f =  for plane mirror.

19. Same rules approximately apply to an extended object, but only for “paraxial rays”: those close to and at small angle with “principal axis” – the line between the center of the mirror (V) and the center of curvature (C). 1/p + 1/q = 1/f, and magnification M  h’/h = -q/p (proven in text from above diagram for concave mirror) Note that the real image (q and p positive) is inverted (M <0). All paraxial rays leaving the tip of the object (h) will (approximately) cross at the tip of the image (h’). (The smaller , the better the approximation.)

20. This is also true if p<f so concave mirror forms a virtual image behind the mirror: 1/q = 1/f – 1/ p < 0 M = h’/h = -q/p > 0: virtual image is upright (not inverted).