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.

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 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”.

This is true even when the point object is not directly in front of the mirror Point source p Plane of mirror q 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). image

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

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

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

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

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

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

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

Light from top of head travels path 5-4-3 to eyes 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.

Convex Spherical Mirror

1/p + 1/q = 2/R = 1/f If p = , 1/q = 1/f – 1/p = 1/f – 0 q = f As p decreases, 1/q = 1/f – 1/p, q increases. When p = f, 1/q = 0, q = . (i.e. reverse the arrows in the top picture).

If p = f, 1/q = 1/f – 1/p = 0 q =  If p < f, 1/q = 1/f – 1/ p < 0  q < 0: A virtual image is formed behind the mirror.

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. 1/p + 1/ q = 1/ = 0 q = -p

Problem: Where is the position of the image if an object is p = 12 cm in front of a Convex mirror with |R| = 10 cm Concave mirror with R = 10 cm Concave mirror with R = 30 cm

Problem: Where is the position of the image if an object is p = 12 cm in front of a Convex mirror with |R| = 10 cm R = -10 cm, f = R/2 = -5 cm 1/q = 1/f – 1/p = -1/5cm – 1/12cm = -0.283/cm q = -3.53 cm (virtual, behind mirror)

Problem: Where is the position of the image if an object is p = 12 cm in front of a Convex mirror with |R| = 10 cm R = -10 cm, f = R/2 = -5 cm 1/q = 1/f – 1/p = -1/5cm – 1/12cm = -0.283/cm q = -3.53 cm (virtual, behind mirror) b) Concave mirror with R = 10 cm f = R/2 = 5 cm 1/q = 1/f – 1/p = 1/5cm – 1/12cm = +0.117/cm q = + 8.57 cm (real, in front of mirror)

Problem: Where is the position of the image if an object is p = 12 cm in front of a Convex mirror with |R| = 10 cm R = -10 cm, f = R/2 = -5 cm 1/q = 1/f – 1/p = -1/5cm – 1/12cm = -0.283/cm q = -3.53 cm (virtual, behind mirror) b) Concave mirror with R = 10 cm f = R/2 = 5 cm 1/q = 1/f – 1/p = 1/5cm – 1/12cm = +0.117/cm q = + 8.57 cm (real, in front of mirror) c) Concave mirror with R = 30 cm f = R/2 = 15 cm 1/q = 1/f – 1/p = 1/15cm – 1/12cm = -0.0167/cm q = - 60 cm (virtual, behind mirror)

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.