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.

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). 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.) 1/p + 1/q = 1/f, and magnification M  h’/h = -q/p Note that the real image (q and p positive) is inverted (M <0).

1/p + 1/q = 1/f, and magnification M  h’/h = -q/p Proof: Center of mirror is vertical so its normal is horizontal: Similar triangles with vertex V: |h’/q| = tan q = h/p  h’/h = -q/p  Ray going through C (center of curvature) is normal to surface so reflects back on itself: Similar triangles with vertex C h/(p-R) = tan  = |h’/(R-q)| (R-q)/(p-R) = -h’/h = q/p R/q – 1 = 1 – R/p R/p + R/q = 2 1/p + 1/q = 2/R = 1/f 

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).

Convex Mirror (q, R, f < 0)   Still have 1/p + 1/q = 2/R = 1 /f and h’/h = -q /p

Proof for Convex Mirror   Proof: Center of mirror is vertical so its normal is horizontal: h/p = tan  = h’/|q|  M = h’/h = |q|/p = -q/p  Ray going toward C (center of curvature) is normal to mirror, so reflects back on itself: h/(|R| + p) = tan  = h’/(|R|-|q|) h’/h = (|R|-|q|) / (|R| + p) |q|/p = (|R|-|q|) / (|R| + p)  |R|/|q| -1 = |R|/p + 1 1/p – 1 /|q| = -2/|R|; but q,R < 0 1/p + 1/q = 2/R = 1/f 

Convex Mirror 1/p – 1 /|q| = -1/|f| 1/|q| = 1/p + 1/|f| |q| < p h’ = h(|q|/p) < h Side view mirror in car is a convex mirror, so images are smaller than the object. Even though the location of the image is closer than the object (|q| < p) the fact that it is smaller causes brain to interpret the image as being further away.

Paraxial Approximation: We are measuring all distances from center of mirror, (ignoring the curvature of the mirror in measuring distances), so our expressions only hold for rays near the principle axis. That is, the mirror equations only hold exactly if h << R. However, problems may be given in which h is not << R (to keep the math and the figure easy).

Problems 1) Concave mirror with R = 20 cm. Object, 3 cm high, is at p = 50 cm. Where is image, how large is it, and is it real or virtual?

1/q = 1/f – 1/p = 1/10cm – 1/50cm = 0.08/cm q = 12.5 cm; real image 1) Concave mirror with R = 20 cm. Object, 3 cm high, is at p = 50 cm. Where is image, how large is it, and is it real or virtual? f = R/2 = 10 cm; h = 3 cm 1/p + 1/q = 1/f 1/q = 1/f – 1/p = 1/10cm – 1/50cm = 0.08/cm q = 12.5 cm; real image h’ = Mh = -(q/p) h = 3 cm (-12.5/50) = - 0.75 cm (inverted) Numbered rays are “characteristic rays” to be discussed later.

2) Concave mirror with R = 20 cm. Object, 3 cm high, is at p = 5 cm 2) Concave mirror with R = 20 cm. Object, 3 cm high, is at p = 5 cm. Where is image, how large is it, and is it real or virtual?

1/q = 1/f – 1/p = 1/10cm – 1/5cm = - 0.1/cm 2) Concave mirror with R = 20 cm. Object, 3 cm high, is at p = 5 cm. Where is image, how large is it, and is it real or virtual? f = R/2 = 10 cm; h = 3 cm 1/q = 1/f – 1/p = 1/10cm – 1/5cm = - 0.1/cm q = - 10.0 cm; virtual image h’ = Mh = -(q/p)h = -(-10/5)(3 cm)= + 6 cm (upright) Numbered rays are “characteristic rays” to be discussed later.

1/q = 1/f – 1/p = -1/10cm – 1/5 cm = - 3/10 cm 3) Convex mirror with |R|= 20 cm. Object, 3 cm high, is at p = 5 cm. Where is image, how large is it, and is it real or virtual? f = -|R|/2 = - 10 cm; h = 3 cm 1/q = 1/f – 1/p = -1/10cm – 1/5 cm = - 3/10 cm q = - 3.33 cm; virtual image h’ = Mh = - (q/p)h = - (-3.33/5)(3cm) = + 2 cm (upright) Numbered rays are “characteristic rays” to be discussed later.

Mirrors Object distance = p, image distance = q, radius of curvature = R, focal length = f 1/p + 1/q = 2/R = 1/f Lateral Magnification: M = image size/object size = h’/h = -q/p 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. Although these mirror equations are generally more accurate and easier to use, it is informative to analyze images from “ray diagrams”.

Draw h, p, f and C (at R) to scale; mirror can be arbitrarily large (but mirror doesn’t need to be spherical in drawing). (Above example is for a concave mirror, with R, f >0.) Since all paraxial rays from h will be reflected to its image h’, only need to consider two such rays to find h’ and q, but we will consider three: a) Ray from h “through” f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f. 3) These reflected rays will intersect at the tip of h’, so h’ and q can be measured. (If they are diverging after reflecting from mirror, extend them to back of mirror to find virtual image.) “Through”: Ray may actually pass through f (C), e.g. when p>f (p>R), or seem to come from f (C), e.g. f>p (R>p), or be heading toward f and C, when f and C are negative.

Concave mirror with p > R . q h’ Real, inverted image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Concave mirror with p > R . q h’ Real, inverted image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f [Not in text: there is a 4th characteristic ray (but you need a protractor to draw it: Since the center of the mirror is perpendicular (normal) to the principle axis, a ray from h to the center of the mirror reflects making the same angle with the principle axis.] Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Concave mirror with f < p < R extend size of mirror if needed h q h’ Real, inverted image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Concave mirror with f < p < R extend size of mirror if needed h q h’ Real, inverted image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f [4th characteristic ray: Ray from h to center of mirror reflects making the same angle with the principle axis.] Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Concave mirror with p < f Diverging reflected rays extended behind mirror h’ q Virtual, upright image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Concave mirror with p < f Diverging reflected rays extended behind mirror h’ q Virtual, upright image a) Ray from h through f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f [4th characteristic ray: Ray from h to center of mirror reflects making the same angle with the principle axis.] Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Diverging reflected rays extended behind mirror Convex Mirror Diverging reflected rays extended behind mirror h’ q Virtual, upright image a) Ray from h “through” f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Diverging reflected rays extended behind mirror Convex Mirror Diverging reflected rays extended behind mirror h’ q Virtual, upright image a) Ray from h “through” f reflects parallel to principal axis. b) Ray from h “through” C is radial, so reflects back on itself. c) Ray from h parallel to principal axis will reflect “through” f [4th characteristic ray: Ray from h to center of mirror reflects making the same angle with the principle axis.] Intersection is tip of image. h’ and q can then be measured. If drawn accurately to scale, will find 1/q = 1/f – 1/p and h’ = -(q/p)h.

Note: Because of paraxial ray approximation, mirror equations only hold if h << R. However, problems may be given in which h is not << R (to keep the math and the figure easy). Therefore, horizontal distances should be measured from the vertex (center) of the mirror. Similarly, in your diagrams, minimize the curvature of the mirror, or even draw it as “straight”, so reflecting surface lines up with mirror’s vertex. vertex