1. Geometric Optics
Figure Mirrors with convex and concave spherical surfaces. Note that θr = θi for each ray.

2. Image Formation by Spherical Mirrors
Spherical Mirrors: Are shaped like sections of a sphere.
May reflect on the inside (concave) or outside (convex).
Figure Mirrors with convex and concave spherical surfaces. Note that θr = θi for each ray.

3. Reflected rays don’t converge to the same point.
Approximation: Rays coming from far away objects are approximately parallel.
Consider such parallel rays striking a concave mirror. The law
of reflection holds at each point where a ray strikes the mirror.
But, if a mirror has large curvature,
parallel rays striking it will not all
converge at exactly the same place.
This is called spherical aberration & it causes a blurred image.
Figure If the object’s distance is large compared to the size of the mirror (or lens), the rays are nearly parallel. They are parallel for an object at infinity (∞).
Reflected rays don’t converge to the same point.

4. The distance between the focal point & the mirror center
The Principle Axis is the straight line perpendicular to the mirror surface at it’s center. If the incoming rays are parallel to the principle axis
& if the mirror is small compared to it’s radius of curvature, r, so that the
reflected ray makes a small angle with respect to the incident ray, then, reflected rays will converge with each other at approximately a single point, called The Focal Point F.
Focal length  f.
Radius of
curvature  r
Figure Rays parallel to the principal axis of a concave spherical mirror come to a focus at F, the focal point, as long as the mirror is small in width as compared to its radius of curvature, r, so that the rays are “paraxial”—that is, make only small angles with the horizontal axis.
The distance between the focal point & the mirror center
 The focal length, f.

5. Using simple geometry, and assuming that the angle θ is small, it can be shown that the focal length is half the radius of curvature.
Spherical aberration can be avoided by using a parabolic reflector; these are more difficult & expensive to make, & so are used only when necessary, such as in research telescopes.

6. Image Formation by Spherical Mirrors
Ray diagrams are used to determine where an image will be. For mirrors, three key rays, each beginning on the object, are used:
A ray parallel to the axis; after reflection it passes through the focal point.
A ray through the focal point; after reflection it is parallel to the axis.
A ray perpendicular to the mirror; it reflects back on itself.
See the figures on the next slide!

7. Three key rays, each beginning on the object, are used:
Ray 1: Goes out from O'
parallel to the axis &
reflects through F.
Ray 2: Goes through F &
then reflects back parallel
to the axis
Ray 3: Chosen perpendicular
to the mirror & so must
reflect back on itself &
go through the center of
curvature C
image
Figure Rays leave point O’ on the object (an arrow). Shown are the three most useful rays for determining where the image I’ is formed. [Note that our mirror is not small compared to f, so our diagram will not give the precise position of the image.] l

8. The intersection of these 3 rays gives the position of the image of that point on the object. To get a full image, do the same with other points (2 points are enough for many purposes).
A relation between the object distance dO, the image distance di, & the mirror focal length f can be obtained from simple geometry. The result is:
This is called
The Mirror Equation
Since light passes through the image, the image in this case is called a Real Image. It can be seen on a piece of paper at point I or on a photographic film there. Contrast: For a plane mirror, no light passes through the image, so it is called a Virtual Image.
Figure Diagram for deriving the mirror equation. For the derivation, we assume the mirror size is small compared to its radius of curvature.
Note: In your book do & di are denoted as so & si

9. Again using geometry, the magnification m (the ratio of the image height, hi to the object height hO) is found to be:
The negative sign indicates that the image is inverted. This object is between the center of curvature and the focal point, and its image is larger, inverted, and real.

10. 2. m < 0, so the image is inverted.
Example: Image in a concave mirror. A diamond ring of height hO = 1.50 cm is placed a distance dO = 20.0 cm from a concave mirror with radius of curvature r = 30.0 cm. Calculate (a) the position of the image, and (b) its size.
Solution: The focal length f = (½)r = 15.0 cm.
(a) Mirror Equation gives:
(1/di) = (1/f) - (1/dO)
So, di = 60.0 cm
(b) The magnification is then:
m = - (di/dO) = -3.0
Solution: a. Using the mirror equation, we find di = 60.0 cm.
b. Using the magnification equation, we find M = and hi = -4.5 cm.
Note
1. di > 0, so the image is real (on the same side as the object).
2. m < 0, so the image is inverted.

11. Conceptual Example: Reversible rays.
If the object in this figure is placed where the image is, where will the new image be?
Figure goes here.
Solution: The equations, and the physical setup, are symmetric between the image and the object. The new image will be where the old object was.

12. If an object is outside the center of curvature of a concave mirror, its image will be inverted, smaller, and real.
Figure You can see a clear inverted image of your face when you are beyond C (do > 2f), because the rays that arrive at your eye are diverging. Standard rays 2 and 3 are shown leaving point O on your nose. Ray 2 (and other nearby rays) enters your eye. Notice that rays are diverging as they move to the left of image point I.

13. Example: Object closer to concave mirror.
A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm.
(a) Draw a ray diagram to locate (approximately) the position of the image.
(b) Find the position of the image & the magnification.
Figure Object placed within the focal point F. The image is behind the mirror and is virtual, [Note that the vertical scale (height of object = 1.0 cm) is different from the horizontal (OA = 10.0 cm) for ease of drawing, and reduces the precision of the drawing.]
Example 32–6. Solution: a. The figure shows the ray diagram and the image; the image is upright, larger in size than the object, and virtual.
b. Using the mirror equation gives di = cm. Using the magnification equation gives M =

14. Convex Mirrors
Exactly the same type of analysis can holds. It is found that the mirror equation still holds in this case, if the radius of curvature r & the focal length f are considered to be negative. Also, the image is always virtual, upright, and smaller
than the object.
Figure Convex mirror: (a) the focal point is at F, behind the mirror; (b) the image I of the object at O is virtual, upright, and smaller than the object.

15. Problem Solving: Spherical Mirrors
Draw a ray diagram; the image is where the rays intersect.
Apply the mirror and magnification equations.
Sign conventions: If the object, the image, or the focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. The magnification is positive if image is upright. It is negative otherwise.
Check that your solution agrees with the ray diagram.

16. Example Convex rearview mirror
An external rearview car mirror is convex with a radius of curvature r = 16.0 m. Find the location of the image and its magnification for an object dO = 10.0 m from the mirror.
Solution: The ray diagram for a convex lens appears in Figure 32-19b. A convex mirror has a negative focal length, giving di = -4.4 m and M = The image is virtual, upright, and smaller than the object.