1. Chapter 32Light: Reflection and Refraction
Chapter 32 opener. Reflection from still water, as from a glass mirror, can be analyzed using the ray model of light. Is this picture right side up? How can you tell? What are the clues? Notice the people and position of the Sun. Ray diagrams, which we will learn to draw in this Chapter, can provide the answer. See Example 32–3. In this first Chapter on light and optics, we use the ray model of light to understand the formation of images by mirrors, both plane and curved (spherical). We also begin our study of refraction—how light rays bend when they go from one medium to another—which prepares us for our study in the next Chapter of lenses, which are the crucial part of so many optical instruments.

2. 32-2 Reflection; Image Formation by a Plane Mirror
Example 32-1: Reflection from flat mirrors.
Two flat mirrors are perpendicular to each other. An incoming beam of light makes an angle of 15° with the first mirror as shown. What angle will the outgoing beam make with the second mirror?
Solution. The rays are drawn in figure 32-5b. The outgoing ray from the first mirror makes an angle of 15° with it, and an angle of 75° with the second mirror. The outgoing beam then makes an angle of 75° with the second mirror (and is parallel to the incoming beam).

3. Problem 4
4. (II) A person whose eyes are 1.64 m above the floor stands 2.30 m in front of a vertical plane mirror whose bottom edge is 38 cm above the floor, Fig. 32–46. What is the horizontal distance x to the base of the wall supporting the mirror of the nearest point on the floor that can be seen reflected in the mirror?

4. 32-3 Formation of Images by Spherical Mirrors
Spherical mirrors are shaped like sections of a sphere, and may be reflective on either the inside (concave) or outside (convex).
Figure Mirrors with convex and concave spherical surfaces. Note that θr = θi for each ray.

5. Spherical Mirrors
Concave
Like you are
in a cave
Convex

6. 32-3 Formation of Images by Spherical Mirrors
Rays coming from a faraway object are effectively parallel.
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 (∞).

7. 32-3 Formation of Images by Spherical Mirrors
Parallel rays striking a spherical mirror do not all converge at exactly the same place if the curvature of the mirror is large; this is called spherical aberration.
Figure Parallel rays striking a concave spherical mirror do not intersect (or focus) at precisely a single point. (This “defect” is referred to as “spherical aberration.”)

8. 32-3 Formation of Images by Spherical Mirrors
If the curvature is small, the focus is much more precise; the focal point is where the rays converge.
Center of curvature:
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.
:Focal point
:Focal length

9. Spherical Mirrors: Concave
Parallel ray goes through f
Ray through center reflects back
Ray through f comes out parallel
object
image C f
Image:
Upright or upside down
Real or virtual
Bigger or smaller

10. Spherical Mirrors: Convex
Radius of
Curvature R
Center of
Curvature
image
object f C
Principle axis
Focal point
f=R/2
Image:
Upright or upside down
Real or virtual
Bigger or smaller
Parallel ray goes through f
Ray through center reflects back
Ray through f comes out parallel

11. 32-3 Formation of Images by Spherical Mirrors
The intersection of the three rays gives the position of the image of a point on the object. To get a full image, we can do the same with other points (two points suffice for many purposes).
Because the light passes through the image itself, this is called “Real image”.

12. 32-3 Formation of Images by Spherical Mirrors
Geometrically, we can derive an equation that relates the object distance, image distance, and focal length of the mirror:
Figure Diagram for deriving the mirror equation. For the derivation, we assume the mirror size is small compared to its radius of curvature.

13. 32-3 Formation of Images by Spherical Mirrors
Using geometry, we find that the focal length is half the radius of curvature:
The More curved is the mirror, the worse is the approximation and the more blurred is the image. Spherical aberration can be avoided by using a parabolic reflector; these are more difficult and expensive to make, and so they are used only when necessary, such as in research telescopes.

14. Magnification: Magnification = image height / object height
= - image distance (di) / object distance (d0)
m and hi are negative if the image is upside down. di d0
di is negative if the image is virtual

15. Magnification: = - image distance (di) / object distance (d0) -di do
Magnification = image height / object height
= - image distance (di) / object distance (d0) d0
-di do di
Negative m = upside down
Negative di = virtual (behind mirror)

16. 32-3 Formation of Images by Spherical Mirrors
Example 32-7: Convex rearview mirror.
An external rearview car mirror is convex with a radius of curvature of 16.0 m. Determine the location of the image and its magnification for an object 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.

17. Problem 26
26. (II) A shaving or makeup mirror is designed to magnify your face by a factor of 1.35 when your face is placed 20.0 cm in front of it. (a) What type of mirror is it? (b) Describe the type of image that it makes of your face. (c) Calculate the required radius of curvature for the mirror.

18. 32-3 Formation of Images by Spherical Mirrors
We use ray diagrams to determine where an image will be. For mirrors, we use three key rays, all of which begin on the object:
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.

19. Sign Conventions for Mirrors
Mirror Equation
Sign Conventions for Mirrors
Quantity
Positive “+”
Negative “-”
Object distance d0
Real
Virtual
Image distance, di
Focal length, f
Concave: f=r/2
Convex: f=-r/2
Magnification, m
Upright
Upside down
f≠d0+di

20. 32-4 Index of Refraction
In general, light slows somewhat when traveling through a medium. The index of refraction of the medium is the ratio of the speed of light in vacuum to the speed of light in the medium:

21. 32-5 Refraction: Snell’s Law
Light changes direction when crossing a boundary from one medium to another. This is called refraction, and the angle the outgoing ray makes with the normal is called the angle of refraction.
Figure Refraction. (a) Light refracted when passing from air (n1) into water (n2): n2 > n1. (b) Light refracted when passing from water (n1) into air (n2): n2 < n1.

22. 32-5 Refraction: Snell’s Law
The angle of refraction depends on the indices of refraction, and is given by Snell’s law:

23. 32-5 Refraction: Snell’s Law
Refraction is what makes objects half-submerged in water look odd.
Figure Ray diagram showing why a person’s legs look shorter when standing in waist-deep water: the path of light traveling from the bather’s foot to the observer’s eye bends at the water’s surface, and our brain interprets the light as having traveled in a straight line, from higher up (dashed line).

24. 32-5 Refraction: Snell’s Law
Example 32-8: Refraction through flat glass.
Light traveling in air strikes a flat piece of uniformly thick glass at an incident angle of 60, as shown. If the index of refraction of the glass is 1.50, (a) what is the angle of refraction θA in the glass; (b) what is the angle θB at which the ray emerges from the glass?
Solution: a. Applying Snell’s law gives sin θA = 0.577, or θA = 35.3°.
b. Snell’s law gives sin θB = 0.866, or θB = 60°. The outgoing ray is parallel to the incoming ray.

25. Problem 43
43. (II) A light beam strikes a 2.0-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.0-cm-thick piece of glass for which What is the distance D in Fig. 32–48?