Chapter 31 Maxwell’s Equations and Electromagnetic Waves Chapter 31 opener. These circular disk antennas, each 25 m in diameter, are pointed to receive radio waves from out in space. Radio waves are electromagnetic (EM) waves that have frequencies from a few hundred Hz to about 100 MHz. These antennas are connected together electronically to achieve better detail; they are a part of the Very Large Array in New Mexico searching the heavens for information about the Cosmos. We will see in this Chapter that Maxwell predicted the existence of EM waves from his famous equations. Maxwell’s equations themselves are a magnificent summary of electromagnetism. We will also examine how EM waves carry energy and momentum.

31-8 Energy in EM Waves; the Poynting Vector Energy is stored in both electric and magnetic fields, giving the total energy density of an electromagnetic wave: Each field contributes half the total energy density:

31-8 Energy in EM Waves; the Poynting Vector This energy is transported by the wave. Figure 31-15. Electromagnetic wave carrying energy through area A.

31-8 Energy in EM Waves; the Poynting Vector The energy transported through a unit area per unit time is called the intensity: Its vector form is the Poynting vector:

31-8 Energy in EM Waves; the Poynting Vector Typically we are interested in the average value of S: .

31-8 Energy in EM Waves; the Poynting Vector Example 31-6: E and B from the Sun. Radiation from the Sun reaches the Earth (above the atmosphere) at a rate of about 1350 J/s·m2 (= 1350 W/m2). Assume that this is a single EM wave, and calculate the maximum values of E and B. Solution: The rate at which radiation arrives is the average value of S; we can use equation 31-19a to find E0 and B0. E0 = 1.01 x 103 V/m. B0 = 3.37 x 10-6 T.

31-9 Radiation Pressure In addition to carrying energy, electromagnetic waves also carry momentum. This means that a force will be exerted by the wave. The radiation pressure is related to the average intensity. It is a minimum if the wave is fully absorbed: and a maximum if it is fully reflected:

31-9 Radiation Pressure Example 31-7: Solar pressure. Radiation from the Sun that reaches the Earth’s surface (after passing through the atmosphere) transports energy at a rate of about 1000 W/m2. Estimate the pressure and force exerted by the Sun on your outstretched hand. Solution: Estimate P = S/c = 3 x 10-6 N/m2. If your hand is about 10 cm by 20 cm, or 0.02 m2, this translates to a force of 6 x 10-8 N.

31-9 Radiation Pressure Example 31-8: A solar sail. Proposals have been made to use the radiation pressure from the Sun to help propel spacecraft around the solar system. (a) About how much force would be applied on a 1 km x 1 km highly reflective sail, and (b) by how much would this increase the speed of a 5000-kg spacecraft in one year? (c) If the spacecraft started from rest, about how far would it travel in a year? Solution: a. Since it is highly reflective, the pressure is 2S/c, so the total force is 6 N. (Clearly, you would want to start your journey far from the Earth so its gravity doesn’t interfere!) b. The acceleration is about 1.2 x 10-3 m/s2. Using standard kinematics with constant acceleration (assuming the sail doesn’t get far enough away from the Sun that the radiation pressure decreases noticeably) gives the final speed as 4 x 104 m/s. c. Again using standard kinematics with constant velocity, the distance is 6 x 1011 m. This is about four times the Sun-Earth distance.

Summary of Chapter 31 Maxwell’s equations are the basic equations of electromagnetism:

Summary of Chapter 31 Electromagnetic waves are produced by accelerating charges; the propagation speed is given by The fields are perpendicular to each other and to the direction of propagation.

Summary of Chapter 31 The wavelength and frequency of EM waves are related: The electromagnetic spectrum includes all wavelengths, from radio waves through visible light to gamma rays. The Poynting vector describes the energy carried by EM waves:

Chapter 32 Light: 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.

Units of Chapter 32 The Ray Model of Light Reflection; Image Formation by a Plane Mirror Formation of Images by Spherical Mirrors Index of Refraction Refraction: Snell’s Law

Units of Chapter 32 Visible Spectrum and Dispersion Total Internal Reflection; Fiber Optics Refraction at a Spherical Surface

32-1 The Ray Model of Light Light very often travels in straight lines. We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics. Figure 32-1. Light rays come from each single point on an object. A small bundle of rays leaving one point is shown entering a person’s eye.

32-2 Reflection; Image Formation by a Plane Mirror Law of reflection: the angle of reflection (that the ray makes with the normal to a surface) equals the angle of incidence. Figure 32-2. Law of reflection: (a) Shows a 3-D view of an incident ray being reflected at the top of a flat surface; (b) shows a side or “end-on” view, which we will usually use because of its clarity.

32-2 Reflection; Image Formation by a Plane Mirror When light reflects from a rough surface, the law of reflection still holds, but the angle of incidence varies. This is called diffuse reflection. Figure 32-3. Diffuse reflection from a rough surface.

32-2 Reflection; Image Formation by a Plane Mirror With diffuse reflection, your eye sees reflected light at all angles. With specular reflection (from a mirror), your eye must be in the correct position. Figure 32-4. A narrow beam of light shines on (a) white paper, and (b) a mirror. In part (a), you can see with your eye the white light reflected at various positions because of diffuse reflection. But in part (b), you see the reflected light only when your eye is placed correctly mirror reflection is also known as specular reflection. (Galileo, using similar arguments, showed that the Moon must have a rough surface rather than a highly polished surface like a mirror, as some people thought.)

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).

32-2 Reflection; Image Formation by a Plane Mirror What you see when you look into a plane (flat) mirror is an image, which appears to be behind the mirror. Figure 32-7. Formation of a virtual image by a plane mirror.

32-2 Reflection; Image Formation by a Plane Mirror This is called a virtual image, as the light does not go through it. The distance of the image from the mirror is equal to the distance of the object from the mirror.

32-2 Reflection; Image Formation by a Plane Mirror Example 32-2: How tall must a full-length mirror be? A woman 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how close must its lower edge be to the floor, if she is to be able to see her whole body? Assume her eyes are 10 cm below the top of her head. Solution: At the minimum height, light from her feet strikes the bottom edge of the mirror and reflects into her eyes; light from the top of her head strikes the top edge of the mirror and reflects into her eyes. Geometry then shows that the mirror must be 80 cm high, with its bottom 75 cm off the floor.

32-2 Reflection; Image Formation by a Plane Mirror Conceptual Example 32-3: Is the photo upside down? Close examination of the photograph on the first page of this Chapter reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Show why the reflection is not the same as the real scene by drawing a sketch of this situation, showing the Sun, the camera, the branch, and two rays going from the Sun to the camera (one direct and one reflected). Is the photograph right side up? Solution: Draw a sketch of the situation – the camera, sun, and branch are all above the water. Light goes directly from the sun to the camera, and also bounces off the water and to the camera. In order for the sun to appear below the branches in one image and through them in the other, the branches must be between the sun and the camera. The picture is upside down.

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 32-10. Mirrors with convex and concave spherical surfaces. Note that θr = θi for each ray.

32-3 Formation of Images by Spherical Mirrors Rays coming from a faraway object are effectively parallel. Figure 32-12. 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 (∞).

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 32-13. 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.”)

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. Figure 32-14. 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.

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

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.

32-3 Formation of Images by Spherical Mirrors Figure 32-15. 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.]

32-3 Formation of Images by Spherical Mirrors The intersection of these three rays gives the position of the image of that point on the object. To get a full image, we can do the same with other points (two points suffice for may purposes).

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 32-16. Diagram for deriving the mirror equation. For the derivation, we assume the mirror size is small compared to its radius of curvature.

32-3 Formation of Images by Spherical Mirrors We can also find the magnification (ratio of image height to object height): 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.

32-3 Formation of Images by Spherical Mirrors Example 32-4: Image in a concave mirror. A 1.50-cm-high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.0 cm. Determine (a) the position of the image, and (b) its size. Solution: a. Using the mirror equation, we find di = 60.0 cm. b. Using the magnification equation, we find M = -3.00 and hi = -4.5 cm.

32-3 Formation of Images by Spherical Mirrors Conceptual Example 32-5: Reversible rays. If the object in this figure is placed where the image is, where will the new image be? Figure 32-16 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.

32-3 Formation of Images by Spherical Mirrors If an object is outside the center of curvature of a concave mirror, its image will be inverted, smaller, and real. Figure 32-18. 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.

32-3 Formation of Images by Spherical Mirrors Example 32-6: 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) Determine the position of the image and the magnification analytically. Figure 32-17. 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 = -30.0 cm. Using the magnification equation gives M = +3.00.

32-3 Formation of Images by Spherical Mirrors For a convex mirror, the image is always virtual, upright, and smaller. Figure 32-19. 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.

32-3 Formation of Images by Spherical Mirrors 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, image, or focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. Magnification is positive if image is upright, negative otherwise. Check that your solution agrees with the ray diagram.

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 = +0.44. The image is virtual, upright, and smaller than the object.

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:

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 32-21. 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.

32-5 Refraction: Snell’s Law Refraction is what makes objects half-submerged in water look odd. Figure 32-22. 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).

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

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.

32-5 Refraction: Snell’s Law Example 32-9: Apparent depth of a pool. A swimmer has dropped her goggles to the bottom of a pool at the shallow end, marked as 1.0 m deep. But the goggles don’t look that deep. Why? How deep do the goggles appear to be when you look straight down into the water? Solution: The ray diagram appears in Figure 32-25. Refraction causes the goggles to appear to be less deep than they actually are. Snell’s law plus a small-angle approximation (sin θ ≈ tan θ ≈ θ) gives d’ ≈ d/n1 = 0.75 m.

32-6 Visible Spectrum and Dispersion The visible spectrum contains the full range of wavelengths of light that are visible to the human eye. Figure 32-26. The spectrum of visible light, showing the range of wavelengths for the various colors as seen in air. Many colors, such as brown, do not appear in the spectrum; they are made from a mixture of wavelengths.

32-6 Visible Spectrum and Dispersion The index of refraction of many transparent materials, such as glass and water, varies slightly with wavelength. This is how prisms and water droplets create rainbows from sunlight. Figure 32-28. Index of refraction as a function of wavelength for various transparent solids. Figure 32-29. White light dispersed by a prism into the visible spectrum.

32-6 Visible Spectrum and Dispersion This spreading of light into the full spectrum is called dispersion. Figure 32-20. (a) Ray diagram explaining how a rainbow (b) is formed.

32-6 Visible Spectrum and Dispersion Conceptual Example 32-10: Observed color of light under water. We said that color depends on wavelength. For example, for an object emitting 650 nm light in air, we see red. But this is true only in air. If we observe this same object when under water, it still looks red. But the wavelength in water λn is 650 nm/1.33 = 489 nm. Light with wavelength 489 nm would appear blue in air. Can you explain why the light appears red rather than blue when observed under water? Solution: Evidently our eyes respond to the frequency of the light (which does not change) rather than its wavelength. We refer to wavelengths as they are much more easily measured.

32-7 Total Internal Reflection; Fiber Optics If light passes into a medium with a smaller index of refraction, the angle of refraction is larger. There is an angle of incidence for which the angle of refraction will be 90°; this is called the critical angle:

32-7 Total Internal Reflection; Fiber Optics If the angle of incidence is larger than this, no transmission occurs. This is called total internal reflection. Figure 32-31. Since n2 < n1, light rays are totally internally reflected if the incident angle θ1 > θc, as for ray L. If θ1 < θc, as for rays I and J, only a part of the light is reflected, and the rest is refracted.

32-7 Total Internal Reflection; Fiber Optics Conceptual Example 32-11: View up from under water. Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool. Figure 32-32. (a) Light rays, and (b) view looking upward from beneath the water (the surface of the water must be very smooth). Example 32–11. Solution: The critical angle for an air-water interface is 49°, so the person will see the upwards view compressed into a 49° circle.

32-7 Total Internal Reflection; Fiber Optics Binoculars often use total internal reflection; this gives true 100% reflection, which even the best mirror cannot do. Figure 32-33. Total internal reflection of light by prisms in binoculars.

32-7 Total Internal Reflection; Fiber Optics Optical fibers also depend on total internal reflection; they are therefore able to transmit light signals with very small losses. Figure 32-34. Light reflected totally at the interior surface of a glass or transparent plastic fiber.

32-8 Refraction at a Spherical Surface Rays from a single point will be focused by a convex spherical interface with a medium of larger index of refraction to a single point, as long as the angles are not too large. Figure 32-37. Diagram for showing that all paraxial rays from O focus at the same point I (n2 > n1).

32-8 Refraction at a Spherical Surface Geometry gives the relationship between the indices of refraction, the object distance, the image distance, and the radius of curvature:

32-8 Refraction at a Spherical Surface For a concave spherical interface, the rays will diverge from a virtual image. Figure 32-38. Rays from O refracted by a concave surface form a virtual image (n2 > n1). Per our conventions, R < 0, di < 0, do > 0.

32-8 Refraction at a Spherical Surface Example 32-12: Apparent depth II. A person looks vertically down into a 1.0-m-deep pool. How deep does the water appear to be? Solution: Using equation 32-8 gives di = -0.75 m.

32-8 Refraction at a Spherical Surface Example 32-13: A spherical “lens.” A point source of light is placed at a distance of 25.0 cm from the center of a glass sphere of radius 10.0 cm. Find the image of the source. Solution: Use equation 32-8 twice (once at each interface); the first image is at -90.0 cm, and the final image is at 28 cm.

Summary of Chapter 32 Light paths are called rays. Index of refraction: Angle of reflection equals angle of incidence. Plane mirror: image is virtual, upright, and the same size as the object. Spherical mirror can be concave or convex. Focal length of the mirror:

Summary of Chapter 32 Mirror equation: Magnification: Real image: light passes through it. Virtual image: light does not pass through.

Summary of Chapter 32 Law of refraction (Snell’s law): Total internal reflection occurs when angle of incidence is greater than critical angle: