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.

ConcepTest 32.4a Refraction I Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster? 1) medium 1 2) medium 2 3) both the same 1 air air 2

ConcepTest 32.4a Refraction I Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster? 1) medium 1 2) medium 2 3) both the same The greater the difference in the speed of light between the two media, the greater the bending of the light rays. 1 air air 2 Follow-up: How does the speed in air compare to that in #1 or #2?

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 32-13: A spherical “lens.” A point source of light is placed at a distance of 25.0 cm from the center of a glass (n=1.5) 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:

Chapter 33 Lenses and Optical Instruments Chapter 33 opener. Of the many optical devices we discuss in this Chapter, the magnifying glass is the simplest. Here it is magnifying part of page 886 of this Chapter, which describes how the magnifying glass works according to the ray model. In this Chapter we examine thin lenses in detail, seeing how to determine image position as a function of object position and the focal length of the lens, based on the ray model of light. We then examine optical devices including film and digital cameras, the human eye, eyeglasses, telescopes, and microscopes.

Units of Chapter 33 Thin Lenses; Ray Tracing The Thin Lens Equation; Magnification Combinations of Lenses Lensmaker’s Equation Cameras: Film and Digital The Human Eye; Corrective Lenses Magnifying Glass

Units of Chapter 33 Telescopes Compound Microscope Aberrations of Lenses and Mirrors

33-1 Thin Lenses; Ray Tracing Thin lenses are those whose thickness is small compared to their radius of curvature. They may be either converging (a) or diverging (b). Figure 33-2. (a) Converging lenses and (b) diverging lenses, shown in cross section. Converging lenses are thicker in the center whereas diverging lenses are thinner in the center. (c) Photo of a converging lens (on the left) and a diverging lens (right). (d) Converging lenses (above), and diverging lenses (below), lying flat, and raised off the paper to form images.

33-1 Thin Lenses; Ray Tracing Parallel rays are brought to a focus by a converging lens (one that is thicker in the center than it is at the edge). Figure 33-3. Parallel rays are brought to a focus by a converging thin lens.

33-1 Thin Lenses; Ray Tracing A diverging lens (thicker at the edge than in the center) makes parallel light diverge; the focal point is that point where the diverging rays would converge if projected back. Figure 33-5. Diverging lens.

33-1 Thin Lenses; Ray Tracing The power of a lens is the inverse of its focal length: Lens power is measured in diopters, D: 1 D = 1 m-1.

33-1 Thin Lenses; Ray Tracing Ray tracing for thin lenses is similar to that for mirrors. We have three key rays: This ray comes in parallel to the axis and exits through the focal point. This ray comes in through the focal point and exits parallel to the axis. This ray goes through the center of the lens and is undeflected.

33-1 Thin Lenses; Ray Tracing Figure 33-6. Finding the image by ray tracing for a converging lens. Rays are shown leaving one point on the object (an arrow). Shown are the three most useful rays, leaving the tip of the object, for determining where the image of that point is formed.

33-1 Thin Lenses; Ray Tracing Conceptual Example 33-1: Half-blocked lens. What happens to the image of an object if the top half of a lens is covered by a piece of cardboard? Top half eliminated Bottom half eliminated Image complete; brightness lower Solution: The image is unchanged (follow the rays); only its brightness is diminished, as some of the light is blocked.

33-1 Thin Lenses; Ray Tracing For a diverging lens, we can use the same three rays; the image is upright and virtual. Figure 33-8. Finding the image by ray tracing for a diverging lens.

33-2 The Thin Lens Equation; Magnification The thin lens equation is similar to the mirror equation: Figure 33-9. Deriving the lens equation for a converging lens.

33-2 The Thin Lens Equation; Magnification The sign conventions are slightly different: The focal length is positive for converging lenses and negative for diverging. The object distance is positive when the object is on the same side as the light entering the lens (not an issue except in compound systems); otherwise it is negative. The image distance is positive if the image is on the opposite side from the light entering the lens; otherwise it is negative. The height of the image is positive if the image is upright and negative otherwise.

Mirrors and Lenses Summary of sign conventions:

33-2 The Thin Lens Equation; Magnification The magnification formula is also the same as that for a mirror: The power of a lens is positive if it is converging and negative if it is diverging.

33-2 The Thin Lens Equation; Magnification Problem Solving: Thin Lenses Draw a ray diagram. The image is located where the key rays intersect. Solve for unknowns. Follow the sign conventions. Check that your answers are consistent with the ray diagram.

33-2 The Thin Lens Equation; Magnification Example 33-2: Image formed by converging lens. What are (a) the position, and (b) the size, of the image of a 7.6-cm-high leaf placed 1.00 m from a +50.0-mm-focal-length camera lens? Solution: The figure shows the appropriate ray diagram. The thin lens equation gives di = 5.26 cm; the magnification equation gives the size of the image to be -0.40 cm. The signs tell us that the image is behind the lens and inverted.

33-2 The Thin Lens Equation; Magnification Example 33-3: Object close to converging lens. An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size. Figure 33-12. An object placed within the focal point of a converging lens produces a virtual image. Example 33–3. Solution: a. The thin lens equation gives di = -30 cm and m = 3.0. The image is virtual, enlarged, and upright. b. See the figure.

33-2 The Thin Lens Equation; Magnification Example 33-4: Diverging lens. Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens, on the same side as the object? Solution: Since the lens is diverging, the focal length is negative. The lens equation gives do = 100 cm.

33-3 Combinations of Lenses In lens combinations, the image formed by the first lens becomes the object for the second lens (this is where object distances may be negative). The total magnification is the product of the magnification of each lens.

33-3 Combinations of Lenses Example 33-5: A two-lens system. Two converging lenses, A and B, with focal lengths fA = 20.0 cm and fB = 25.0 cm, are placed 80.0 cm apart. An object is placed 60.0 cm in front of the first lens. Determine (a) the position, and (b) the magnification, of the final image formed by the combination of the two lenses. Figure 33-14. Two lenses, A and B, used in combination. The small numbers refer to the easily drawn rays. Solution: a. Using the lens equation we find the image for the first lens to be 30.0 cm in back of that lens. This becomes the object for the second lens - it is a real object located 50.0 cm away. Using the lens equation again we find the final image is 50 cm behind the second lens. b. The magnification is the product of the magnifications of the two lenses: +0.500. The image is half the size of the object and upright.

33-4 Lensmaker’s Equation This useful equation relates the radii of curvature of the two lens surfaces, and the index of refraction, to the focal length: Figure 33-16. Diagram of a ray passing through a lens for derivation of the lensmaker’s equation.

33-4 Lensmaker’s Equation Example 33-7: Calculating f for a converging lens. A convex meniscus lens is made from glass with n = 1.50. The radius of curvature of the convex surface is 22.4 cm and that of the concave surface is 46.2 cm. (a) What is the focal length? (b) Where will the image be for an object 2.00 m away? Solution: Using the lensmaker’s equation gives f = 87 cm. Then the image distance can be found: di = 1.54 m.

33-5 Cameras: Film and Digital Basic parts of a camera: Lens Light-tight box Shutter Film or electronic sensor Figure 33-18. A simple camera.

33-5 Cameras: Film and Digital Camera adjustments: Shutter speed: controls the amount of time light enters the camera. A faster shutter speed makes a sharper picture. f-stop: controls the maximum opening of the shutter. This allows the right amount of light to enter to properly expose the film, and must be adjusted for external light conditions. Focusing: this adjusts the position of the lens so that the image is positioned on the film.

33-5 Cameras: Film and Digital Example 33-8: Camera focus. How far must a 50.0-mm-focal-length camera lens be moved from its infinity setting to sharply focus an object 3.00 m away? Solution: Using the lens equation gives the image distance as 50.8 mm; since the focal length of the lens is 50.0 mm (where an object at infinity would be in sharp focus) the lens needs to move 0.8 mm away from the infinity setting.

33-6 The Human Eye; Corrective Lenses The human eye resembles a camera in its basic functioning, with an adjustable lens, the iris, and the retina. Figure 33-25. Diagram of a human eye.

33-6 The Human Eye; Corrective Lenses Figure 33-26 goes here. Most of the refraction is done at the surface of the cornea; the lens makes small adjustments to focus at different distances. Figure 33-26. Accommodation by a normal eye: (a) lens relaxed, focused at infinity; (b) lens thickened, focused on a nearby object.

33-6 The Human Eye; Corrective Lenses Near point: closest distance at which eye can focus clearly. Normal is about 25 cm. Far point: farthest distance at which object can be seen clearly. Normal is at infinity. Nearsightedness: far point is too close. Farsightedness: near point is too far away.

33-6 The Human Eye; Corrective Lenses Nearsightedness can be corrected with a diverging lens. Figure 33-27a. Correcting eye defects with lenses: (a) a nearsighted eye, which cannot focus clearly on distant objects, can be corrected by use of a diverging lens

33-6 The Human Eye; Corrective Lenses And farsightedness with a diverging lens. Figure 33-27b. (b) a farsighted eye, which cannot focus clearly on nearby objects, can be corrected by use of a converging lens.

33-6 The Human Eye; Corrective Lenses Example 33-12: Farsighted eye. Sue is farsighted with a near point of 100 cm. Reading glasses must have what lens power so that she can read a newspaper at a distance of 25 cm? Assume the lens is very close to the eye. Solution: Using do = 25 cm and di = -100 cm gives f = 0.33 m. The lens power is +3.0 D.

33-6 The Human Eye; Corrective Lenses Example 33-13: Nearsighted eye. A nearsighted eye has near and far points of 12 cm and 17 cm, respectively. (a) What lens power is needed for this person to see distant objects clearly, and (b) what then will be the near point? Assume that the lens is 2.0 cm from the eye (typical for eyeglasses). Solution: The lens should put a distant object at the far point of the eye. Using do = ∞ and di = -15 cm (since the eye is 2 cm from the lens) gives f = -0.15 m. The lens power is -6.7 D.

33-6 The Human Eye; Corrective Lenses Vision is blurry under water because light rays are bent much less than they would be if entering the eye from air. This can be avoided by wearing goggles. Figure 33-31. (a) Under water, we see a blurry image because light rays are bent much less than in air. (b) If we wear goggles, we again have an air–cornea interface and can see clearly.

33-7 Magnifying Glass A magnifying glass (simple magnifier) is a converging lens. It allows us to focus on objects closer than the near point, so that they make a larger, and therefore clearer, image on the retina. Figure 33-33a. Leaf viewed (a) through a magnifying glass. The eye is focused at its near point

33-7 Magnifying Glass The power of a magnifying glass is described by its angular magnification: If the eye is relaxed (N is the near point distance and f the focal length): If the eye is focused at the near point:

33-10 Aberrations of Lenses and Mirrors Spherical aberration: rays far from the lens axis do not focus at the focal point. Figure 33-41. Spherical aberration (exaggerated). Circle of least confusion is at C. Solutions: compound-lens systems; use only central part of lens.

33-10 Aberrations of Lenses and Mirrors Distortion: caused by variation in magnification with distance from the lens. Barrel and pincushion distortion: Figure 33-42. Distortion: lenses may image a square grid of perpendicular lines to produce (a) barrel distortion or (b) pincushion distortion.

33-10 Aberrations of Lenses and Mirrors Chromatic aberration: light of different wavelengths has different indices of refraction and focuses at different points. Figure 33-43. Chromatic aberration. Different colors are focused at different points.

33-10 Aberrations of Lenses and Mirrors Solution: Achromatic doublet, made of lenses of two different materials Figure 33-44. Achromatic doublet.

Summary of Chapter 33 Lens uses refraction to form real or virtual image. Converging lens: rays converge at focal point. Diverging lens: rays appear to diverge from focal point. Power is given in diopters (m-1):

Summary of Chapter 33 Thin lens equation: Magnification:

Summary of Chapter 33 Camera focuses image on film or electronic sensor; lens can be moved and size of opening adjusted (f-stop). Human eye also makes adjustments, by changing shape of lens and size of pupil. Nearsighted eye is corrected by diverging lens. Farsighted eye is corrected by converging lens.

Summary of Chapter 33 Magnification of simple magnifier: Telescope: objective lens or mirror plus eyepiece lens. Magnification: