At the interface between two materials, where light can be reflected or refracted. Within a material, where the light can be scattered or absorbed.

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1 At the interface between two materials, where light can be reflected or refracted. Within a material, where the light can be scattered or absorbed. The eye sees by focusing a diverging bundle of rays from each point on the object. Objects: these can be either Self luminous these emit radiation; e.g. the sun, flames, light bulbs. Reflective most objects reflect the light from self luminous sources. Light rays from an object are emitted in all directions, but we are not aware of them unless they enter the eye. For example, a laser beam traveling through a room it s impossible to see a laser beam from the side unless dust particles in the air scatters a few of the rays towards your eye. We consider two idealised sets of light rays: Point source (figure ): emits diverging rays in all directions e.g. rays from a light bulb. Parallel bundle (figure 73) : a collection of parallel light rays e.g. rays from a laser beam or from a very very distant light source. Ray diagrams: rays originate from every point on an object and travel outward in all directions. We illustrate this using a ray diagram which shows only a few rays leaving the top and bottom of the object (see, for example, figure ). Aperture: this is the hole through which light passes. For example, consider the camera obscura (figures 75 and ): 127

2 Figure 73: Point sources and parallel bundles represent idealised light sources. Figure 74: A ray diagram simplifies the situation by showing only a few rays. Each point on the object emits light rays in all directions, but only a very few of these pass through the aperture and reach the back wall. Geometry tells us that the image is upside down. However, a finite aperture size allows several rays from the same point 60 Knight, Figure 23.3, page Knight, Figure 23.4, page Knight, Figure 23.5, page

3 Figure 75: Camera obscura: general viewpoint. Figure 76: Camera obscura: ray diagram. 129

4 on the object to pass through at slightly different angles. Thus, the image on the wall is slightly blurred and out of focus. Maximum sharpness is achieved by making the hole smaller and smaller this makes the image dimmer and dimmer, and also we get close to the 1 mm limit. Image size is not the same as the object size. We define the magnification M as the ratio We therefore see that M = image height object height = h i h o. (4.1) M > 1: image larger than object. M < 1: image smaller than object. For example, in figure 76, using similar triangles we find that if d o and d i are, respectively, the distance to the object and the depth of the camera obscura, then (4.1) becomes Realistically, d i < d o and so M < 1. M = h i h o = d i d o. (4.2) 4.2 Reflection Reflection from a flat, smooth surface such as a mirror or polished metal, is called specular reflection. Figure shows a bundle of light rays reflecting from a mirror-like surface. Both the incident and reflected rays are in a plane that is perpendicular to the reflective surface (the so-called normal). 63 Knight, Figure 23.7(a), page

5 Figure 77: Specular reflection of light. This system is illustrated in figure 77: The plane of the page is the plane of incidence and reflection. The reflective surface extends into and out of the page. A single light ray represents the entire collection of parallel rays. We define The angle of incidence, θ i, as the angle between the incident ray and the normal. The angle of reflection, θ r, as the angle between the reflected ray and the normal. 131

6 These two angles are connected via the following law: Law of Reflection: The incident and reflected rays are in the same plane normal to the surface, and the angle of reflection is equal to the angle of incidence, i.e. θ i = θ r. (4.3) Note: in optics, we always use the angle between the ray and the normal, and not the angle between the ray and the surface. Example: 64 a dressing mirror on a closet door is 1.5 m tall. The bottom is 0.5 m above the floor. A bare light bulb hangs 1.0 m from the closet door, 2.5 m above the floor. Calculate the length of the streak of reflected light across the floor. Solution: consider the situation shown in figure We treat the bulb as being a point source of light i.e. it emits rays which are diverging and consider only the two rays that strike the top and bottom edges of the mirror. If the normal is perpendicular to the mirror, then using the law of reflection (4.3), we have that by trigonometry, ( ) 0.5 θ 1 = tan 1 = 26.6, 1.0 ( ) 2.0 θ 2 = tan 1 = The distances to the points where the rays strike the floor are then 64 Knight, Ex. 23.1, page Knight, Figure 23.8, page 718 l 1 = 2.0 tan θ 1 = 4.00 m, l 2 = 0.5 tan θ 2 = 0.25 m. 132

7 Figure 78: Pictorial representation of the light rays reflecting from a mirror. 133

8 Hence, the length of the light streak is l 1 l 2 = 3.75 m. Diffuse reflection: this occurs on rough surfaces i.e. ones that have surface irregularities. At each point on the surface, the law of reflection is obeyed, but reflected rays off the irregular surface travel in random directions (figure ). Figure 79: Diffuse reflection from an irregular surface. Wavelengths of visible light are 0.5 µm. Therefore, any surface with scratches or other irregularities of > 1 µm in size will cause diffuse reflection. By contrast, mirrors, etc. which reflect specularly have irregularities of < 1 µm in size. e.g. Hubble Space Telescope. Diffuse reflection is more prevalent than specular reflection and indeed, diffuse reflection is how you see your hand, your friend, etc. 66 Knight, Figure 23.9, page

9 Plane Mirror: consider the image of an object in a mirror (figure ); Figure 80: Light rays reflecting from a plane mirror. A ray from point source P on an object is reflected off a mirror, obeying the law of reflection. Geometrically, we can trace the reflected ray back to point P behind the mirror. The law of reflection, θ i = θ r can be used to show that P is the same distance s behind the mirror as P is a distance s in front of it, i.e. s = s. (4.4) The location of P is independent of θ i, and so all reflected rays travel along lines that pass through the same point P. 67 Knight, Figure 23.10(b), page

10 As in figure 81, 68 whereas the original rays diverged from point P, the reflected rays appear to diverge from point P. Figure 81: Light rays reflecting from a plane mirror. For a plane mirror, the image distance s to point P is equal to the object distance s, as in (4.4). If the rays diverge from an object point P and interact with a mirror such that the reflected rays diverge from a point P and appear to come from P, then P is called a virtual image of P. The image is virtual in the sense that no rays actually leave P, but as far as your eye is concerned, the light rays behave exactly as if the light really originated from P. For an extended object, such as figure 82, Knight, Figure 23.10(c), page Knight, Figure 23.11, page

11 Figure 82: Light rays from an object reflecting from a plane mirror. Each point on the object from which the rays strike the mirror has a corresponding image point an equal distance on the opposite side of the mirror. The eye captures and focuses the rays from each point of the image in order to see the full image in the mirror. Note: rays from each point on an object spread out in all directions and strike every point on the mirror. Only very few rays enter your eye, but the other rays are very real and may be seen by other observers. Rays from points P and Q enter your eye after reflecting from different areas of the mirror this is why you can t always see the full image of 137

12 an object in a very small mirror. 4.3 Refraction What happens when a light ray is incident upon a smooth boundary between two transparent materials e.g. air and glass? Recall that Part of the light reflects from the boundary, obeying the law of reflection. Part of the light is transmitted into the second medium. However, this transmitted ray changes direction as it crosses the boundary. The transmission of light from one medium to another, but with a change in direction, is called refraction. Reflection from the boundary between two transparent media is usually weak: typically, 5% of the light is reflected at the boundary. 95% of the light is transmitted across the boundary. Note: we recall that a light wave does not need a medium to propagate, but it can propagate through a material. While the medium does affect the speed of light, we recall that the particles of the medium do not oscillate as the wave passes through it it is the electromagnetic field that oscillates. Although an infinite number of rays are incident on the boundary, we simplify the situation by focusing on a single light ray: Consider a light ray incident on a boundary between medium 1 and medium 2, as in figure Knight, Figure 23.15(b), page

13 Figure 83: Refraction of light rays. Let the angle of incidence between the incident ray and the normal be θ 1 in medium 1, and θ 2 in medium 2. The angle θ 2 on the transmitted side is called the angle of refraction. The relationship between θ 1 and θ 2 is given by Snell s law of refraction: sin θ 1 sin θ 2 = n 2 n 1, (4.5) where n 1 and n 2 are the refractive indices of medium 1 and medium 2, respectively. Note: Snell s law does not distinguish between the incident and reflected rays: We could equally well formulate the problem as a light ray traveling from 139

14 medium 1 to medium 2 (figure ). In such circumstances, θ 2 is the angle of incidence and θ 1 is the angle of refraction. Figure 84: Refraction of light rays. Recalling the definition of refractive index n as n = c v, (4.6) where v is the speed of light in the medium such that v c gives n 1. In figure 83, as the ray moves from medium 1 to medium 2, where n 2 > n 1, it bends towards the normal. 71 Knight, Figure 23.15(c), page

15 In figure 84, as the ray moves from medium 2 to medium 1, where n 1 > n 2, it bends away from the normal. This general conclusion follows from Snell s law: A ray traveling from a medium of lower refractive index to one of higher refractive index bends towards the normal. A ray traveling from a medium of higher refractive index to one of lower refractive index bends away from the normal. Example: an underwater diver sees the sun 55 above the horizontal. How high is the sun above the horizon to a fisherman in a boat directly above the diver? Solution: consider figure 85; the ray that arrives at the diver 55 above the horizontal is refracted into the water at an angle θ water = 35. Using Snell s law at the water air boundary, if Figure 85: Fisherman diver system. then n air sin θ air = n water sin θ water 141

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