Department of Physics United States Naval Academy Lecture 36: Spherical Refracting Surfaces & Thin Lenses Spherical Refracting Surfaces: A single spherical surface that refracts light can form an image. The object distance p, the image distance i, and the radius of curvature r of the surface are related by n n n − n 1 + 2 = 2 1 p i r where n1 is the index of refraction of the material where the object is located and n2 is the index of refraction on the other side of the surface. Sign Convention for refracting surfaces: When the object faces a convex refracting surface, the radius of curvature r is positive. When it faces a con- cave surface, r is negative. Be careful: This is just the reverse of the sign convention for mirrors. Thus, for refracting surfaces, real images form on the side of a refracting surface that is opposite the object [Fig (a) and (b)], and virtual images form on the same side as the object [Fig (c) - (f)]. Thin Lenses: A lens is a transparent object with two refracting surfaces whose central axes coincide. In this section, we will consider two types of lenses: a lens that causes light rays initially parallel to the central axis to converge is called a converging lens (or convex lens). If, instead, it causes such rays to diverge, the lens is a diverging lens (or concave lens). A converging or convex lens can form a real image (if the object is outside the focal point) or a virtual image (if the object is inside the focal point). A diverging or concave lens can form only a virtual image. • The focal length f is positive for convex lenses and negative for concave lenses. • The image distance i is positive for real images and negative for virtual images. For an object in front of a lens, object distance p and image distance i are related to the lens’s focal length f , index of refraction n, and radii of curvature r1 and r2 by 1 1 1 1 1 + = = (n − 1) − p i f r1 r2 The magnitude of the lateral magnification m of an object is the ratio of the image height h0 to object height h, h0 jmj = h and is related to the object distance p and image distance i by i m = − p Note: For a system of lenses with a common central axis, the image produced by the first lens acts as the object for the second lens, and so on, and the overall magnification is the product of the individual magnifications. If m < 0 the image is inverted, and if jmj > 1 the image is large than the object. © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Images: In geometric optics, an object is a "source" of light. The rules for image forma- tion are all based on the reflection and refraction angles for light rays passing through interfaces with different index of refraction (n). Those rules are: • A reflected ray leaves the surface at the same angle from which it arrives • A refracted ray is bent toward the normal if entering a regions of higher n, and away from the normal if entering a region of lower n. Real and Virtual images: 1.A real image is formed at point I if rays leaving the object converge on I. A screen at I shows an image of the object. Real image correspond to a positive image distance I. 2.A virtual image is formed at I if rays leaving the object appear to diverge from i.A screen cannot show an image of the object, but an observer looking toward I sees the object. Virtual image correspond to a negative image distance i. Sign conventions 1. Object distance p is (always) positive if object O is on the same side of the incoming or incident light. This is usually the case but not always.1 2. Image distance i is positive if image I is on the same side as the outgoing or re- fracted light, in which case the image is real and inverted; if i is negative, the image is on the side of the incoming light, is virtual and upright. 3. Radius of curvature r is positive if the center of curvature C is on the side of the outgoing light. 4. Focal length f is positive ( f > 0) if the focus is real, so that the incident parallel rays converge; f < 0 if parallel rays diverge from a virtual focus. Note: f and r always have the same sign; f = r=2. Rules for locating images by drawing rays 1. A ray that is initially parallel to the central axis of the lens will pass through focal point F2 (ray 1 in Fig. 34-16a). 2. A ray that initially passes through focal point F1 will emerge from the lens parallel to the central axis (ray 2 in Fig. 34-16a). 3. A ray that is initially directed toward the center of the lens will emerge from the lens with no change in its direction (ray 3 in Fig. 34-16a) because the ray encounters the two sides of the lens where they are almost parallel. 1Object distance might be negative when multiple mirrors or lenses are involved © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Problem 1: CH34-Q09 The figure shows four thin lenses, all of the same material, with sides that either are flat or have a radius of curvature of magnitude 10 cm. Without written calculation, rank the lenses according to the magnitude of the focal length, greatest first. © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Problem 1: CH34-P32 An object O stands on the central axis of a spherical refracting surface. For this situation, each problem in the table below refers to the index of refraction n1 where the object is located, (a) the index of refraction n2 on the other side of the refracting surface, (b) the object distance p, (c) the radius of curvature r of the surface, and (d) the image distance i. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R) or virtual (V) and (f) on the same side of the surface as object O or on the opposite side, for lines 34 and 37. © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Problem 2: CH33-P68 A real inverted image I of an object O is formed by a particular lens (not shown); the object-image separation is d = 40:0 cm, measured along the central axis of the lens. The image is just half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens? © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Problem 3: CH33-P66 Object O stands in front of a thin lens, on the central axis. For this situation, the object distance p = +18 cm, index of refraction of the lens n = 1:60, radius of the nearer lens surface r1 = −27 cm, and radius of the farther lens surface r2 = +24 cm. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, de- termine whether the image is (c) real (R) or virtual (V), (d) inverted (I) from object O or noninverted (NI), and (e) on the same side of the lens as object O or on the opposite side. © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD Problem 4: CH33-P82 In the figure below, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1, a diverging lens with a focal length of f1 = 6:0 cm, is mounted within the boxed region closer to O, which is at object distance p1 = +8 cm. Lens 2, a converging lens with a focal length of f2 = 6:0 cm, is mounted within the farther boxed region, at distance d = 12 cm. Find (a) the image distance i2 for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object O or noninverted (NI), and (e) on the same side of lens 2 as object O or on the opposite side. © 2018 Akaa Daniel Ayangeakaa, Ph.D., Department of Physics, United States Naval Academy, Annapolis MD.
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