Hong Kong Polytechnic University Optics I AP 219

Hong Kong Polytechnic University Optics I (AP 219) Dr. Haitao Huang (黄海涛) Department of Applied Physics, Hong Kong Poly. U Tel: 27665694; Office: CD 602 Lecture Notes can be downloaded from http: //ap. polyu. edu. hk/apahthua Text Books: Optics, A. H. Tunnacliffe and J. G. Hirst, (Association of British Dispensing Opticians 1996) Optics & Vision, L. S. Pedrotti and F. L. Pedrotti, (Prentice-Hall 1998) Optics 1 ----by Dr. H. Huang, Department of Applied Physics 1

Hong Kong Polytechnic University Reference Books: Introduction to Optics, 2 nd Edition, F. L. Pedrotti and L. S. Pedrotti (Prentice Hall 1993) Optics, 4 th Edition, E. Hecht (Addison Wesley ) Optics, 3 rd Edition, A. Ghatah (Mc. Graw Hill 2005) Fundamentals of Physics, 7 th Edition, D. Halliday, R. Resnick, and J. Walker (John Wiley & Sons 2005) University Physics, 11 th Edition, H. D. Young and R. A. Freedman (Pearson 2004) Physics for Scientists and Engineers, R. D. Knight (Pearson 2004) Physics for Scientists and Engineers, 3 rd Edition, D. C. Giancoli, (Prentice Hall 2000) College Physics, A. Giambattista, B. Mc. Richardson, and R. C. Richardson, (Mc. Graw Hill 2004) Advanced Studies: The Feynman Lectures on Physics, Definitive Edition, R. P. Feynman, R. B. Leighton, and M. Sands (Addison-Wesley 2006) Principles of Optics, 7 th Edition Expanded, M. Born et al. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 2

Hong Kong Polytechnic University Assessment Weighting: Final Exam: 60% (A minimum grade D is required) The final exam will be in the form of open-book test. You are allowed to take at most one book plus the lecture notes. Mid-term test: 20% (1. 5 hour on 20/10/2006) Course work: 40% Lab reports: 10% Homework: 10% Features: • Many physics concepts will be learned. • Plenty of exercises and examples are provided in each chapter. • Course related experiments are provided to help students get a better understanding on course materials. • The Power. Point of each chapter can be downloaded from my website. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 3

Hong Kong Polytechnic University Introduction Duality in Nature: Rays of light can be regarded as streams of (very small) particles emitted from a source of light and traveling in straight lines. Geometrical Optics: when wavelength can be neglected as compared with dimension of the relevant components in the optical system. Light also takes a wave motion, spreading out from a light source in all directions and propagating through an all-pervasive medium. Light can be viewed as the electromagnetic radiation in a particular region of spectrum. Physical Optics: when wave character can be ignored. Index of Refraction: Optics 1 ----by Dr. H. Huang, Department of Applied Physics 4

Hong Kong Polytechnic University Introduction Law of Reflection When a ray of light is reflected at an interface dividing two uniform media, the reflected ray remains within the plane of incidence includes the incident ray and the normal to the point of incidence. Law of Refraction (Snell’s Law) When a ray of light is refracted at an interface dividing two uniform media, the transmitted ray remains within the plane of incidence and the sine of the angle of refraction is directly proportional to the sine of the angle of incidence. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 5

Hong Kong Polytechnic University Introduction Huygens’ Principle Wavefront: the locus of the points which are in the same phase. Example: a small stone dropped in a calm pool of water. Circular ripples spread out from the point of impact, each point on the circumference of the circle (whose center is at the point of impact) oscillates with the same amplitude and same phase and thus we have a circular wavefront. Each point of a wavefront is a source of secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. The envelope of these wavelets gives the shape of the new wavefront. The ray is always perpendicular to the wavefront. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 6

Hong Kong Polytechnic University Introduction Optics 1 ----by Dr. H. Huang, Department of Applied Physics 7

Hong Kong Polytechnic University Introduction Fermat’s Principle The Fermat’s Principle is also called the “principle of least time”. The field of geometrical optics can be studied by using Fermat’s principle which determines the path of the rays. The ray will correspond to that path for which the time taken is an extreme in comparison to nearby paths, i. e. , it is either a minimum or a maximum or stationary. The mathematical form of the Fermat’s principle is, where n is the refractive index and the integration is done along the path. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 8

Hong Kong Polytechnic University Introduction Principle of Reversibility: Any actual ray of light in an optical system, if reversed in direction, will retrace the same path backward. This principle is directly derived from the Fermat’s principle. Reflection in Plane Mirrors Specular reflection: the reflection from a perfectly smooth surface which obeys the law of reflection. Specular reflection from the xy-plane: the reflected ray PQ remains within the plane of incidence, making equal angles with the normal at P. If the direction of the incident ray is described by its unit vector, r 1=(x, y, z), then the reflected ray is r 2=(x, y, z) Corner reflector: Three successive reflections from orthogonal planes. r 1= (x, y, z) r 2= (-x, -y, -z) Optics 1 ----by Dr. H. Huang, Department of Applied Physics 9

Hong Kong Polytechnic University Introduction A virtual image of a point object S is formed at point S since the rays of light appear to come from that point. SA=S A The image is the same size as the object. The image is reversed, laterally inverted and upright. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 10

Hong Kong Polytechnic University Introduction Refraction through Plane Surfaces: For rays making a small angle with the normal to the surface, a reasonably good image can be located. In this approximation, we only allow paraxial rays to form the image. Since n 1>n 2, the virtual image S of the point object appears to be nearer than it actually is. A critical angle of incidence C is reached when the angle of refraction reaches 90°. For angles of incidence larger than this critical angle, the incident ray experiences total internal reflection, as shown. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 11

Hong Kong Polytechnic University Introduction Example: A swimmer has dropped her goggles in the shallow end of a pool, marked as 1. 0 m deep. But the goggles don’t look that deep. How deep do the goggles appear to be when you look straight down into the water? Solution: n 1=1. 00 for air and n 2=1. 33 for water For small angle sin =tan , Example: Light strikes a flat piece of uniform thick glass at an incident angle of 60°. If the index of refraction of the glass is 1. 50, (a) what is the angle of refraction A in the glass; (b) what is the angle B at which the ray emerge from the glass? Solution: n 1=1. 00 and n 2=1. 50 A=35. 2° B=60° Optics 1 ----by Dr. H. Huang, Department of Applied Physics 12

Hong Kong Polytechnic University Geometrical Optics Imaging by an Optical System An optical system includes any number of reflecting and/or refracting surfaces, of any curvature, that may alter the direction of rays leaving an object. The region may include any number of intervening media, but we assume each individual medium is homogenous and isotropic, and so characterized by its own refractive index. The rays from the object point O spread out radially in all directions in real object space. In real object space the rays are diverging and spherical wavefronts are expanding. Suppose that the optical system redirects these rays in to the real image space, the wavefronts are contracting and the rays are converging to a common image point I. From Fermat’s principle, the rays are said to be isochronous. Further, by the principle of reversibility, if I is the object point, the image point will be O. The points O and I are said to be conjugate points for the optical system. Both points O and I can be imaginary. Reflecting or refracting surfaces that form perfect images are called Cartesian surfaces. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 13

Hong Kong Polytechnic University Geometrical Optics Spherical Mirror: Spherical mirrors may be either concave or convex relative to an object point O, depending on weather the center of curvature C is on the same or opposite side of the surface. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 14

Hong Kong Polytechnic University Geometrical Optics Reflection at a Spherical Mirror: One ray originating at O is normal to the spherical surface at its vertex V and the other is incident arbitrary at P. The intersection of the two rays (extended backward) determines the image point I conjugate to O. The image is virtual, located behind the mirror surface. Similarly, for a concave mirror, we have 1. The object and image distances are negative to the left of the vertex and positive to the right. 2. The radius of curvature R is positive when C is to the right of V (convex mirror) and negative when C is to the left of V (concave mirror). 3. Vertical dimensions are positive above the horizontal axis and negative below. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 15

Hong Kong Polytechnic University Geometrical Optics Focal Length: For an object at infinity, the image distance is the focal length, The focal length is positive for a convex mirror and negative for a concave mirror. The mirror equation is, Lateral magnification: We assign a (+) magnification to the case where the image has the same orientation as the object and a ( ) magnification when the image is inverted relative to the object. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 16

Hong Kong Polytechnic University Geometrical Optics Graphic diagram for spherical mirrors: (a) Rays parallel to the principal axis are reflected though F; (b) Rays passing through F are reflected parallel to the principal axis; (c) Rays passing through C is reflected back along the same path. Real object Image position Nature of image Image size At F Real, inverted Very small Between and C Between F and C Real, inverted Smaller At C Real, inverted Same Between C and F Between C and Real, inverted Larger At F At — — Between F and A Behind mirror Virtual, upright Larger Optics 1 ----by Dr. H. Huang, Department of Applied Physics 17

Hong Kong Polytechnic University Geometrical Optics Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave spherical mirror, each of 10 cm focal length. Determine the position and nature of the image in each case. Solution: (a) Convex mirror: f=10 cm and s= 20 cm. From mirror equation, we get s =6. 67 cm The image is virtual, 6. 67 cm to the right of the mirror vertex, and is erect and 1 cm high. (b): Concave mirror: f= 10 cm and s= 20 cm. From mirror equation we obtain s = 20 cm and, therefore, m= 1 The image is real, 20 cm to the left of the mirror vertex, and is inverted and the same size as the object. Example: A convex rearview car mirror has a radius of curvature of 16 m. Determine the location of the image and its magnification for an object 10. 0 m from the mirror. Solution: The focal length f=R/2=8. 0 m; the object distance s= 10. 0 m. s =4. 4 m; Optics 1 ----by Dr. H. Huang, Department of Applied Physics 18

Hong Kong Polytechnic University Geometrical Optics Example: A 1. 50 -cm-high diamond ring is placed 20. 0 cm from a concave mirror whose radius of curvature is 30. 0 cm. Determine (a) the position of the image, and (b) its size. Solution: The focal length f=R/2= 15. 0 cm; the object distance s= 20. 0 cm. s = 60. 0 cm; ; hi=mho= 4. 5 cm Example: A 1. 00 -cm-high object is placed 10. 0 cm from a concave mirror whose radius of curvature is 30. 0 cm. Determine the position of the image and the magnification. Solution: The focal length f=R/2= 15. 0 cm; the object distance s= 10. 0 cm. s =30. 0 cm; Optics 1 ----by Dr. H. Huang, Department of Applied Physics 19

Hong Kong Polytechnic University Geometrical Optics Homework: 1. Show that if two plane mirrors meet at an angle , a single ray reflected successively from both mirrors is deflected through an angle 2 independent of the incident angle. Assume <90° and that only two reflections, one from each mirror, take place. 2. Show with ray diagrams that the magnification of a concave mirror is less than 1 if the object is beyond the center of curvature C, and is greater than 1 if it is within this point. 3. A beam of light in air strikes a slab of crown glass (n=1. 52) and is partially reflected and partially refracted. Find the angle of incidence if the angle of reflection is twice the angle of refraction. 4. A light beam strikes a piece of glass at a 60. 00° incident angle. The beam contains two wavelengths, 450. 0 nm and 700. 0 nm, for which the index of refraction of the glass is 1. 4820 and 1. 4742, respectively. What is the angle between the two refracted beams? 5. Fermat’s principle states that “light travels between two points along that path which requires the least time, as compared to other nearby paths. ” From Fermat’s principle derive (a) the law of reflection and (b) the law of refraction (Snell’s law) Optics 1 ----by Dr. H. Huang, Department of Applied Physics 20

Hong Kong Polytechnic University Geometrical Optics Curvatures and Vergences: The curvature is usually expressed in reciprocal of radius. Its unit is diopter (m -1). Sign Convention: The curvature is negative if the wave front is diverging and it is positive if the wave front is converging. Specifically we define vergence mirror equation: Mirror power: we have The initial vergence S of the incident wave front at the mirror is modified by the mirror power P to produce the final vergence S of emergent wave front at the mirror. Magnification: Optics 1 ----by Dr. H. Huang, Department of Applied Physics 21

Hong Kong Polytechnic University Geometrical Optics Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave spherical mirror, each of 10 cm focal length. Determine the position and nature of the image in each case. Solution: (a) Convex mirror: P= 1/f= 10 D and S=1/s= 5 D. Then S =S+P= 15 D. So we get s = 1/S =6. 67 cm m=S/S =1/3 The image is virtual, 6. 67 cm to the right of the mirror vertex, and is erect and 1 cm high. (b): Concave mirror: P= 1/f=10 D and S=1/s= 5 D. Then S =S+P=5 D, therefore, s= 1/S = 20 cm m=S/S = 1 The image is real, 20 cm to the left of the mirror vertex, and is inverted and the same size as the object. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 22

Hong Kong Polytechnic University Geometrical Optics Refraction at a Spherical Surface: Employing the same sign convention for mirrors, Lateral magnification: Optics 1 ----by Dr. H. Huang, Department of Applied Physics 23

Hong Kong Polytechnic University Geometrical Optics Example: A real object is positioned in air, 30 cm from a convex spherical surface of radius 5 cm. To the right of the interface, the refractive index is 1. 33. What is the image distance and lateral magnification of the image? Solution: s= 30 cm and R=5 cm. This gives s =40 cm which means the image is real. indicating an inverted image. As shown in the figure, suppose now the second medium is only 10 cm thick, forming a thick lens, with a second, concave spherical surface, also of radius 5 cm. The refraction by the first surface will be unaffected by this change. Inside the lens, before a real image is formed, rays are intercepted and again refracted by the second surface to produce a different image. We have s 2=30 cm and R 2= 5 cm. So, s 2 =9 cm. It is a real image. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 24

Hong Kong Polytechnic University Geometrical Optics Vergence in Refraction: Vergence of incident wave front arriving at the spherical surface: Vergence of emerging wave front leaving the spherical surface: Power of spherical surface: Then we have and Example: Repeat the example on previous slide. For the first spherical surface, we have s= 30 cm, R=5 cm, n 1=1 and n 2=1. 33. S=n 1/s= 3. 33 D; P=(n 2 n 1)/R=6. 67 D; S =S+P=3. 33 D; s =n 2/S =0. 4 m m=S/S = 1 For the second spherical surface, we have s 2=30 cm, R= 5 cm, n 1=1. 33 and n 2=1. S 2=n 1/s 2=4. 44 D; P=(n 2 n 1)/R=6. 67 D; S 2 =S 2+P=11 D; s 2=n 2/S 2 =0. 09 m m=S 2/S 2 =0. 4 Optics 1 ----by Dr. H. Huang, Department of Applied Physics 25

Hong Kong Polytechnic University Geometrical Optics Thin Lenses: 1 st refracting surface 2 nd refracting surface For thin lens, s 1 =s 2, then Thin lens equation: Focal length: This is the lensmaker’s equation. simplified lens equation: Optics 1 ----by Dr. H. Huang, Department of Applied Physics 26

Hong Kong Polytechnic University Geometrical Optics Ray Tracing: can be used to locate the image of an object. Real image: the rays after refraction actually intersect. Virtual image: the rays do not intersect after refraction. The focal length is positive for converging lenses and negative for diverging lenses. Lateral Magnification: Example: What is (a) the position and (b) the size of a large 7. 6 -cm-high flower placed 1. 00 m from a +50. 0 -mm-focal length camera lens? Solution: f=5. 00 cm; s= 100 cm; ho=7. 6 cm s =5. 26 cm; Optics 1 ----by Dr. H. Huang, Department of Applied Physics 27

Hong Kong Polytechnic University Geometrical Optics Example: Find and describe the intermediate and final images produced by a two-lens system as show in the figure. Both are 15 -cm focal length lenses and their separation is 60 cm. Let the object RO 1 be 25 cm from the first lens. Solution: 1 st lens f 1=15 cm, s 1= 25 cm s 1 =37. 5 cm The first image is real, 37. 5 cm to the right of the first lens, inverted and 1. 5 times the size of the object. For 2 nd lens, f 2= 15 cm, s 2= (60 s 1 )= 22. 5 cm s 2 = 9 cm, The final image is virtual, 9 cm to the left of the second lens and inverted. The overall magnification is m=m 1 m 2= 0. 6. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 28

Hong Kong Polytechnic University Geometrical Optics Vergence of the Thin Lens: The power of the lens as a whole is just the sum of the powers of the two surfaces. The lateral magnification is Example: Repeat the previous one. Solution: 1 st lens: f 1=15 cm, s 1= 25 cm, S 1=n 1/s 1= 4 D, P 1=n 1/f 1=6. 67 D S 1 =S 1+P 1=2. 67 D; s 1 =n 1/S 1 =0. 375 m; m 1=S 1/S 1 = 1. 5 2 nd lens: f 2= 15 cm, s 2= 22. 5 cm, S 2=n 1/s 2= 4. 44 D, P 2=n 1/f 2= 6. 67 D S 2 =S 2+P 2= 11. 11 D; s 2 =n 2/S 2 = 0. 09 m; m 2=S 2/S 2 =0. 4 Example: A convex meniscus lens is made from glass with n=1. 50. The radii of the two surfaces are 22. 4 cm and 46. 2 cm, respectively. (a) What is the focal length? (b) Where will it focus an object 2. 00 m away? Solution: R 1=22. 4 cm and R 2=46. 2 cm. (a) Using the lensmaker’s equation, we can easily get f=87 cm. (b) s= 2 m, therefore, s =1. 54 m. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 29

Hong Kong Polytechnic University Geometrical Optics Thin Lenses in Contact: The vergence approach is of particular importance and interest to optometrists. Consider two thin lenses in contact Since s 1 =s 2, so several thin lenses in contact Newtonian Equation for the Thin Lens: Optics 1 ----by Dr. H. Huang, Department of Applied Physics 30

Hong Kong Polytechnic University Geometrical Optics Example: An object is placed 10 cm from a 15 -cm-focal-length converging lens. Determine the image position and size. Solution: f=15 cm and s= 10 cm; s = 30 cm Example: Where must a small insect be placed if a 25 -cm-focal-length diverging lens is to form a virtual image 20 cm in front of the lens? Solution: f= 25 cm and s = 20 cm s= 100 cm Example: To measure the focal length of a diverging lens, a converging lens is placed in contact with it as shown. The Sun’s rays are focused by this combination at a point 28. 5 cm behind the lenses. If the converging lens has a focal length f. C of 16. 0 cm, what is the focal length f of the diverging lens? Assume both lenses are thin and the space between them is negligible. Solution: fd= 36. 5 cm Optics 1 ----by Dr. H. Huang, Department of Applied Physics 31

Hong Kong Polytechnic University Geometrical Optics Homework: 1. It is desired to magnify reading materials by a factor of 2. 5 when a book is placed 8. 0 cm behind a lens. (a) What type of lens is needed for this? (b) What is the power of the lens in diopters? 2. Two 27. 0 -cm-focal-length converging lenses are placed 16. 5 cm apart. An object is placed 35. 0 cm in front of one. Where will the final image formed by the second lens be located? What is the total magnifications? 3. A prescription for a corrective lens calls for +2. 50 D. The lensmaker grinds the lens from a “blank” with n=1. 56 and a performed convex front surface of radius of curvature of 20. 0 cm. What should be the radius of curvature of the other surface? 4. Two identical, thin, plano-convex lenses with radii of curvature of 15 cm are situated with their curved surfaces in contact at their centers. The intervening space is filled with oil of refractive index 1. 65. The index of the glass is 1. 50. Determine the focal length of the combination. (Hint: Think of the oil layer as an intermediate thin lens. ) 5. A sphere 5 cm in diameter has a small scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where does the scratch appear and what is its magnification? Assume n=1. 50 for the glass. Solve by both image location and vergence models. Optics 1 ----by Dr. H. Huang, Department of Applied Physics 32