ROINN NA FISICE Department of Physics

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1 ROINN NA FISICE Department of 1.1

2 Astrophysics Telescopes Profs Gabuzda & Callanan 1.2

3 Astrophysics Faraday Rotation Prof. Gabuzda 1.3

4 Laser Spectroscopy Cavity Enhanced Absorption Spectroscopy Prof. Ruth 1.4

5 Quantum Drs. Pelucchi and Ruschhaupt 1.5

6 Optoelectronics Prof. McInerney Dr. Kelleher Lasers and Non-Linear 1.6

7 Photonic Systems Prof. Townsend Dr. Gunning Fibre Optic Communications 1.7

8 Integrated Photonics Peters 1.8

9 Geometric 3 Laws of Geometric : 1) Law of Rectilinear Propagation Light goes in a straight line through a homogeneous medium 2) Law of Reflection Describes how light reflects 3) Law of Refraction Describes how light acts moving from one medium into another 1.9

10 The Shadow a.) Umbra total Darkness Source Obstacle Screen Penumbra Partial Darkness b.) Obstacle Source Screen 1.10

11 Geometric Path It is the geometrical distance s measured along a path between any two points A and B. B s = න ds A where ds is the differential unit of length ds = dx 2 + dy 2 + dz

12 Index of Refraction The refractive index characterizes the apparent speed of light on the medium of propagation. n v c speedof speedof light in vacuum light in medium In homogeneous medium: n = constant In isotropic but inhomogeneous medium: n varies with (x, y, z) In anisotropic medium, n also varies with polarization 1.12

13 Optical Path Length The time dt it takes for light to propagate a geometrical length ds is: dt = ds v = n c ds To go from A to B it takes: T = න A The optical path length L is defined as B Λ = ct = න nds A Λ has units of length (e.g. mm) but intrinsically represents a time of propagation as well B dt = 1 c න A B nds 1.13

14 Geometric Wavefront Def: Given rays emanating from a single point source, a geometrical wavefront is a surface that is the locus of constant optical path length from the source (i.e. wavefronts are equiphase surfaces). If the source is located at X o, Y o, Z o and light leaves at t o, the wavefront at time t satisfies. Λ r = c t t 0 r = න n x, y, z ds

15 Wavefronts 1 2 A In an homogeneous media, the wavefront emanating from a single point is a sphere to all rays emanating from A When A, the light rays become, the surface becomes a plane. 1.15

16 Huygens Principle (1678) In 1678 Christian Huygens expressed his intuitive convictions that each point on a wavefront can be considered to be a secondary spherical wavelet, and the wavefront at a later instant can be found by constructing the envelope of these wavelets. We now recognize this concept as a description of the convolution of an optical disturbance with the impulse response of the diffraction process. a.) Spherical wavefront emanating form a point source. b.) Propagation of an arbitrary extended wavefront. c.) Truncated plane wave. * 1.16

17 Huygens Principle Christian Huygens ( ) Every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets. 1.17

18 Huygens Principle What good is it? B C AB is a wavefront at T1 A θ i θ r D CD and DE are wavefronts at T2 E θ t ΔT = T2 T1 V i ΔT = BD = AC V t ΔT = AE 1.18

19 Huygens Principle 1.19

20 Hero Hero of Alexandria (region of 0 AD) The actual path taken by light in going from some point S to a point P via a reflecting surface was the shortest possible one. This explains reflection 1.20

21 Hero and Reflection w h θ i θ r x L x = 0 L = h 2 + x 2 + h 2 + w x

22 Fermat Pierre de Fermat ( ) The actual path between two points taken by a beam of light is the one which is traversed in the least time. This explains both reflection and refraction Hero used the shorted distance, Fermat time. 1.22

23 Fermat and Refraction w h θ i Remember: n = c v v i = c n i x b t = h2 + x 2 v i + b2 + w x 2 v t t x = 0 θ t 1.23

24 Reflection Revisited A w B h θ i θ r x Isn t the shortest path between A & B the straight one? Thus Fermat s principle needs to be restated to say, that light takes a Path that results in a locally stationary, or an extremum: Maximum, Minimum or Stationary 1.24

25 OPL Reflection Revisited Λ n i = w Λ n i = h 2 + x 2 + h 2 + w x 2 Optical Path Length OPL h=5 w= X 1.25

26 Extremum An extremum path can be a minimum, a maximum, or a stationary path. Perfect Lens 1.26

27 Geometric The Law of Rectilinear Propagation Light goes straight in a homogeneous media The Law of Reflection The angle of incidence equals the angle of reflection. The Law of Refraction The angle of refraction obeys Snell s Law n i sin θ i = n t sin θ t 1.27

28 Applications of Fermat We will apply Fermat s Theorem to consider the ideal shape of various mirrors and lenses. We are interested in the stationary condition. i.e. don t take a derivative to find a minima/maxima! This is where all paths have the same optical path length Λ. 1.28

29 Mirrors What is a perfect mirror? One point is perfectly imaged onto plane wave. (As used in a lighthouse) According to Fermat: 2A + C = const = C + x + x 2 + y 2 A 1.29

30 Mirrors What is a perfect mirror? One point is perfectly imaged onto another point. According to Fermat: const = d + c + d c = 2d = (x + c) 2 + y 2 + x c 2 + y 2 origin B c c d A 1.30

31 Mirrors origin B c c A 1.31

32 Lenses What is a perfect lens? Light from a point source focused in glass According to Fermat: n 1 d + x 2 + y 2 + n 2 L x 2 + y 2 = n 1 d + n 2 L This unfortunately does not simplify! d L But, there are some special conditions that do simplify x 1.32

33 Lenses What is a perfect lens? Light from a point is converted into a plane wave in glass According to Fermat: d n 1 n 2 L n 1 d + n 2 L = n 1 d + x 2 + y 2 + n 2 L x x 1.33

34 Lenses 1.34

35 Lenses What is a perfect lens? Light from a plane wave is focused in glass According to Fermat: n 1 d + n 1 x + n 2 L x 2 + y 2 = n 1 d + n 2 L L n 1 x n

36 Curvature A measure of how fast the angle changes: i.e. For a general curve: Thus, to solve, we first do the following: Now: ds = dx 2 + dy 2 x = x t ; y = y(t) κ = dφ dt κ = dφ ds dt ds = dφ dt ds dt So: ds dt = dx dt 2 + dy dt

37 Curvature We write: To solve for: Thus: = d dt κ = dφ dt dx dt dφ dt 2 + dy dt 2 = tan φ = dy dx = dy dt dx dt dφ dt x 2 + y 2 = y x d dt tan φ = 1 + tan2 φ dφ dt y x = y x x 2 + y x = x y x y x

38 Curvature dφ dt = tan 2 φ = y x 2 x y x y x 2 x y x y x 2 = x y x y x 2 + y 2 So: κ = x y x y x 2 + y 2 x 2 + y 2 = x y x y x 2 + y

39 Curvature of a circle A circle is described by the following equation: x 2 + y 2 = R 2 For a general curve: So, we can write for the circle: x = x t ; y = y(t) x = R sin t, y = R cos t x = R cos t, y = R sin t x = R cos t, y = R sin t κ = x y x y x 2 + y = R2 sin 2 t + cos 2 t R 3 sin 2 t + cos 2 t 3 2 = 1 R 1.39

40 Terminology Curvature: κ = 1 R Take a parabola: At x = y = 0, the Curvature: y = x 2 x = t ; y = t 2 κ = x y x y x 2 + y x = 1 x = 0 y = 2t y 0 = 0 y = 2 So for this special case: κ = y =

41 Terminology Curvature: κ = 1 R Take a parabola: y = x 2 x = t ; y = t 2 At x = y = 0, the Curvature: κ = 2 = 1 R (for a circle) Therefore a Circle of Radius 0.5 approximates this parabola: How well does a circle approximate other Cartesian surfaces? Parabola Ellipse Hyperbola y = x

42 Comparison of Surfaces Cartesian Surfaces Parabola Circle 2 Ellipse 1.5 Hyperbola

43 Comparison of Surfaces Cartesian Surfaces Parabola 0.12 Circle 0.1 Ellipse 0.08 Hyperbola

44 Reality Check Normally our optical axis is in the x-direction Here, the surface is described by: The curvature at x = y = 0: y = t; x = ct 2 0,0 x = cy 2 x κ = x y x y x 2 + y = 1 R = d2 x dt 2 = 2c 1.44

45 Spherical Optical Elements While high quality aspherical optics are very common today, this is a very recent phenomenon. Earlier it was only possible to make high quality spherical optics. How does one use a spherical mirror or lens, and what are the drawbacks? 1.45

46 Spherical Lenses - Fermat l 0 s 0 φ R s i l i Use law of cosines: c 2 = a 2 + b 2 2ab cos C And: cos 180 φ = cos φ l o 2 = s o + R 2 + R 2 2R s o + R cos φ l 2 i = s i R 2 + R 2 + 2R s i R cos φ Λ = n 1 l o + n 2 l i φ Λ =

47 Spherical Lenses - Fermat l 0 φ R l i s 0 s i 1.47

48 Spherical Lenses - Snell α s o φ α θ i n 1 n 2 h R φ s i n 1 sin θ i θt β = n 2 sin θ t θ i = α + φ; θ t = 180 β 180 φ = φ β n 1 sin α + φ = n 2 sin φ β n 1 sin α cos φ + cos α sin φ = n 2 sin φ cos β cos φ sin β 1.48

49 Spherical Lenses - Snell 1.49

50 Spherical Lenses Using Fermat and Snell we calculated: s 0 n 1 n R 2 φ s i Time for some discussion n 1 s o + n 2 s i = 1 R n 2 n

51 Spherical Lenses - Comments R f = focal length s o s i n 1 + n 2 s i = 1 R n 2 n 1 s i = f i = n 2 n 2 n 1 R s 0 R s i n 1 s o + n 2 = 1 R n 2 n 1 s o = f o = n 1 n 2 n 1 R 1.51

52 Spherical Lenses - Comments s i < 0 h R n 1 s 0 + n 2 s i = 0 1 s o = n 2 n 1 1 s i s 0 This goes back to Snell s Law: h = tan θ s 1 sin θ 1 = h n 2 = n 2 tan θ o s i n 1 n 2 n 2 sin θ 1 n 2 1 s 0 = -s i n 1 s o n 2 s o = 1 R n 2 n 1 R = s o 1.52

53 Thin Lens d s i1 s o1 s i2 s o2 n 1 s o1 + n 2 s i1 = 1 R 1 n 2 n 1 n 2 s o2 + n 1 s i2 = 1 R 2 n 1 n 2 Note: s i1 < 0 s o2 = s i1 + d 1.53

54 Thin Lens Thin lens as: d 0 1 s o + 1 s i = n 2 n 1 n 1 1 R 1 1 R

55 Thin Lens Equation 1 s o + 1 s i = n 2 n 1 n 1 1 R 1 1 R 2 s o or s i 1 f = n 2 n 1 n 1 1 R 1 1 R 2 The Lensmakers Formula Therefore: 1 s o + 1 s i = 1 f The Gaussian Lens Formula 1.55

56 Lens Sign Conventions O s o θ i V R θ t φ C s i 1 s o + 1 s i = 1 f β P The following rules must be followed in using this equation: 1. Assume that light propagates from left to right. Object distance s o is positive when point O is to the left of point V. 2. Image distance s i is positive when P is to the right of V (real image) and negative when to the left of V (virtual image). 3. Lens radius of curvature R is positive for C to the right of V (convex), negative for C to left of V (concave). 1.56

57 Spherical Mirrors θ = α + φ 2θ = α + β θ φ α α s o θ β s i R φ 1 s o + 1 s i = 2 R 1.57

58 Mirrors continued Let s examine the equation: 1 s o + 1 s i = 2 R When light comes from infinity: s i = 2 R 1 And: f = 2 R f = R 2 1 s o + 1 s i = 1 f : the same as for a lens. θ θ s o R s i = f 1.58

59 Mirror Sign Conventions The following rules must be followed in using this equation: 1. Assume that light propagates from left to right. Object distance s o is positive when point O is to the left of point V. 2. Image distance s i is positive when I is to the left of V (real image) and negative when to the right of V (virtual image). 3. Mirror radius of curvature R is positive for C to the right of V (convex), negative for C to left of V (concave). R O α V I C s o s i 1.59

60 Simple Rules for Image Construction Use two (or more) rays to construct an image Light ray parallel to the principal axis refracted through the focal point Light ray passing straight through center of lens refracted undeviated Light ray passing through focal point refracted parallel to the principal axis. y o object x o f f x i s o image y i si 1.60

61 Magnification f>0 object y o x o f f x i 1 s o + 1 s i = 1 f s o image y i s i y i f = y o s o f = y o x o y i s i = y o s o y o f = y i s i f = y i x i Magnification: M T = y i = f = s i = x i y o x o s o f f 2 = x o x i 1.61

62 Magnification f<0 object 1 y o s o + 1 s i = 1 f f i y i x o s o x i s i f o y i f = y o s o f = y o x o y i s i = y o s o y o f = y i s i f = y i x i Magnification: M T = y i = f = s i = x i y o x o s o f As before f 2 = x o x i 1.62

63 Lens Image Summary Convex Lens Object Image Location Type Location Orientation Relative Size >s 0 >2f Real f<s i <2f Inverted Minified s o =2f Real s i =2f Inverted Same Size f<s o <2f Real >s i >2f Inverted Magnified s o =f ± s o <f Virtual s i >s o Erect Magnified Concave Lens Object Image Location Type Location Orientation Relative Size Anywhere Virtual s i < f Erect Minified 1.63

64 Mirror Image Summary Convex Mirror Object Image Location Type Location Orientation Relative Size Anywhere Virtual s i < f Erect Minified Concave Mirror Object Image Location Type Location Orientation Relative Size >s 0 >2f Real f<s i <2f Inverted Minified s o =2f Real s i =2f Inverted Same Size f<s o <2f Real >s i >2f Inverted Magnified s o =f ± s o <f Virtual s i >s o Erect Magnified 1.64

65 Magnifying Glass 1.65

66 Definitions Aperture Stop Field Stop Image Plane The larger the aperture the more light collected. Flux r 2 D 2 D D D 1.66

67 Camera Lenses f Image Area f 2 The larger the focal length, the larger the image size. f 1.67

68 Definitions Flux Density D2 f 2 Relative Aperture = D f f# = f D An exposure time will require a certain amount of optical energy. Exposure = Flux Density Exposure Time Exposure Time f#

69 Smartphone vs Camera Lens Samsung S7 EF-S 17-55mm Maximum f/# f/1.7 Maximum f/# f/2.8 What is the difference? 1.69

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