Ray optics! 1. Postulates of ray optics! 2. Simple optical components! 3. Graded index optics! 4. Matrix optics!!

Transcription

1 Ray optics! 1. Postulates of ray optics! 2. Simple optical components! 3. Graded index optics! 4. Matrix optics!!

2 From ray optics to quantum optics! Ray optics! Wave optics! Electromagnetic optics! Quantum optics!

3 Ray optics is the simplest (yet powerful) theory of light! Light is described by rays that travel in different optical media in accordance with a set of geometrical rules. Ray optics is therefore also called geometrical optics.! Use it when the dimensions involved are much greater than the wavelength of light. Ray optics is the limit of wave optics when the wavelength is infinitesimally small.

4 What is a ray?! A light ray can be defined by two co-ordinates: y! y θ! z Ray bundles energy density is proportional to the density of rays

5 Ray tracing! Computers are capable of tracing the behavior of many light rays illuminating objects. This allows the creation of realistic 3D graphics.! from the simple to the very complex.

6 Ray tracing software! There are a few industry standard optical design programs such as Zemax and Code V which are used to design modern, complex optical systems.!

7 1. Postulates of ray optics! Light travels in the form of rays They are emitted by light sources and can be observed when they reach an optical detector.!! An optical medium is characterized by a quantity n > 1, called the refractive index.! n = c 0 c d c = nd c 0 c 0 = speed of light in free space c = speed of light in the medium. Optical pathlength: the time taken by light to travel a distance d.

8 The refractive index is characteristic of each medium! medium! speed of light (km/s)! refractive index! vacuum! ! 1.00! air! ! ! water! ! 1.33! glass! ! 1.52! diamond! ! 2.42!

9 Postulates of ray optics! The optical pathlength between points A and B is! optical pathlength = B A n(r)ds n(r) = refractive index at position r = (x,y,z)!! The time taken by light to travel from A to B is proportional to the pathlength.!!

10 Postulates of ray optics! Fermat s principle Optical rays traveling between A and B foliow a path such that the or the time of travel between the two points is an extremum relative to neighboring paths:!!!!!! B δ n(r)ds = 0 A Light rays travel along the path of least time! Actually we don t need to use Fermat s principle all the time. Three simple rules are applicable for most basic optical components.

11 Rule #1: Propagation in a homogeneous medium! Light travels in straight lines. Shadows are perfectly defined.!

12 Rule #2: Reflection from a mirror! The reflected ray lies in the plane of incidence; the angle of reflection equals the angle of incidence.!

13 Rule #3: Reflection and refraction at the boundary between two media! The refracted ray lies in the plane of incidence; the angle of refraction θ 2 is related to the angle of incidence θ 1 by Snells law:! n 1 sinθ 1 = n 2 sinθ 2 (from Fermat s principle: see FP Ex ) Those are the three rules. Now let s study some examples of simple optical components.!

14 2. Simple optical components! A Mirrors! B Planar boundaries! C Spherical boundaries and lenses! D Light guides! y R < 0 R > 0 θ > 0 θ < 0 θ > 0 θ < 0 z concave convex

15 Relevant demonstrations! Tracing Rays Reflected from a Spherical Mirror! Reflections in an Elliptical Region! Paraxial Approximation and the Mirror! Total Internal Reflection! Lens Aberrations! Light Rays in a Lens! Lensmakers Equation!!

16 (A) Planar mirrors! Rays originating from a point P 1 appear to originate at the image point P 2!

17 (A) Paraboloidal mirrors! All incoming rays parallel to the axis are focused to a single point called the focus.! Parabolic mirrors play an important role in solar energy and other fields e.g. astronomy, illumination...!

18 (A) Elliptical mirrors! All rays emitted from one focus are imaged onto the other focus: all optical paths are equal!! A typical application is to collect light from a source and focus it at a single point.!

19 (A) Spherical mirrors! These mirrors are a lot easier to manufacture than paraboloidal or elliptical ones; but they only work well close to the axis.! This is the domain of paraxial optics.!

20 In the paraxial limit:! sinθ tanθ θ ( θ 2 ) + θ 1 2y ( R) Paraxial equations for spherical mirrors! z 1 z 2 R Focal ( ) f = ( R ) length:! = 1 z 1 z 2 f Every point in the object plane has a corresponding point in the image plane. Rays coming from will be focused at the focal length.!

21 (B) Planar boundary! Two media of different refractive indices n 1 and n 2 :! n 1 sinθ 1 = n 2 sinθ 2 For paraxial rays the Snell equation may be linearized:! n 1 θ 1 n 2 θ 2 n 1 < n 2 n 1 > n 2

22 Total internal reflection (TIR)! For internal refraction (n 1 > n 2 ) there is a critical angle at which no refraction occurs.! θ c = sin 1 n 2 n 1 e.g. medium-to-air (n~1.0):!! water (n~1.33): θ c = 48.7º! glass (n~1.5): θ c = 41.8º!!

23 (B) Prisms! Prism of apex angle α and refractive index n:! θ d = θ α + sin 1 ( n 2 sin 2 θ )sinα sinθ cosα (n 1)α

24 (C) Spherical boundaries! Just apply Snell s law at the boundary + geometrical considerations to determine the relevant angles to the surface normal (= radius vector).! θ 2 n 1 n 2 θ 1 y f Every point in the object plane has a corresponding point in the image plane.!! Rays coming from will be focused at the focal length:! f = n 2 n 2 n 1 R

25 (C) Lenses! A spherical lens can be considered a succession of two spherical (or one spherical and one flat) surfaces with radii R 1 and R 2.! What is the sign of R 1, R 2? Different applications may require different types of lenses.!

26 (C) Lenses! The calculations can be greatly simplified if we assume that the lens is thin (y 1 =y 2 )! θ 2 = θ 1 y f the focal length is given by! 1 f = (n 1) 1 & 1 R 1 R 2 ) ( 0.5& 1 1 R 1 R 2 ) ( biconvex (R 1 = R 2 ):! f R plano-convex (R 2 = ):! f 2R

27 Imaging formation by a thin lens! Particular cases of interest:!! z! 1 = f z 2 = z 1 = f 2 z 2 = f 2 M = z 2 z 1 = 1 z 2 = f z 1 = Imaging equation! 1 f = z 1 z 2 Magnification! y 2 = z 2 z 1 y 1

28 (D) Light guides! Optical guiding by total internal refraction (n 2 <n 1 ):! Condition for guiding:! θ > θ c = sin 1 n 2 n 1

29 Exercise! Numerical aperture and angle of acceptance of an optical fiber! show that the numerical aperture is given by:! NA = sinθ a = n 1 2 n 2 2

30 3. Graded index optics! A graded index material such as a GRIN lens has a refractive index that varies with position: n = n(r).!

31 Graded index optics! In these media it is no longer possible to use the 3 Rules we must use Fermat s Principle.! B δ n(r) ds = 0 This leads to the Ray Equation. A d " ds n dr = n # ds & Ray Equation

32 The paraxial ray equation! We can describe the trajectory of a ray by two functions x(z) and y(z) such that! ds = dz 1+ (dx / dz) 2 + (dy / dz ) 2 In the paraxial approximation ds dz, the ray equation can be simplified to! d " dz n dx # dz n & x, d " dz n dy # dz n & y Paraxial Ray Equation Given n = n(x,y,z), these two partial equations may be solved for the trajectory x(z) and y(z).!

33 The paraxial ray equation - examples! d " dz n dx # dz n & x, d " dz n dy # dz n & y a) n=constant:! d 2 x dz 2 = d 2 y dz 2 = 0 the trajectories are straight lines! b) n=n(y):! d " dz n dy # dz = dn & dy d 2 y dz 2 = 1 n(y ) dn(y ) dy

34 Example 1: slab with a parabolic index profile! n 2 (y) = n 2 ( o 1 α 2 y 2 ) d 2 y dz 2 α2 y y(z) = y 0 cosαz + θ 0 α sinαz θ(z) = dy dz = y 0 α sinαz + θ 0 cosαz A glass slab with this profile has the commercial name of SELFOC

35 Homework: using a GRIN slab as a lens! f 1 n 0 α sinαd AH tan αd / 2 ( ) n 0 α Show that a SELFOC slab of length d < π/2a acts as a cylindrical lens (a lens with focusing power in the y-z plane) by deriving the expressions for the focal length f and the principal point H.!! More information:! Index_lenses.pdf!

36 Example 2: graded-index fiber! This is a generalization of the previous example to x and y:! [ ( )] n 2 (y) = n o 2 1 α 2 x 2 + y 2 d 2 x dz 2 α2 x, d 2 y dz 2 α2 y The solutions are periodic (2π/α)! x(z) = θ x 0 α sinαz y(z) = θ y 0 α sinαz + y 0 cosαz

37 Exercise: numerical aperture of the graded-index fiber! Show that in the paraxial approximation the numerical aperture is:! NA sinθ a n 0 aα Compare with the previous results for the step-index fiber, assuming that n 1 = n 0 n 2 = n 0 1 α 2 a 2 n ( α2 a 2 )

38 4. Matrix optics! It is a technique for tracing paraxial rays! Rays are described by position and angle! The output ray is related to the input ray by two algebraic equations! y 2 = Ay 1 + Bθ 1 θ 2 = Cy 1 + Dθ 1 y 2 θ 2 The ray-transfer (or ABCD) matrix = A B C D y 1 θ 1

39 The ray-transfer matrix! What is the meaning of the ABCD?! y 2 = # y & # 2 ( y y 1 + y & 2 ( θ 1 θ 1 1 # θ 2 = θ & # 2 ( y y 1 + θ & 2 ( θ 1 θ 1 1 y 2 θ 2 # y 2 y 1 & ( = A B C D θ 2 & ) y 1 ( & & y 2 θ 1 ) ( y 1 θ 1 θ 2 θ 1 ) (

40 a) Ray-matrix for free space propagation! If y 1 and θ 1 are the position and slope upon entering, let y 2 and θ 2 be the position and slope after propagating from z = 0 to d.! y 1, θ 1! y 2, θ 2! y 2 = y 1 + d θ 1 θ 2 = θ 1 0! " M space = 1 d # 0 1 & d! Rewriting this in matrix notation:! # y 2 θ 2 & ( = # 1 d &# 0 1 ( y 1 θ 1 & (

41 b) Ray-matrix for reflection from a mirror! At the interface we have: y 2 = y 1 θ 2 = θ 1 θ 2! θ 1! y 1! y 2! In matrix form: # y 2 θ 2 & ( = # 1 0 &# 0 1 ( y 1 θ 1 & " ( M mirror = 1 0 # 0 1 &

42 c) Ray-matrix for refraction at a planar boundary! At the interface we use Snell s law in the paraxial approximation: θ 1! y 1! y 2! θ 2! y 2 = y 1 n 2 θ 2 = n 1 θ 1 n 1! n 2! # In matrix form: y 2 θ 2 & ( = # 1 0 &# 0 n 1 ( y 1& " n 2 θ ( M boundary = n 1 # n 2 &

43 d) Ray-matrix for refraction at a spherical boundary! As we saw before: y 2 = y 1 θ 1! y 1! θ 2! θ 2 n 1 θ n 1 y 1, f = 2 f n 2 n 2 n 1 R n 1! n 2! In matrix form: # y 2 θ 2 & ( = # 1 0 n 2 n 1 n 1 n 2 R n 2 &# ( y 1& θ ( M spher. bound. = 1 # 1 0 & n 1 ( n 2 R n 2 n 2 n 1

44 e) Ray-matrix for transmission through a thin lens! Again, as we saw before: y 2 = y 1 θ 1! θ 2! y 1! y 2! θ 2 = θ 1 y f 1 f = (n 1) 1 In matrix form: # y 2 θ 2 ( R 1 1 ) R 2 & ( = # 1 0 &# 1 f 1 ( y 1 θ 1 & # ( M lens = 1 0 & 1 f 1 (

45 f) Ray-matrix for reflection from a spherical mirror! Again, as we saw before: y 2 = y 1 ( θ 2 ) θ 1 + y f θ 2! θ 1! <<! y 1! y 2! ( f = R ) 2 R! In matrix form: # y 2 θ 2 & ( = # 1 0 &# 2 R 1 ( y 1 θ 1 & " ( M spherical mirror = # R 1 &

46 All ray-matrices for imaging components are similar! focal length ray matrix Spherical boundary f = n 2R # 1 0 n n 2 n 1 1 n 2 n 1 n 2 R n 2 & ( Spherical mirror Spherical lens (also the general expression) ( f = R ) 2 f = f " # R 1 & # 1 0& 1 f 1 (

47 Matrices of cascaded optical components! This is typically the case of interest. A real optical system normally has more than one single component! This is the main reason why matrices are so useful.!... M 1 M 2 M N The total ray transfer matrix of the system is (note the order):! " M system = M N M 2 M 1 = A N # C N B N D N & " A 2 B 2 # C 2 D 2 & " A 1 B 1 # C 1 D 1 &

48 Exercise: collimation using a lens! Calculate the matrix and the output ray for an input ray starting at y 1 =0 and at a distance f from a lens.! # M lens = 1 0 & 1 f 1 ( " M space = 1 f # 0 1 & # M system = 1 0 & 1 f 1 ( # 1 f & 0 1 ( = # 1 f & 1 f 0 ( f y 2 = y 1 + fθ 1 = fθ 1 θ 2 = 1 f y 1 = 0

49 Periodic optical systems! = a cascade of identical unit systems. If M=[ABCD] is the matrix for each unit system, then! # y m θ m & ( = # A B & C D ( m # y 0 θ 0 & ( y m +1 = Ay m + Bθ m θ m +1= = Cy m + Dθ m This is solved for y m by eliminating θ m and obtaining a difference equation.! y m+2 = 2by m+1 F 2 y m F 2 = AD BC = det[m] b = (A + D) / 2 y m = y 0 F m exp ( ±imϕ) ϕ = cos 1 ( b /F )

50 Condition for stability! It can be shown that a general solution for a system starting and ending in air (n=1) is given by! y m = y max sin( mϕ + ϕ ) 0 For y m to be a harmonic function, φ must be real (condition of stability):! b A + D 1

51 Exercise! Determine the condition of stability for an optical resonator composed of two spherical mirrors R 1, R 2, separated by d.!