PH880 Topics in Physics

Transcription

1 PH880 Topics in Physics Modern Optical Imaging (Fall 2010)

2 The minimum path principle n(x,y,z) Γ Γ has the minimum optical path length, compared to the alternative paths. nxyzdl (,, ) Γ

3 Thelaw of reflection θ i = θ r The law of refraction (a.k.a Snell s law) n sin θ = n 1 i 2 sin θ t reflected transmitted θ r θ t θ i incident n 1 n 2

4 Overview of week 2 Monday Paraxial approximation Ray tracing /Matrix formulism NA / Field Stop/ Window Wednesday Fourier Optics

5 Paraxial approximation 1. Existence of an optical axis (i.e. perfect alignment!) optical axis 2. small angle to the optical axis (1 st order Tayler) (up to θ << 1 rad, ~ degrees) sinθ θ tanθ θ θ cosθ 1 (sometimes, 1 ) 2 θ 1+ θ

6 Paraxial approximation 3. Simple geometry Ignore distance

7 Sign convention for geometric optics 1. Light travels to right 2. A radius of curvature is positive if the surface is convex towards the left 3. Longitudinal distances are positive if pointing to the right 4. Lateral distances are positive if pointing up 5. Ray angles are positive if the ray direction is obtained by rotating the +z axis counterclockwise through an acute angle x α R optical axis +z

8 (paraxial) Ray tracing Input ray output ray n' n x 0 α 0 x 1 α 1 Optical System For an arbitrary ray entering, determine output ray, upon the optical system

9 Ray tracing : propagation thru. distant D D α 2 α 1 x 2 x 1 optical axis x = x + Dα α = α 2 1

10 Ray tracing : refraction at spherical surface n 1 n 2 α 2 α 1 x x 2 1 x α = x 2 1 n 1 n n = α1 x1 n2 n 2 R Power of the surface

11 Ray tracing : an example (onespherical surface) Objective: find α 2 and x 2 from α 0, x 0 propagation + refraction + propagation D 01 D 12 α 2 α 0 x 2 x 0 α 0 X non paraxial ray High angle (large error) R

12 Ray tracing : an example translation + refraction + translation n n' x 0 α 0= α 1 α' 1 =α 2 x 2 x 1 x' 1 D 01 D 12 x 0 α 0 1. Starting ray: location direction x1 = x0 + D01α 0 2. Propagation through D 01 : α1 = α0 3. Refraction : x2 = x1 n1 1 n2 n1 α = α x n2 n 2 R x2 = x1+ D12α 0 4. Propagation through D 12 : α2 = α1

13 Ray tracing : an example translation + refraction + translation n n' x 0 α 0= α 1 α' 1 =α 2 x 2 x 1 x' 1 5. Put together: D 01 D 12 Note: calculating ray tracing is very tedious, and there s an easier way

14 Power of surfaces Positive power bends rays inwards (converging) 1 n R>0 P n right n left n 1 1(+) R = R (+) > 0 n 1 nright nleft 1 n ( ) P = > R R ( ) R<0 0

15 Power of surfaces Negative power bends rays outwards (diverging) 1 n P n right n left n 1 1(+) R = R (-) < 0 R<0 n 1 nright nleft 1 n ( ) P = < R R ( + ) R>0 0

16 Focal length & Power f (+) P n (+) = > f (+) 0 1 n f ( ) P n (+) = < f (-) 0 n 1 * Unit of power : diopters, 1 D1 D=1 m 1 1

17 (paraxial) Ray tracing with matrix n 2 n 1 x 1 α 1 α 2 x 2 Optical System n2α 2 n1α1 M11 M12 n1α1 M x = 2 x = 1 M21 M 22 x 1

18 Matrix formulation n2α 2 n1α1 M11 M12 n1α1 = M = x 2 x 1 M21 M 22 x 1 1. Propagation through distant D in medium with n M 1 0 = D 1 n 2. Refraction by spherical surface with power P M 1 P = 0 1

19 Ray tracing : an example (revisited) n n' x 0 α 0= α 1 x 1 x' 1 α' 1 =αα 2 x 2 D 01 D 12

20 Ray tracing : an example (revisited) n n' x 0 α 0= α 1 x 1 x' 1 α' 1 =αα 2 x 2 D 01 D 12 n' α2 translation refraction translation n α0 x = 2 by D 12 w/ power P by D 01 x 0

21 Ray tracing : an example (revisited) n n' x 0 α 0= α 1 x 1 x' 1 α' 1 =αα 2 x 2 D 01 D 12 n' α2 translation refraction translation n α0 x = 2 by D 12 w/ power P by D 01 x 0 Solving

22 Thin lens in air n = n=1 n n n=1 + R R ( ) Refraction from the 1 st surface + refraction from the 2 nd surface ignore space in between, i.e. thin lens. approx

23 Thin lens in air n=1 n n n=1 + R R ( ) P 1 = n 1 R P 2 = 1 n R ' 1 P 1 P 1 ( P + P ) M = M 2 M 1 = = Consequence: Lens maker s formula n 1 1 n 1 1 Pthin lens = P1 + P2 = + = ( n 1) R R ' R R'

24 Thin lens and focal length, f Object at f f = P 1 thin lens

25 Thick lens in air = n n=1 n n n=1 + R R ( ) + Refraction (left) + translation (middle) + refraction (right)

26 Thick lens in air P = n 1 R n n=1 n n n=1 + d R R ( ) d l n + P ' = 1 R n ' M 1 P' P 1 P' l ( P+ P' PP' l) = M3M2M1 = 0 1 l 1 = 0 1 l 1 Pl

27 Matrix Ray tracing n 2 α 1 n 1 x 2 x 1 Optical System α 2 n2α 2 n1α1 M11 M12 n1α1 = M = x x M M x Power = M12 Imaging condition : M = 21 0 (x 2 is independent of a 1 ) x Lateral magnification = x α2 Angular magnification = 2 = M 22 1 M 11 21= 0 α M 21= 0 1 = M

28 Example: thin lens system f = 10 cm f = 10 cm x 5 cm z=?

29 Example: thin lens system f = 10 cm f = 10 cm x 5 cm z=? translation positive translation negative 0?? by z lens = by 5 cm lens x 0 x = z x 3 z x 2 20 z = 30 cm

30 Example: finding image plane f = 10 cm f = 5 cm 15 cm 10 cm z=?

31 Example: finding image plane f = 10 cm f = 5 cm 15 cm 10 cm z=? translation lens translation lens translation α1 α 2 by z f=5 cm by 10 cm f=10 cm = by 15 cm x x α1 α z.1z = x x 1 2 Imaging condition : M 21 = 0 z = 4 cm

32 Field & Aperture Field ofview (FoV, field) is size ofimage Field Stop : physical element limits FoV Large FoV Small FoV Aperture : the angle of acceptance of the imaging system Large Aperture Small Aperture

33 Numerical Aperture (NA) n: medium refractive index θ NA = n sin θ = max 1 Speed (f/#) = 2NA e.g. f/8 means (f/#)=8 Aperture Stop: a physical elements that limits NA Physical meaning: NA limits the optical energy (or information) that can enter the system Thus, NA limits i optical resolution. (higher h NA higher h resolving li power)

34 Multi lens imaging system Object plane 1 st image plane 2 nd image plane

35 Multi lens imaging system Object plane 1 st image plane 2 nd image plane You can find the image planes using ray tracing (or using Matrix)

36 Multi lens imaging system: Field stop Field Stop Entrance Window Imaging backward Imaging Field Stop Imaging forward Exit Window

37 Multi lens imaging system: Aperture stop Aperture Stop Entrance Imaging Aperture Imaging Exit Pupil backward Stop forward Pupil

38 Multi lens imaging system: all together Entrance Window Entrance Field Aperture Exit Pupil Stop Stop Pupil Only physical components! Exit Window

39 2f & 4f optical imaging system 2fimaging system f 2f 2f 4f imaging system f 1 f 2 f 1 f 1 f 2 f 2

40 Effects of NA in microscopic imaging (4f ) Low NA resolution = λ 2NA High NA Objective lens Tube lens

41 Effects of f/# in photographic image deep depth of field shallow depth of field (due to more optical information, i.e. better axial resolving power)

42 Why we need Fourier Optics Simple object, lens & rays can be described by Geometric optics f f Inmodern optical imaging, you need to consider optical field (amplitude + phase), complex optical elements (grating, your own device), etc. Fourier Optics makes it much easier

43 Reading list (Week2 Day1) (see the webpage for the links) B. Masters, "Ernst Abbe and the Foundation of Scientific Microscopes," Optics and Photonics News 18(2), (2007) : History of E. Abbe, who contributed to scientific microscopic contribution and diffraction resolution (from NA) E. G. Putten and A. P. Mosk, The information age in optics: Measuring the transmission matrix, Physics 3, 22 (2010)