Physics 42200 Waves & Oscillations Lecture 32 – Geometric Optics Spring 2013 Semester Matthew Jones Thin Lens Equation First surface: Second surface: Add these equations and simplify using 1 and → 0: 1 1 1 1 1 (Thin lens equation) Thick Lenses • Eliminate the intermediate image distance, • Focal points: – Rays passing through the focal point are refracted parallel to the optical axis by both surfaces of the lens – Rays parallel to the optical axis are refracted through the focal point – For a thin lens, we can draw the point where refraction occurs in a common plane – For a thick lens, refraction for the two types of rays can occur at different planes Thick Lens: definitions First focal point (f.f.l.) Primary principal plane First principal point Second focal point (b.f.l.) Secondary principal plane Second principal point Nodal points If media on both sides has the same n, then: N1=H 1 and N2=H 2 Fo Fi H1 H2 N1 N2 - cardinal points Thick Lens: Principal Planes Principal planes can lie outside the lens: Thick Lens: equations Note: in air ( n= 1) 1 1 1 x x = f 2 1 1 1 (n −1)d + = o i effective = ()n −1 − + l l s s f focal length : l o i f R1 R2 nl R1R2 ( − ) ( − ) = − f nl 1 dl = − f nl 1 dl Principal planes: h1 h2 nl R2 nl R1 ≡ yi = − si = − xi = − f Magnification: MT yo so f xo Thick Lens Calculations 1. Calculate focal length 1 1 1 1 1 2. Calculate positions of principal planes 1 1 3. Calculate object distance, , measured from principal plane 4. Calculate image distance: 1 1 1 5. Calculate magnification, / Thick Lens: example Find the image distance for an object positioned 30 cm from the vertex of a double convex lens having radii 20 cm and 40 cm, a thickness of 1 cm and nl=1.5 1 1 1 + = f f so si f 30 cm so si 1 1 1 (n −1)d 1 1 0.5⋅1 1 = ()n −1 − + l l = 0.5 − + l − ⋅ ⋅ f R1 R2 nl R1R2 20 40 1.5 20 40 cm f = 26 .8 cm s = 30 cm + 0.22 cm = 30.22 cm 26 .8⋅0.5⋅1 o h = − cm = 0.22 cm 1 1 1 1 − 40 ⋅1.5 + = ⋅ ⋅ 30 .22 cm si 26 .8cm = − 26 .8 0.5 1 = − h2 cm 0.44 cm = 20 ⋅1.5 si 238 cm Compound Thick Lens Can use two principal points (planes) and effective focal length f to describe propagation of rays through any compound system Note : any ray passing through the first principal plane will emerge at the same height at the second principal plane 1 1 1 d = For 2 lenses (above): = + − H11 H1 fd f2 f f1 f2 f1 f2 = H22 H2 fd f1 Example: page 246 Ray Tracing • Even the thick lens equation makes approximations and assumptions – Spherical lens surfaces – Paraxial approximation – Alignment with optical axis • The only physical concepts we applied were – Snell’s law: sin sin – Law of reflection: (in the case of mirrors) • Can we do better? Can we solve for the paths of the rays exactly? – Sure, no problem! But it is a lot of work. – Computers are good at doing lots of work (without complaining) Ray Tracing • We will still make the assumptions of – Paraxial rays – Lenses aligned along optical axis • We will make no assumptions about the lens thickness or positions. • Geometry: Ray Tracing • At a given point along the optical axis, each ray can be uniquely represented by two numbers: – Distance from optical axis, – Angle with respect to optical axis, • If the ray does not encounter an optical element its distance from the optical axis changes according to the transfer equation : – This assumes the paraxial approximation sin Ray Tracing • At a given point along the optical axis, each ray can be uniquely represented by two numbers: – Distance from optical axis, – Angle with respect to optical axis, • When the ray encounters a surface of a material with a different index of refraction, its angle will change according to the refraction equation : – Also assumes the paraxial approximation Ray Tracing • Geometry used for the refraction equation: sin / / Matrix Treatment: Refraction At any point of space need 2 parameters to fully specify ray: distance from axis ( y) and inclination angle ( α) with respect to the optical axis. Optical element changes these ray parameters. Refraction: note: paraxial approximation α = α − nt1 t1 ni1 i1 D1 yi1 Reminder: = ⋅α + yt1=yi1 yt10 n ii 11 y i 1 A B α Aα + By ≡ C D y Cα + Dy n α 1 - D n α Equivalent matrix t1 t1 = 1 i1 i1 representation: yt1 0 1 yi1 ≡ ri1 - input ray = ≡ R - refraction matrix rt1 R1ri1 1 ≡ rt1 - output ray Matrix: Transfer Through Space Transfer: α = α + ⋅ ni2 i2 nt1 t1 0 yt1 y = d ⋅α + y yi2 i2 21 t1 t1 yt1 α α Equivalent matrix ni2 i2 1 0nt1 t1 presentation: = yi2 d21 nt1 1 yt1 ≡ rt1 - input ray ≡ T - transfer matrix r = T r 21 i2 21 t1 ≡ ri2 - output ray System Matrix y y y i2 i2 i1 yt1 ri1 rt1 ri2 rt2 ri3 rt3 R1 T21 T32 R3 = = = rt1 R1ri1 ri2 T21 rt1 T21 R1ri1 Thick lens ray transfer: = rt2 R2T21 R1ri1 System matrix: = = A R2T21 R1 rt2 A ri1 Can treat any system with single system matrix Thick Lens Matrix d n = l A R2T21 R1 y y 1 - D i2 i2 = yi1 y R t1 0 1 1 0 Reminder: T = A B a b Aa + Bc Ab + Bd d n 1 ≡ C D c d Ca + Dc Cb + Dd 1 1 1 1 1 - D 2 1 01 - D1 A = 0 1 dl nl 10 1 system matrix of thick lens D d D D d − 2 l + 1 2 l For thin lens dl=0 1 - D1 - D 2 - nl nl A = 1 - 1/ f d D d = l 1− 1 l A 0 1 nl nl Thick Lens Matrix and Cardinal Points n (1− a ) V H = i1 11 1 1 − a12 n (a −1) V H = i2 22 2 2 − a12 D d D D d 1− 2 l - D - D - + 1 2 l 1 2 a a nl nl 11 12 A = = dl D1dl a a 1− 21 22 nl nl ni1 nt2 in air 1 a = − = − = − effective focal length 12 a12 fo fi f Matrix Treatment: example rI rO = rI T1A lT2rO α α nI I 1 0a11 a12 1 0nO O = yI d I 2 nI 1a21 a22 d1O nO 1 yO (Detailed example with thick lenses and numbers: page 250) Mirror Matrix note : R< 0 −1 − 2n / R M = 0 1 α α n r n i = M yr yi = yr yi α = − α − n r n i 2ny i / R α = −α − r i 2yi / R.