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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 1 1 1 1 (Thin lens equation) 1 and → 0: 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 Fo Fi H1 H2 N1 N2 - cardinal points If media on both sides has the same n, then: N1=H1 and N2=H2 Thick Lens: Principal Planes Principal planes can lie outside the lens: Thick Lens: equations Note: in air (n=1) 1 1 1 xo xi = f 2 + = so si f effective focal length: Principal planes: Magnification: 1 1 1 (nl − 1)d l = (nl − 1) − + f R R n R R 2 l 1 2 1 f (nl − 1)d l f (nl − 1)d l h1 = − h2 = − nl R2 nl R1 yi si xi f MT ≡ =− =− =− yo so f xo Thick Lens Calculations 1. Calculate focal length 1 1 1 1 1 2. Calculate positions of principal planes ℎ ℎ 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 + = so si f f 30 cm so f si 1 1 1 (nl − 1)d l 1 0.5 ⋅ 1 1 1 = (nl − 1) − + = 0.5 − + cm f R R n R R 20 − 40 1 . 5 ⋅ 20 ⋅ 40 2 l 1 2 1 f = 26.8 cm so = 30cm + 0.22cm = 30.22 cm 26.8 ⋅ 0.5 ⋅ 1 h1 = − cm = 0.22cm 1 1 1 + = − 40 ⋅ 1.5 30.22cm si 26.8cm 26.8 ⋅ 0.5 ⋅ 1 h2 = − cm = −0.44cm si = 238 cm 20 ⋅ 1.5 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 For 2 lenses (above): Example: page 246 1 1 1 d = + − f f1 f 2 f1 f 2 H11H1 = fd f 2 H 22 H 2 = fd f1 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 – Law of reflection: sin (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: nt1α t1 = ni1αi1 − D 1 yi1 yt1=yi1 Equivalent matrix representation: yt1 = 0 ⋅ ni1α i1 + yi1 note: paraxial approximation Reminder: A B α Aα + By ≡ C D y C α + Dy nt1α t1 1 - D 1 ni1αi1 = y t 1 0 1 yi 1 ≡ ri1 - input ray rt1 = R1ri1 ≡ R1 - refraction matrix ≡ rt1 - output ray Matrix: Transfer Through Space Transfer: ni 2αi 2 = nt1α t1 + 0 ⋅ yt1 yi2 yi 2 = d 21 ⋅ α t1 + yt1 yt1 Equivalent matrix presentation: 0 nt1α t1 ni 2αi 2 1 = yi 2 d 21 nt1 1 yt1 ≡ rt1 - input ray ri 2 = T21rt1 ≡ T21 - transfer matrix ≡ ri2 - output ray System Matrix yi2 yi1 y t1 ri1 rt1 R1 ri2 T21 rt1 = R1ri1 Thick lens ray transfer: yi2 rt2 ri3 T32 rt3 R3 ri 2 = T21rt1 = T21R1ri1 rt 2 = R 2T21R1ri1 System matrix: A = R 2T21R1 Can treat any system with single system matrix rt 2 = A ri1 Thick Lens Matrix d A = R 2T21R1 nl yi2 yi1 y t1 yi2 Reminder: A B a b Aa + Bc Ab + Bd ≡ C D c d Ca + Dc Cb + Dd 1 - D 2 1 A = 0 1 d l nl D 2d l 1 − nl A = dl n l 0 1 - D 1 1 0 1 D 1D 2d l - D1 - D 2 - + nl D 1d l 1− nl 1 - D R = 0 1 1 0 T = d n 1 1 1 1 system matrix of thick lens For thin lens dl=0 1 - 1/ f A = 1 0 1 Thick Lens Matrix and Cardinal Points ni1 (1 − a11 ) V1H1 = − a12 ni 2 (a22 − 1) V2 H 2 = − a12 D 2d l 1 − nl A = dl n l D 1D 2d l - D1 - D 2 - + nl a11 a12 = Dd a21 a22 1− 1 l nl ni1 nt 2 a12 = − =− fo fi in air a12 = − 1 f effective focal length Matrix Treatment: example rI rO rI = T1A l T2rO nIα I 1 = y I d I 2 nI 0 a11 a12 1 1 a21 a22 d1O nO 0 nOαO 1 yO (Detailed example with thick lenses and numbers: page 250) Mirror Matrix note: R<0 − 1 − 2n / R M = 1 0 nα r nαi = M yr yi yr = yi nα r = −nαi − 2nyi / R α r = −αi − 2 yi / R