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Download Waves & Oscillations Physics 42200 Spring 2014 Semester Lecture 27 – Geometric Optics
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Physics 42200 Waves & Oscillations Lecture 27 – Geometric Optics Spring 2014 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 (front focal length) Primary principal plane First principal point Second focal point (back focal length) 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 Lenses and Principal Planes • For a single refracting surface, we measured and with respect to the vertex (ie, the surface of the lens) • For a thick lens, we need to define and with respect to the principal planes. • We need to calculate where they are, but it makes the algebra simpler. • We are not going to derive the following formula… Thick Lens: equations Note: in air (n=1) 1 1 1 xo xi = f 2 + = so si f effective focal length: Principal planes: 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 =− =− =− Magnification: M T ≡ 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 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 Sign convention: R > 0 − 1 2n / R M = 1 0 nα r nαi = M yr yi yr = yi nα r = −nα i + 2nyi / R α r = −α i + 2 yi / R