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Unformatted text preview: n = 1 . 500 and the radius of the lens is R = 0 . 500 mm . For this case, calculate the focal distance f in millimeters for the values of h given below. Express your answers accurately to the nearest 0 . 0001 mm . Put your ±nal answers in the spaces provided. h ( mm ) f ( mm ) 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 Fiber Optic Ball Lens Five equations and fve unknowns sin θ 11 = n sin θ 21 Snell’s law at frst sur±ace sin θ 11 = h/R ( h, R ) triangle θ 21 = θ 12 ( R, R ) equilateral triangle n sin θ 12 = sin θ 22 Snell’s law at second sur±ace → θ 22 = θ 11 sin ( πθ 11 ) R + f = sin 2( θ 11θ 21 ) R ( R, R + f ) triangle , law o± sines Solving R + f R = sin ( πθ 11 ) sin 2( θ 11θ 21 ) f = R ± sin θ 11 sin 2( θ 11θ 21 )1 ² f = R ± h/R sin 2[ sin1 ( h/R )sin1 ( h/nR )]1 ² For n = 1 . 5 and R = 0 . 500 mm h ( mm ) f ( mm ) 0.05 0.2479 0.10 0.2416 0.15 0.2310 0.20 0.2158 0.25 0.1958 0.30 0.1704 0.35 0.1389 0.40 0.1001 0.45 0.0522...
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 Spring '08
 Gaylord
 refractive index, Geometrical optics, Optic Ball Lens

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