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UF PHY 2054 - Quiz

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PHY2054: Chapter 231Chapter 23 QuizÎWhich of the following is nota principal ray? (1) 1 (2) 2 (3) 3 (4) 4 (5) All are principal raysPHY2054: Chapter 232Concave and Convex Mirrors2Rf=PHY2054: Chapter 233Images from MirrorsfChqMhp′≡=−111pqf+=LocationSizeq > 0 along reflected rayPHY2054: Chapter 234Mirror Example111133.380 50qq+= ⇒=fC123¾ 2 cm tall object (h = 2), 80 cm from mirror (p = 80)¾ Mirror: 100 cm radius of curvature concave towards object (R = +100)()133.31.67801.67 3.33qMphh=− =− =−′=− =−Roughly agrees with ray diagram (good to check it)PHY2054: Chapter 235Images Formed by Refraction12nqhMhnp′==−12 21nn nnpqR−+=PHY2054: Chapter 236Image from Flat Refractive SurfaceR = ∞121nqMnp=−=+12 210nn nqppqn+=⇒=−Same size virtual image!PHY2054: Chapter 237Example: Image in Pool10.751.333qpp=− ≅−n1= 1.333n2= 1.021nqpn=−So depth of image is about3/4 depth of objectPHY2054: Chapter 238Atmospheric Refraction¾ Refraction in atmosphere makes sun always appear higher in sky¾ Can see the sun even when it is below the horizon!PHY2054: Chapter 239Atmospheric Refraction Causes Various MiragesPHY2054: Chapter 2310Approaching Car on Hot RoadPHY2054: Chapter 2311Floating Iceberg! Cold Air, Warm WaterPHY2054: Chapter 2312“Squashed” Sun Setting Over OceanPHY2054: Chapter 2313Thin LensesPHY2054: Chapter 2314Thin LensesffConverging LensRays parallel to axis refractand pass through focal pointf > 0ffDiverging LensRays parallel to axis refract andappear to emerge from focal pointf < 0PHY2054: Chapter 2315Location and Size of Image (Recall Mirror)hqMhp′≡=−111pqf+=PHY2054: Chapter 2316Finding Image from Lens: Example11115cm30 10qq+= ⇒=+¾ f = 10 cm¾ p = 30 cm¾ h = 4 cm150.5302cmqMph=− =− =−′=−PHY2054: Chapter 2317Principal Ray Diagrams for Lensesq > 0PHY2054: Chapter 2318Principal Ray Diagrams for Lenses (cont)q < 0PHY2054: Chapter 2319Principal Ray Diagrams for Lenses (cont)q < 0PHY2054: Chapter 2320Calculating Focal Length for Thin Lenses()121111nfRR⎛⎞=− −⎜⎟⎝⎠Lensmaker’s equationSign conventionR < 0 R > 0PHY2054: Chapter 2321Calculating Focal Length Example1110.5 0.110 10f⎛⎞=−=⎜⎟−⎝⎠1110.5 0.110 10f⎛⎞=−=⎜⎟−⎝⎠Let n = 1.5, R = 101110.5 0.0510f⎛⎞=−=⎜⎟∞⎝⎠So f1= f2/21 has twice the “power” of 2PHY2054: Chapter 2322Quiz on LensesÎ All the lenses below have either flat sides or radius of curvature R. Rank in descending order the value of 1/f (lens power), i.e. from most positive to most negative (1) A, C, B, E, D (2) C, A, D, B, E (3) B, A, C, E, D (4) D, E, B, C, A (5) C, A, E, D, BABCD EFlat()121111nfRR⎛⎞=− −⎜⎟⎝⎠Each term adds to lens powerPHY2054: Chapter 2323Quiz on MirrorsÎAn upright object is located in front of a convex mirror a distance greater than the focal length. The image formed by the mirror is:  (1) real, inverted, and smaller than the object  (2) virtual, inverted, and larger than the object  (3) real, inverted, and larger than the object  (4) real, erect, and larger than the object  (5) virtual, erect, and smaller than the object 111qfp=−f < 0, so q < 0M = −q/p <


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