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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