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Transcript
Physics 208 Exam 1 Review Mon. Feb. 18, 2008 1 Exam covers Ch. 21.5-7, 22-23,25-26, Lecture, Discussion, HW, Lab Exam 1 is Wed. Feb. 20, 5:30-7 pm, 2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306), 180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308) Chapter 21.5-7, 22 Chapter 23 Reflection, refraction, and image formation Chapter 25 Waves, interference, and diffraction Electric charges and forces Chapters 26 Mon. Feb. 18, 2008 Electric fields 2 Quick Quiz An electric dipole is in a uniform electric field as shown. The dipole Dipole A. accelerates left B. accelerates right C. stays fixed D. accelerates up E. none of the above Mon. Feb. 18, 2008 3 Electric torque on dipoles Remember torque? r F , magnitude rF sin r Here there are two torques, both into page: F qE, r s /2 qE s /2sin r F qE, r s /2 qE s /2sin Total torque is sum of these qsE sin p qs Mon. Feb. 18, 2008 Torque on dipole in uniform field p E 4 Dipole in non-uniform field A permanent dipole is near a positive point charge in a viscous fluid. The dipole will A. rotate CW & move toward charge B. rotate CW & move away C. rotate CCW & move toward + D. rotate CCW & move away E. none of the above Mon. Feb. 18, 2008 5 Properties of waves Wavelength, frequency, propagation speed related as f v Phase relation In-phase: crests line up 180˚ Out-of-phase: crests line up with trough Time-delay leads to phase difference Path-length difference leads to phase difference Mon. Feb. 18, 2008 6 Chapter 22: Waves & interference Path length difference and phase Two slit interference Alternating max and min due to path-length difference Phase change on reflection different path length -> phase difference. π phase change when reflecting from medium with higher index of refraction Interference in thin films Different path lengths + reflection phase change Mon. Feb. 18, 2008 7 Path length difference Path length difference d Phase difference = (d/)2 radians Constructive for 2n phase difference L Shorter path Light beam Longer path Foil with two narrow slits Mon. Feb. 18, 2008 Recording plate 8 Question You are listening to your favorite radio station, WOLX 94.9 FM (94.9x106 Hz) while jogging away from a reflecting wall, when the signal fades out. About how far must you jog to have the signal full strength again? (assume no phase change when the signal reflects from the wall) Hint: wavelength = (3x108 m/s)/94.9x106 Hz A. 3 m B. 1.6 m C. 0.8 m D. 0.5 m Mon. Feb. 18, 2008 =3.16 m d-x x d path length diff = (d+x)-(d-x)= 2x Destructive 2x=/2x=/4 Constructive make 2x=x=/2 x increases by /4 = 3.16m/4=0.79m 9 Two-slit interference Mon. Feb. 18, 2008 10 Two-slit interference: path length L y Constructive int: Phase diff = 2m, m 0,1,2 Path length diff = m, m 0,1,2 Phase diff = 2 (m 1/2), m 0,1,2 diff = (m 1/2), m 0,1,2 Path length Path length difference d sin d y /L d sin 2 d Phase difference 2 y /L Destructive int. Mon. Feb. 18, 2008 11 Reflection phase shift Possible additional phase shift on reflection. Start in medium with n1, reflect from medium with n2 n2>n1, 1/2 wavelength phase shift n2<n1, no phase shift Difference in phase shift between different paths is important. Mon. Feb. 18, 2008 12 Thin film interference air 1/2 wavelength phase shift from top surface reflection Reflecting from n2 air: n1=1 t n2>1 air/n air: n1=1 Extra path length No phase shift from bottom interface Reflecting from n1 Extra phase shift needed for constructive interference is m 1/2air /n 2t m 1/2air /n Mon. Feb. 18, 2008 13 Diffraction from a slit Each point inside slit acts as a source Net result is series of minima and maxima Similar to two-slit interference. Mon. Feb. 18, 2008 14 Overlapping diffraction patterns Two independent point sources will produce two diffraction patterns. If diffraction patterns overlap too much, resolution is lost. Image to right shows two sources clearly resolved. Angular separation Circular aperture diffraction limited: min 1.22 Mon. Feb. 18, 2008 D 15 Diffraction gratings Diffraction grating is pattern of multiple slits. Very narrow, very closely spaced. Same physics as two-slit interference d sin bright m, m 0,1,2 Mon. Feb. 18, 2008 sin bright m d 16 Chap. 23: Refraction & Ray optics Refraction Ray tracing Types of images Real image: project onto screen Virtual image: image with another lens Lens equation Can locate image by following specific rays Relates image distance, object distance, focal length Magnification Ratio of images size to object size Mon. Feb. 18, 2008 17 Refraction Occurs when light moves into medium with different index of refraction. Light direction bends according to n1 sin 1 n2 sin 2 i,1 r n1 n2 Special case: Total internal reflection 2 Angle of refraction Mon. Feb. 18, 2008 18 Lenses: focusing by refraction F Object P.A. Image F 1) Rays parallel to principal axis pass through focal point. 2) Rays through center of lens are not refracted. 3) Rays through F emerge parallel to principal axis. Here image is real, inverted, enlarged Mon. Feb. 18, 2008 19 Different object positions Object Image (real, inverted) Image (real, inverted) Image (virtual, upright) Mon. Feb. 18, 2008 These rays seem to originate from tip of a ‘virtual’ arrow. 20 Question You have a focused image on the screen, but you want the image to be bigger. Relative to the lens, you should A. Move screen away, move object away B. Move screen closer, move object away C. Move screen away, move object closer D. Move screen closer, move object closer Mon. Feb. 18, 2008 21 Equations Image and object different sizes Image (real, inverted) p Relation between image distance object distance focal length Magnification =M= Mon. Feb. 18, 2008 q 1 1 1 p q f image height q image distance object height p object distance 22 Question You want an image on a screen to be ten times larger than your object, and the screen is 2 m away. About what focal length lens do you need? 1 1 1 p q f A. f~0.1m q=2 m B. f~0.2m C. f~0.5m D. f~1.0m Mon. Feb. 18, 2008 mag=10 -> q=10p->p=0.2m 1 1 1 5.5 f 0.18m 0.2m 2m f 23 Chapter 25: Electric Charges & Forces Triboelectric effect: transfer charge Vector forces between charges Total charge is conserved Add by superposition Drops off with distance as 1/r2 Insulators and conductors Polarization of insulators, conductors Mon. Feb. 18, 2008 24 Charges conductors & insulators Two types of charges, + and Conductor: Like charges repel Unlike charges attract Charge free to move Distributed over surface of conductor Insulator Charges stuck in place where they are put Mon. Feb. 18, 2008 25 Electric force: magnitude & direction Electrical force between two stationary charged particles The SI unit of charge is the coulomb (C ), µC = 10-6 C 1 C corresponds to 6.24 x 1018 electrons or protons ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πeo) -12 C2 / N.m2 eo permittivity of free space = 8.854 x 10 Directed along line joining particles. Mon. Feb. 18, 2008 26 Forces add by superposition Equal but opposite charges are placed near a negative charge as shown. What direction is the net force on the negative charge? A) Left kq1q2 F 2 r - B) Right C) Up D) Down E) Zero Mon. Feb. 18, 2008 + - 27 Chapter 26: The Electric Field Defined as force per unit charge (N/C) Calculated as superposition of contributions from different charges Examples Single charge Electric dipole Line charge, sheet of charge Electric field lines Force on charged particles Mon. Feb. 18, 2008 28 Electric field Fe = qE If q is positive, F and E are in the same direction Example: electric field from point charge Qp=1.6x10-19 C + E r = 1x10-10 m E = (9109)(1.610-19)/(10-10)2 N = 2.91011 N/C Mon. Feb. 18, 2008 (to the right) 29 Pictorial representation of E: Electric Field Lines Mon. Feb. 18, 2008 30 Electric field lines Local electric field tangent to field line Density of lines proportional to electric field strength Fields lines can only start on + charge Can only end on - charge (but some don’t end!). Electric field lines can never cross Mon. Feb. 18, 2008 31 Electric dipole E 1 2p 4eo r 3 On y-axis far from dipole Electric dipole p qs moment +q E -q 1 p 4eo r 3 On x-axis far from dipole Electric field magnitude drops off as 1/r3 Mon. Feb. 18, 2008 32 Quick quiz: continuous charge dist. Electric field from a uniform ring of charge. The magnitude of the electric field on the x-axis y A. Has a maximum at x=0 B. Has a maximum at x= C. Has a maximum at finite nonzero x x D. Has a minimum at finite nonzero x E. Has neither max nor min Mon. Feb. 18, 2008 33 Force on charged particle Electric field produces force qE on charged particle Force produces an acceleration a = FE / m Uniform E-field (direction&magnitude) produces constant acceleration if no other forces Positive charge accelerates in same direction as field Negative charge accelerates in direction opposite to electric field Mon. Feb. 18, 2008 34 Force on a dipole Dipole made of equal + and - charges Force exerted on each charge Uniform field causes rotation Mon. Feb. 18, 2008 Dipole 35 Dipole in non-uniform field A dipole is near a positive point charge in a viscous fluid. The dipole will A. rotate CW & move toward charge B. rotate CW & move away C. rotate CCW & move toward + D. rotate CCW & move away E. none of the above Mon. Feb. 18, 2008 36