Properties of waves. Question. Ch 22, : Waves & interference. Question. Phase difference & interference

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1 Exam Tue. Sep. 9, 5:30-7 pm, 45 Birge Covers.5-7,, 3.-4, 3.7, 4.-5, 6 + lecture, lab, discussion, HW Chap.5-7, Waves, interference, and diffraction Chap 3 Reflection, refraction, and image formation Chap 4 Optical instruments Chap 6 8 / x handwritten note Electric charges and forces sheet (both sides) allowed Properties of waves Wavelength, frequency, propagation speed related as λf = v Phase relation In-phase: crests line up 80 Out-of-phase: crests line up with trough Time-delay leads to phase difference Path-length difference leads to phase difference Two waves of wavelength λ are traveling through a medium with index of refraction n=. One of the waves passes a medium of thickness d=λ and index n=.5.. What is the phase difference between the waves far to the right? A. λ/4 B. λ/ C. λ D. λ n= λ n=.5 Ch,.5-7: Waves & interference Path length difference and phase different path length -> phase difference. Two slit interference Alternating max and min due to path-length difference Phase change on reflection π phase change when reflecting from medium with higher index of refraction Interference in thin films Different path lengths + reflection phase change 3 4 Light beam Phase difference & interference Path length difference d Phase difference = d(π /λ) radians Constructive for πn phase difference Foil with two narrow slits Shorter path Longer path L Recording plate 5 You are listening to your favorite radio station, WOLX 94.9 FM (94.9x0 6 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 = (3x0 8 m/s)/94.9x0 6 Hz A. 3 m B..6 m C. 0.8 m D. 0.5 m λ=3.6 m d-x path length diff = (d+x)-(d-x)= x Destructive x=λ/ x=λ/4 Constructive make x=λ x=λ/ x increases by λ/4 = 3.6m/4=0.79m d x 6

2 Two-slit interference Two-slit interference: path length L y θ 7 Constructive int: Destructive int. Phase diff = πm, m = 0,±,± Path length diff = mλ, m = 0,±,± Phase diff = π(m +/), m = 0,±,± Path length diff = (m +/)λ, m = 0,±,± ( ) Path length difference d sinθ d y /L Phase difference π dsinθ λ πd λ y /L ( ) 8 Reflection phase shift Possible additional phase shift on reflection. Start in medium with n, reflect from medium with n n >n, / wavelength phase shift n <n, no phase shift Difference in phase shift between different paths is important. t Thin film interference λ air air: n = n > λ air /n air: n = Extra path length / wavelength phase shift from top surface reflection Reflecting from n No phase shift from bottom interface Reflecting from n Extra path length needed for constructive interference is ( m +/) ( λ air /n) t = ( m +/) ( λ air /n) 9 0 Thin-film interference example Coated glass in air, coating thickness = 75nm Incident white light nm Glass infinitely thick What color reflected light do you see? Both paths have 80 phase shifts So only path length difference is important t = mλ air /n film m = λ = 660nm Incident light n film =. eye n air = n glass =.5 t=75nm Thin Film Interference II Same coated glass underwater Now only one path has 80 phase shift t = ( m+/ )λ air / n film λ air = tn film /( m +/) = ( 75nm) (.) /( m +/) m=0 gives 30 nm, too long. m= gives 440 nm Color changes underwater! Incident light n glass =.5 eye n air = n film =. n water =.33

3 Diffraction from a slit Overlapping diffraction patterns Each point inside slit acts as a source Net result is series of minima and maxima Similar to two-slit interference. Two independent point sources will produce two diffraction patterns. If diffraction patterns overlap too much, resolution is lost. to right shows two sources clearly resolved. θ Angular separation Angular locations of minima (destructive interference) 3 Circular aperture diffraction limited: θ min =. λ D 4 Diffraction gratings Diffraction grating is pattern of multiple slits. Very narrow, very closely spaced. Same physics as two-slit interference d sinθ bright = mλ, m = 0,, sinθ bright = m λ d Chap. 3-4: Refraction & Ray optics Refraction Ray tracing Can locate image by following specific rays Types of images Real image: project onto screen Virtual image: image with another lens Lens equation Relates image distance, object distance, focal length Magnification Ratio of images size to object size 5 6 Refraction Occurs when light moves into medium with different index of refraction. Light direction bends according to n sinθ = n sinθ θ i, θ r n n θ Angle of refraction Special case: Total internal reflection Total internal reflection Total internal reflection occurs A) at angles of incidence greater than that for which the angle of refraction = 90 B) at angles of incidence less than that for which the angle of refraction = 90 C) at angles of incidence equal to 90 D) when the refractive indices of the two media are matched D) none of the above 7 8 3

4 Lenses: focusing by refraction Different object positions F F P.A. (real, inverted) ) Rays parallel to principal axis pass through focal point. ) Rays through center of lens are not refracted. 3) Rays through F emerge parallel to principal axis. Here image is real, inverted, enlarged (real, inverted) 9 (virtual, upright) These rays seem to originate from tip of a virtual arrow. 0 You have a near point of 5cm. You hold a 5 cm focal length converging lens of focal length a negligible distance from your eye to view a penny more closely. If you hold the penny so that it appears sharp when you focus your eye at infinity (relaxed eye) how many times larger does the penny appear than the best you can do without the converging lens? A. B. 3 C. 4 D. 5 E. 0 s s Relation between image distance object distance focal length Magnification = M = Equations s + s = f and object different sizes (real, inverted) image height object height = s image distance = s object distance You want an image on a screen to be ten times larger than your object, and the screen is m away. About what focal length lens do you need? A. f~0.m B. f~0.m C. f~0.5m D. f~.0m s + s = f s = m mag=0 -> s =0s ->s=0.m 0.m + m = 5.5 = f f = 0.8m 3 Diverging lens Focal length defined to be negative Then thin-lens equation can be used: s + s = f Optical Axis 4 Thurs. Sep. 7, 009 Physics 08, Lecture 5 4

5 p q ive Compound Microscope Chapter 6: Electric Charges & Forces ive lateral mag. Virtual =-q/p~l/f objective image Eyepiece: simple magnifier. Angular Mag.=5cm/p ~5cm/f eyepiece Real, inverted, image Eyepiece Triboelectric effect: transfer charge Total charge is conserved Vector forces between charges Add by superposition Drops off with distance as /r Insulators and conductors Polarization of insulators, conductors Mon. Feb. 4, 008 Physics 08, Lecture Electric force: magnitude & direction Electrical force between two stationary charged particles 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? The SI unit of charge is the coulomb (C ), µc = 0-6 C C corresponds to 6.4 x 0 8 electrons or protons k e = Coulomb constant 9 x 0 9 N. m /C = /(4πε o ) ε o = permittivity of free space = x 0 - C / N. m Directed along line joining particles. A) Left B) Right C) Up D) Down E) Zero kq q F = r 7 8 Can all be approximated by electric dipole. Two opposite charges magnitude q separated by distance s The electric dipole Dipole moment p Vector Points from - charge to + charge Has magnitude qs Force on an electric dipole What is the direction of the force on the electric dipole from the positive point charge? A. Up p B. Down + C. Left D. Right E. Force is zero How does the magnitude of the force depend on p? 5