1. 1 Exam 1 Tue. Sep. 29, 5:30-7 pm, 145 Birge Covers 21.5-7, 22, 23.1-4, 23.7, 24.1-5, 26 + lecture, lab, discussion, HW 8 1/2 x 11 handwritten note sheet (both sides) allowed Chap 21.5-7, 22 Waves, interference, and diffraction Chap 23 Reflection, refraction, and image formation Chap 24 Optical instruments Chap 26 Electric charges and forces

2. 2 Properties of waves Wavelength, frequency, propagation speed related as 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

3. 3 Question Two waves of wavelength λ are traveling through a medium with index of refraction n=1. One of the waves passes a medium of thickness d=λ and index n=1.25.. What is the phase difference between the waves far to the right? A.λ/4 B.λ/2 C.λ D.2 λ n=1 λ n=1.25

4. 4 Ch 22, 21.5-7: 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

5. 5 Phase difference & interference Path length difference d Phase difference = d(2  / ) radians Constructive for 2  n phase difference L Shorter path Longer path Light beam Foil with two narrow slits Recording plate

6. 6 Question You are listening to your favorite radio station, WOLX 94.9 FM (94.9x10 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) A.3 m B.1.6 m C.0.8 m D.0.5 m d x d-x path length diff = (d+x)-(d-x)= 2x =3.16 m Hint: wavelength = (3x10 8 m/s)/94.9x10 6 Hz Destructive  2x= /2  x= /4 Constructive  make 2x=  x= /2 x increases by /4 = 3.16m/4=0.79m

7. 7 Two-slit interference

8. 8 Two-slit interference: path length y L Path length difference Phase difference  Constructive int: Destructive int. Phase diff = Path length diff =

9. 9 Reflection phase shift Possible additional phase shift on reflection. Start in medium with n 1, reflect from medium with n 2 n 2 >n 1, 1/2 wavelength phase shift n 2 <n 1, no phase shift Difference in phase shift between different paths is important.

10. 10 Thin film interference air: n 1 =1 n 2 >1 1/2 wavelength phase shift from top surface reflection t air air /n Extra path length Extra path length needed for constructive interference is No phase shift from bottom interface air: n 1 =1 Reflecting from n 2 Reflecting from n 1

11. 11 Thin-film interference example Coated glass in air, coating thickness = 275nm Incident white light 400-700nm Glass infinitely thick What color reflected light do you see? Both paths have 180˚ phase shifts So only path length difference is important n film =1.2 n glass =1.5 n air =1 Incident light eye t=275nm

12. 12 Thin Film Interference II Same coated glass underwater Now only one path has 180˚ phase shift n film =1.2 n air =1 Incident light eye n glass =1.5 n water =1.33 m=0 gives 1320 nm, too long. m=1 gives 440 nm Color changes underwater!

13. 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. Angular locations of minima (destructive interference)

14. 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:

15. 15 Diffraction gratings Diffraction grating is pattern of multiple slits. Very narrow, very closely spaced. Same physics as two-slit interference

16. 16 Chap. 23-24: 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

17. 17 Refraction Occurs when light moves into medium with different index of refraction. Light direction bends according to  i,1 rr 22 Angle of refraction n1n1 n2n2 Special case: Total internal reflection

18. 18 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

19. 19 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. Lenses: focusing by refraction F F Object Image P.A. Here image is real, inverted, enlarged

20. 20 Different object positions Image (real, inverted) Object Image (virtual, upright) These rays seem to originate from tip of a ‘virtual’ arrow.

21. 21 Question You have a near point of 25cm. 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.2 B.3 C.4 D.5 E.10

22. 22 Equations Magnification = M = Image and object different sizes Image (real, inverted) s s’ Relation between image distance object distance focal length

23. 23 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? A.f~0.1m B.f~0.2m C.f~0.5m D.f~1.0m s’=2 m mag=10 -> s’=10s ->s=0.2m

24. Thurs. Sep. 17, 2009Physics 208, Lecture 5 24 Diverging lens Optical Axis Then thin-lens equation can be used: ObjectImage Focal length defined to be negative

25. Mon. Feb. 4, 2008Physics 208, Lecture 425 Virtual image ‘Object’ Eyepiece Compound Microscope q p Object Real, inverted, image Eyepiece: simple magnifier. Angular Mag.=25cm/p ~25cm/f eyepiece Objective lateral mag. =-q/p~L/f objective Objective

26. 26 Chapter 26: Electric Charges & Forces Triboelectric effect: transfer charge Total charge is conserved Vector forces between charges Add by superposition Drops off with distance as 1/r 2 Insulators and conductors Polarization of insulators, conductors

27. 27 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 10 18 electrons or protons k e = Coulomb constant ≈ 9 x 10 9 N. m 2 /C 2 = 1/(4π  o )  o  permittivity of free space = 8.854 x 10 -12 C 2 / N. m 2 Directed along line joining particles.

28. 28 +- - Equal but opposite charges are placed near a negative charge as shown. What direction is the net force on the negative charge? A)Left B)Right C)Up D)Down E)Zero Forces add by superposition

29. The electric dipole Dipole moment Vector Points from - charge to + charge Has magnitude qs Can all be approximated by electric dipole. Two opposite charges magnitude q separated by distance s

30. Force on an electric dipole What is the direction of the force on the electric dipole from the positive point charge? + A.Up B.Down C.Left D.Right E.Force is zero How does the magnitude of the force depend on ?