Exam 1 Tue. Sep. 29, 5:30-7 pm, 145 Birge

1. Covers 21.5-7, 22, 23.1-4, 23.7, 24.1-5, 26 + lecture, lab, discussion, HW Exam 1 Tue. Sep. 29, 5:30-7 pm, 145 Birge • 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 8 1/2 x 11 handwritten note sheet (both sides) allowed

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. 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? n=1 λ n=1.25 λ/4 λ/2 λ 2λ

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. Shorter path Longer path Phase difference & interference • Path length difference d • Phase difference = d(2π /) radians • Constructive for 2πn phase difference L Light beam Recording plate Foil with two narrow slits

6. x d-x d 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) =3.16 m Hint: wavelength = (3x108 m/s)/94.9x106 Hz 3 m 1.6 m 0.8 m 0.5 m path length diff = (d+x)-(d-x)= 2x Destructive 2x=/2x=/4 Constructive make 2x=x=/2 x increases by /4 = 3.16m/4=0.79m

7. Two-slit interference

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

9. 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.

10. Thin film interference air 1/2 wavelength phase shiftfrom top surface reflection air: n1=1 Reflecting from n2 air/n No phase shift frombottom interface n2>1 t Reflecting from n1 air: n1=1 Extra path length needed for constructive interference is Extra path length

11. eye Incident light nair=1 nfilm=1.2 t=275nm nglass=1.5 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

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

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. Overlapping diffraction patterns Angularseparation • 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.  Circular aperture diffraction limited:

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

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. Special case:Total internal reflection i,1 r n1 n2 2 Angle of refraction Refraction • Occurs when light moves into medium with different index of refraction. • Light direction bends according to

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. Lenses: focusing by refraction Image F P.A. Object 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

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

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? 2 3 4 5 10

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

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? f~0.1m f~0.2m f~0.5m f~1.0m s’=2 m mag=10 -> s’=10s ->s=0.2m

24. Physics 208, Lecture 5 Diverging lens Optical Axis Object Image Focal length defined to be negative Then thin-lens equation can be used:

25. p q Virtual image ‘Object’ Eyepiece CompoundMicroscope Objective Real, inverted, image Object Objective lateral mag. =-q/p~L/fobjective Eyepiece: simple magnifier. Angular Mag.=25cm/p ~25cm/feyepiece Physics 208, Lecture 4

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/r2 • Insulators and conductors • Polarization of insulators, conductors

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 1018 electrons or protons • ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πeo) • eo permittivity of free space = 8.854 x 10-12 C2 / N.m2 • Directed along line joining particles.

28. 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? Left Right Up Down Zero - + -

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

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