Lecture 36Chapter 35 - 36Images & InterferenceReview • Mirrors– Plane – flat mirror– Concave – caved in away from object– Convex – flexed out toward object – Real images on side where object is, virtual images on opposite side – Plane and convex mirrors make only virtual images– Concave mirrors can produce both real and virtual imagesReview • Spherical mirrors have focal point, r is radius of curvature• Find focal length, f from– Object distance p is +– Image distance i is + for real images, - for virtual images– f is + for concave, - for convexfip111=+rf21=Review • Ratio of image’s height h´to object’s height h is called lateral magnification, m• Magnification also equal to hhm′=pim −=• m is + if image has same orientation as object• m is – if image is inverted from object• Plane mirror m =+1Review - Mirrors-++InvertRealEquali = 2fp = 2fConcave+-+SameVirtualBiggerAnywherep < fConcaveAnywherep > 2ff < p < 2fAnywhereObject Location+--SameVirtualSmaller|i| < |f|Convex-++InvertRealSmaller2f > i > fConcave-++InvertRealBiggeri > 2fConcave+1-∞∞∞∞SameVirtualEquali = - pPlaneSignof mSignof iSignof fImageOrient-ationImageTypeImageSizeImage LocationMirror TypeReview • Thin Lenses– Light rays bent by refraction form an image– Converging – lens with convex refracting sides– Diverging – lens with concave sidesReview • Thin Lenses– Real images form on opposite side of lens from object, virtual images on same side– Diverging lens only produces smaller, same orientation, virtual images (like convex mirror)– Converging lens (like concave mirror) can produce both real and virtual images depending on where the object is in relation to the lens’ focal pointReview • Thin lenses have a focal point on each side of lens• Focal length, f same as mirror• Lens maker’s equation –for lens in air, r1is radius of lens surface nearest the object, r2is other surface– r is + for convex surface, - for concave surfacefip111=+−−=2111)1(1rrnfReview • Thin lenses -– Lateral magnification m same as for mirror– For a system of lenses or mirrors the total magnification M is product of each m– Work through system of lenses one by one – use image from one lens as object for next lenspim −=...321mmmM =Review – Thin LensesConverging lens = concave mirrorDiverging lens = convex mirror-++InvertRealEquali = 2fp = 2fConverging+-+SameVirtualBiggerAnywherep < fConvergingAnywherep > 2ff < p < 2fObject Location+--SameVirtualSmaller|i| < |f|Diverging-++InvertRealSmaller2f > i > fConverging-++InvertRealBiggeri > 2fConvergingSignof mSignof iSignof fImageOrient-ationImageTypeImageSizeImage LocationThin Lens TypeLecture 36 (cont.)Chapter 36InterferenceInterference (1)• Light is an EM wave• Interfering light waves combine to enhance or suppress colors in sunlight– Soap bubbles, oil slicks• Interference best evidence that light is a wave• Huygen’s principle – points on wavefront act as point sources of spherical wavelets, at time tnew position of wavefront is tangent to waveletsInterference (2)• Can use Huygen’s principleand geometry to prove Snell’s law (see section 36-2)• Wavelength of light in two different media, 1 and 2, are proportional to 1122sinsinθθnn =12212121sinsinnnvv===θθλλInterference (3)• Frequency of light in medium is same as in vacuum• Wavelength and velocity of light change in a medium and depend on its index of refraction, n• Velocity of light in a medium is always smaller than speed of light in vacuum, c• Wavelength of light in a medium, λnis smaller than in vacuum, λ and related by 12212121sinsinnnvv===θθλλnnλλ=Interference (4)• Phase difference between 2 light waves can change if waves travel through different media with different n• Number of wavelengths in media• Phase difference in terms of λλλ111LnLNn==λλ222LnLNn==)(121212nnLLnLnNN −=−=−λλλInterference (5)• Checkpoint #2 – Rays have same wavelength and initially in phase. A) If 7.6 wavelengths fit within top material and 5.5 fit within bottom, which has greater index of refraction, n ?• Larger n produce smaller λ• Which material has smaller λ?•Smaller λ, more wavelengths in same distanceTop material has greater index of refraction, nnnλλ=Interference (6)• Checkpoint #2 – Rays have same wavelength and initially in phase. B) After material will interference of waves give brightest, bright intermediate, dark intermediate illumination or darkness?• Look at phase difference in terms of λ• Given # of wavelengths for each material • Waves are 2.1 wavelengths out of phase after passing through materials1.25.56.712=−=− NN)(1212nnLNN −=−λInterference (7)• Checkpoint #2 – B) After material will interference of waves give brightest, bright intermediate, dark intermediate illumination or darkness?• If phase difference is an integer # of wavelengths (0,1,2,…) then waves are in phase and have full constructive interference (brightest spot)• Effective phase difference is decimal fraction • Total phase difference = 2.1• Effective phase difference = 0.1Interference (8)• Checkpoint #2 – B) After material will interference of waves give brightest, bright intermediate, dark intermediate illumination or darkness?• If phase difference is 0.5 wavelengths (half a wavelength) then waves are completely out of phase and fully destructive interference (dark spot)• Our effective phase difference of 0.1 is closer to 0 than 0.5 so intermediate bright spot but not the brightest.Interference (9)• For interference pattern to appear waves must have a constant phase difference • If phase difference does not vary with time waves are coherent• Light is produced by emission from individual atoms• Atoms in conventional light (light bulbs, sunlight) are in random phases so light is incoherent• Lasers are designed so atoms emit coherent and monochromatic
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