# Announcements 11/2/12 Prayer Slinkies! (Seth, Ryan, William B, Clement)

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Announcements 11/2/12  Prayer  Slinkies! (Seth, Ryan, William B, Clement)  Progress Report due a week from Saturday Pearls Before Swine From warmup …
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Announcements 11/2/12  Prayer  Slinkies! (Seth, Ryan, William B, Clement)  Progress Report due a week from Saturday Pearls Before Swine From warmup  Extra time on? a.(nothing in particular)  Other comments? a.Do astronomers have to worry about the fact that air doesn't have an n of exactly 1? Clicker question:  A converging lens should be: a. convex b. concave Convex/converging Lens at each interface, Snell’s law is satisfied f if not parallel, they parallel rays converge (image formed) converge to the at a different point focal point Where will image be formed? focus focus Ray diagram just like one for concave mirror, except rays continue to right instead of reflecting back! The equation: 1 1 1 positive q = right Magnification same:   hand side of lens q p q f M  p From warmup  The mirror eqn and the thin lens eqn look identical. How exactly are they different? a.The sign conventions are different, a positive focus means a focus that is behind the lens. The same is true of q. b.(from my answer) Also--how f is calculated is different. For mirrors, it's simply related to the radius of curvature, but for lenses it relates to the curvature as well as the index of refraction of the lens glass. Clicker quiz  Just out of curiosity… have I ever used your warmup answer? a.yes b.no Concave/diverging Lens f parallel rays seem to be coming from the focal point Diverging lenses virtual image! The equation: 1 1 1   (negative f) p q f Negative q: means image on same side of lens as object (virtual) Demos  Demo: “Real image or not?” Class poll: What kind of image is this? a. Real b. Virtual Sign conventions: summary  Assuming that the light rays are coming from the left: quantity positive negative +  p if on the left of the if on the right of the lens/mirror lens/mirror (can only happen in a compound problem) q if on the right of the lens if on the left of the lens (left side of mirror), (right side of mirror), means a real image means a virtual image f if converging: convex if diverging: concave lens/concave mirror lens/convex mirror M if image is right-side up if image is upside-down “Lensmaker’s Eqn” Lens-makers’ eqn: 1  1 1    n  1    f  R1 R2  From warmup  When you look straight down at an object under water, will it appear to be closer to you or farther away than actuality? Why? a. Because you are looking at a flat refracting surface the image of the object forms closer to the surface than the actual object is (figure 36.18). This means the object appears closer to you than it actually is. Clicker Quiz  Is that a real or a virtual image? a. real b. virtual Thick lenses and surfaces n1 n2 n2  n1 Careful with signs!!!   p q R (see table 36.2) Clicker question:  Can a converging lens ever produce a virtual image? a. Yes b. No Clicker question:  What happens if I cover up the upper half of lens? a. Nothing b. Upper half of image disappears c. Lower half of image disappears d. Intensity of image goes down by 50% Clicker question:  Can a diverging lens ever produce a real image? a. Yes b. No Multiple element problems  Image of one becomes object of next  Worked problem: Find qfinal, Mtot f = 20 cm f = -60 cm 60 cm 15 cm Answers: image is 20 cm to right of -60cm lens; M = -0.67
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