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P202 Chap 26 - P202 Spring 2006 Ch 26 Geometrical Optics...

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Unformatted text preview: P202 Spring 2006 Ch 26: Geometrical Optics Homework due at 11:59pm March 19, 2006 Error allowance on all numerical answers is 3% l. A ray of light reflects from a plane mirror with an angle of incidence of [30 —— 50] degrees. If the mirror is rotated by an angle 0 = [10.0 —— 15.0] degrees, through what angle is the reflected ray rotated? Solution: 20 2. Is the image formed by reflection from all three mirrors of a three—dimensional corner reflector upright [radio button] or inverted [radio button]? 13 the image inverted left to right? [choice of yes or no] The easiest way to think about this is to imagine the 3 mirrors being two walls and the floor at the corner ofa room. The image formed by each of the “walls” individually is left—right inverted but upright. Thus the image formed by reflection from the two walls is not L-R inverted ( the second L—R inversion undoes the first one) and upright. Finally, the “floor” creates an image of the image created by the two “walls This does not change the L-R inversion but does make an image that is upside down. Perhaps a better way to think of the problem is to draw three Cartesian axes {x,y,z) as the object with each axis being parallel to the intersection ofa pair of the corner reflector’s three mirrors. The image that involves reflection in all of the three mirrors has each of the Cartesian axes inverted with respect to the object. Thus, whatever direction you call “up" in the object is reversed in the image. Solutions: inverted; no 3. The virtual image produced by a convex mirror is one quarter the size of the object. (a) If the object is d : [30 — 40] cm in front of the mirror, what is the image distance? cm (b) What is the radius of curvature of the mirror? cm Since the image is virtual, di is negative. Solutions: (a) ~d/4 ; (b) 2d/3 4. A submerged scuba diver looks up toward the calm surface of a freshwater lake and notes that the sun appears to be 6 = [30 — 40] degrees from the vertical. The diver’s friend is standing on the shore at the edge of the lake. At what angle above the horizon does the friend see the sun? degrees B‘W SW91:MSWB: "53"‘9 1 319.... 31.;‘(132 5.4. e) ‘9‘, 4-1»le aim-m- L.W~W = a“ " EH be ’ : ‘io- su."(1.3zs...9) Solution: 90 — sin’](l.33 sin 6) [note that sin’l must be evaluated in degrees] 5. A small insect v1ewed through a convex lens - 1 ~ . 1 . [d = 1.0— 1.5] cm from the lens and appears twice its actual size and inverted. (a) What is the focal length of the lens assuming it is made of glass? cm (b) What is the focal length required to achieve the same result if a polycarbonate lens is used? cm J— + is .1. ms‘éi Swat; who»? do twunfid (44'3“: +V¢L f a... 1. - - .L J...=_!... .1 (a) Z—ch/aaed.~2cla—ZA £434 5—9 fl-jélrg 2-,. (b) m {amt Bug-rt ma Manama...» “t It; cam: awaitn ’25:: m. mat-m» M 11;... 5mg cm ta.) Solutions (a) 2 d/3 ; (b) ;2d/3 6. Make-up l shaving mirrors usually have one flat side and one concave side. You find that you can project a magnified image of a light bulb onto the wall of your bathroom if you hold the mirror [u =1.0 — 1.5] m from the bulb and [v = 3.0 -3.5] m from the wall. (a) What is the magnification of the image? (b) Is the image erect [radio button] or inverted [radio button]? (c) What is the focal length of the mirror? m (Ox) mg; —-U-/UL .—. L113” 7,.st “U“ + .L 11" mu” T .L2i Lalf u Solutions: (a) — v/u; (b) inverted; (c) u v / (u + v) 7. A friend tells you that when he holds his eyeglasses [d = 20 — 25] cm above a printed page the print appears the same way up but reduced to [m = 0.60 — 0.70] of its actual size. (a) Is the image of the print real [radio button] or virtual [radio button]? (b) What is the focal length of your friend’s glasses? cm (C) Is you friend nearsighted [radio button] or farsighted [radio button]? (a) Upright images produced by either a convex or a concave lens are always virtual. {bl 4—... J... = J. :12. i = M"! _ a Mal d mcl f .5: Mal 9 4 (10'1“ I) CC) ’3ch \M<l 4m v-Ve. 6a It]. leMr M meats... Cumming, l-EME 0cm uueal t5 com..et «A9114, 5%}: Solutions: (a) virtual; (b) rn d l (m — l); (c) nearsighted 8. A 45° — 90° - 45° prism is placed on a flat table with one side vertical. A horizontal beam of light enters the prism through its inclined face and exits through the vertical face at an angle of [8 = 30 — 40] degrees to the horizontal. What is the refractive index of the prism? 93:45-"82 SM; Lgf: ’Y‘t Swine}. “KC/{AMA ALL '1“. Sum 83 3119“; (“$361) .‘t SWG Vukgafl fags. m [kaVmgL- marshal]: 5.4.6 .....--—1 [—2- 1 -: l ,' L638 : (T, “L :' ______fl-‘ /\E-n ' 2- 2m; Vin '2 ’ h‘-‘— i.-L :41 |— NH. din-1 LLZM"§ Solution: sqrt[{4(sin 9)2+4sin6+2}/2] 1'2. (294nm 0°“: ZnL-l 9. What is the minimum refractive index of a cylindrical optic fiber that transmits all light incident on the flat end of the fiber cut at 90° to the fiber axis? 3W8.— 'Y\QW‘SL (3 W1. 39, {a- {-1) (mi) why m “C: '19.}! 6‘ Jim " “l I {31‘ L- ”wt [HR '1 i k. “ I” (z. 7.? E {HQ ‘E‘wt 1:131. *5 “~21 610‘“ 6g > 9C New m ELL. BC : I .', file-Q “Rx-QCu I'm—1:. llfi IS Vii-t (“0'0“ 01 >" %“ l Q LD' ‘92 ‘3? 7,. , M ' ,_ --. Ci Fm A- ('3 COR Q;- .'.' \f t ”M i M" 3 f} .t ‘ ‘ _. "“t'z" ‘- ”(11 , _ a. flute; Qttgu ”up... at... .r: mo: a: m m a \< arm-tug. fl 5-1 r ”to l. 3; WW J 3:“) Ntv:\‘WUM WM 1 C539; : J‘P‘/"\L _ MM-“ '2 2 ”Tl/WJ WA. 112%“... i 4? Jun “-2. 5 /V1 2» I/n 1- e. it --_~ l “r"- i r: n P l I ta, “n > E. Solution: 1.414 10. The focal length of a lens is inversely proportional to (n — 1) where n is the refractive index of the lens material. The value of n depends on the color of light passing through the lens. For flint glass, the index of refraction is 1.572 for red light and L605 for violet light. Suppose a white object is placed [u = 20 — 25] cm from a flint—glass lens. If the red light reflected from the object forms a sharp image [v = 50 — 60] cm from the lens, at what distance from the lens will the violet image be found? Solution: u v I (1.058 u + 0.058 v) 41v 6:2 I _____ : OM; at 4V - 1e 0 gd}, : ‘iE/I 0‘79“ 45R 0 5m 0 605‘ 71»: Violet 0H”. 31“?“ J 14 .L.; i-J. ‘ .1 ‘fv VU ‘h u. J- : “3.9.5."? m L: 10$wang _, .1 W J (-1 u ’1. IR u _l_ x l as”? u. 4 O (‘ 3 57’ .53 V4 m.” H a} (I- ‘ ._ LR JR . \IV- __,____M ,. m- ...
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