P202 Chap 271

# P202 Chap 271 - P202 Spring 2006 Ch 27 Optical Instruments...

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Unformatted text preview: P202 Spring 2006 Ch 27: Optical Instruments Homework due at 11:59pm March 26, 2006 Error allowance on all numerical answers is 3% 1. You take a picture of a poster on the wall of a room and maximum distance you can get from the picture is [d = 2.8 — 3.2 ] m. The poster is 0.8 m wide and 1.2 m tall and you want it to ﬁt in the 24 x 36 mm frame of the ﬁlm in your camera. What is the largest focal length lens that you can use? mm feoLpLUec‘ waét‘tmiu; :--36/I>.0c:> ,1: ~gxlo'z“ laws We wag/s, so 414‘: 3r to’zclo dictation: pass... it; 51m = a: 4,. so = Loss, t—e do= A/LOZ d (it.- 3x19'ael/I432 is L .- (~03 t-‘axto‘ - 353‘; i «If 34' at" n. m -- A as «l = 28228 cl M Solution: 29.13 (1 You are taking a photograph of 100 yard dash. A shutter speed of 125 at f/5.6 produces a properly exposed image, but the runners’ legs are blurred. Your camera has fstops of2, 2.8, 4, 5.6, 8, 11 and 16. (a) What f stop should you use? (b) What is the shortest exposure you can use and still get a properly exposed image? To “freeze” the runners ’ leg motion you need to increase the shutter speed i. e. decrease the exposure time. T 0 ensure that the same amount of light reaches the ﬁlm. you will need to increase the diaphragm opening i. e. decrease the f-number as much as possible. The minimum f—stop is 2, so this is the answer to part (a). Since you have decreased the f—stop by a factor of 2 *sqrt(2) (i. e. 2. 8), you have increased the diaphragm area by a factor of (2 *sqrt(2))2 = 8. This allows you to reduce the exposure time by the same factor i. e. to 1/1000 3. The question gives the shutter speed as the inverse of the exposure time (not uncommon) so the answer to (b) is 1000. Solution: (a) 2; (b) 1000 With unaided vision a person can focus only on objects that lie between distances of [df= 5.0 — 6.0] m and [dn = 0.5 e 0.6] m. Assume that the person wears bifocal glasses 2 cm from their eyes. (a) What is the refractive power of the main part of the lens of their glasses? diopters (b) What is the refractive power of the bifocal part of each lens, assuming that reading material is held at 25 cm from the eye? diopters (a) For the main part of the glasses, we need a lens forms an image at the far point of an object at inﬁnity. This is a concave lens with a focal length equal to the far point distance minus the distance between the glasses and the eyes. Thus f =—(df— 0.02). T he lens power is just the inverse of this. (b) For the reading glasses, we need a lens that produces an image at the near point of an object at 25 cm from the eye. This is a convex lens that produces a virtual, upright image. So, do = 0.25 — 0.02 = 0.23 m and di = — (dn — 0.02).. Applying the lens equation gives 1xf= 1/023 — I/(dn — 0.02) diopters Solution: (a) -1/(df- 0.02); (b) 4.3478 — l/(dn — 0.02) . Your friend’s eye glasses have a refractive power of [D = 2.5 — 5.0] diopters. When you look at your friend’s eyes, by what factor are they magniﬁed, assuming that the glasses are 2 cm from the eyes? T he focal length of your ﬁ'iend ’s glasses is 100/D cm. The object distance is 2 cm. Applying the lens equation gives an image distance of 200/(2D ~ 100) cm. Magnification is given by m = — di/do = - 100/(2D - 100) Solution: 100/(100 - 2D) . A converging lens with a focal length of[ f= 4.0 v 6.0] cm is to the left of an identical lens. When an object is placed [(1 = 12.0 — 15.0] cm to the left of the first lens, the ﬁnal image is the same size and oﬁentation as the object. What is the separation between the lenses? cm Note that since d > f, the image formed by the first lens is inverted. To obtain a ﬁnal image that is upright, the second lens must also produce an inverted image of its object (which is the image produCed by the ﬁrst lens) :T/M tax-3°. £81 may} its (.l— lambs ; CA—iwtr Eli-r! :é, 0 Manamxlgmc. less-1 clue gnu" it.an ml «- .—- LLalr ac W5 1C; Li J'w‘d-Ct Lg. {we -_.- “Le. last: . 3:» 04A“ in its £w§6 w; a To in Uf'lt-‘Cllv- i) m {ﬁgust we]... t *lytm ft; 4‘2. and (We was? Lu 61:533. . Twe " \S‘M‘J L “1 Q‘H‘J 161w: J"— 4 L: .1. it Me: wait. 5 (j A/ )l, -‘ (31' C12 Z . iii ‘ “it; 9° Ortega—i). fww Wt <9 ﬂ : Ci-DC-‘£CL 01 cix: 4 .2- ltc x:- tdrl) (1‘45) an; a- “‘— Solution: 2 fd /(d— f) A; A '7: 6. A person with a near point distance of [N = 25 v 35] cm ﬁnds that a magnifying glass held very close to the eye gives an angular magniﬁcation that is [m = 1.2 — 1.8] times larger when the image produced by the magniﬁer is at the near point than when the image is at inﬁnity. What is the focal length of the magnifying glass? cm F0! “11 {Wrxm a'r “Jim (EL; — 1113‘: EL 'l do: \$4,; MN: (MAD/0mm).- N/clo l-Q MN: MCEfL-lij): {+N/g PM it; may at angtJ mw: N/‘F MN ‘ my: l+~}/N=rm ~».£:(M—))‘\J cm. Solution: (m — UN 7. Galileo’s ﬁrst telescope used a convex objective with a focal length of f = [1.5 ~ 1.7] m and a concave eyepiece. When the telescope is used to View a planet and produces an equally distant image, its angular magniﬁcation is + [m : 3 — 3.5]. (a) what is the focal length of the eyepiece? m (b) how far apart are the two lenses? Il‘l ll “we a “4* Maw-Q okra: Lit/4.. ,. ng m. e-‘Iavaeé’ wart-mas A“ Mgr“; cita- V'idiné.‘m\ TL: Hacienda Wc’dL‘mJ “a [a cam i v; 7‘“ “We % rt.- \1/\:_¢'v\£r— h“) A N (ME-Mm maén at. new“ ‘-.,Lh;/r?ai)/(L{/jg°)= {0/le 71M: He]: (go/m. gem“. in ELEGPULLJL 1;: arr-ewe Jig.- -—;‘ ’11“ L ‘iNU‘ 19.» he.» 2.9.44.2 QCMI? ’3‘»: (r (C! a! m I : 40 (M -- I) /H 10. ...\ Solution: (:1) — f/m; (b) f(m-1)/m The Moon has an angular size of 0.5 ° when viewed with unaided vision from Earth. Suppose the Moon is viewed through a telescope with an objective whose focal length is [f0 = 50 — 55] cm and an eyepiece whose focal length is [fe = 20 — 30 mm]. What is the angular size of the Moon as seen through the telescope? ° The angular magniﬁcation produced by a telescope is foﬂe. Thus the angular size of the Moon seen through the telescope is just 0.5 fo/fe degrees. In this case we have to careful because f0 is given in cm and fe is given in mm. Solution: 5 f0 / fe When a lamp is placed [u = 40 a 45] cm to the left of a convex lens a real image is formed 37.5 cm to the right of the lens. The lamp and lens are kept ﬁxed and a concave lens is placed [d = 15 — 20] cm to the right of the convex lens. A real image of the lamp is now formed 35 cm to the right of the concave lens. (a) What is the focal length of the convex lens? cm (b) What is the focal length of the concave lens? cm ‘ 7' [UV CAM ug y if” f -V 4 .J. I. + 5 a m J- . “‘7’ \‘ {I ..: l v ; f .L t i .. 1 ‘4 ll -. * Lilaf—J) 35' {L <2 V“) A s1. j, c. .. . f.= 5m 34-571: _ ; .. j 7 ‘ 1 If.- r h 9' r \ ¢. f '- .Wp a 1* 7‘ n «a ~ “5th :1) , LE‘ rel—24‘“ Solutions: (a) 37.5 u /(u + 37.5); (b) —35 (37.5 — d)/(d — 2.5) The “tube length” of a microscope is deﬁned to be the difference between the image distance for the objective and the focal length of the objective. Many microscopes have a standardized tube length of 160 mm. Suppose such a microscope has an objective with a focal length of [f = 7.0 — 8.0] mm (a) How far from the object should this lens be placed? mm (b) What focal length eyepiece would give an overall magniﬁcation of [m = ~ 50 to * 60]assuming a near point of 25 cm? cm . - m 0 0 x n. ; ex) 1‘3" O'kﬁji‘fafwt ‘50 ' (“Bx "'3" 3&1" Jr “‘ 9539 wu .L k » 0 z .. gm do; {null—1.6:. t 5“" \ 0° MN ‘60 do} “QViX‘C“t““L LL“ ﬁ- Lbda-dw? \io “(ENC/(4° : “(£4‘bi)%-h, ~30; i V” . ‘ ,. ,1 ,1; «.4»: in J”: a bag.‘ as {\4 '- ‘0 .,, Q \ “I ‘5 ‘ ks UM‘C Lin; 0.13:“ c. ‘f (‘3?) I g Rm.“ 6 ﬁnk— e €211: a. g n ‘ K" , A ~ 7. _ C 0. FL «Mr—«V w J. (T. , 4a» . .3 M - i * u k L ,cv AC ' (1 Quebe- ° c‘ NH}; CPA/Qua DJ 1 F’ ‘30 W M - ﬂ m9 6‘\ ( s — a Y. lg?) (n M N ..‘ . ‘ a ) k5. D Solution: (3) f(160 + 0/160; (b) — 4000 x (m t) ...
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