This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 fnumber
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 tax3°. £81 may} its (.l— lambs ; CA—iwtr Elir! :é, 0
Manamxlgmc. less1 clue gnu" it.an ml « .— LLalr ac W5 1C; Li J'w‘dCt 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 : CiDC‘£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
lQ MN: MCEfLlij): {+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* MawQ okra: Lit/4..
,. ng m. e‘Iavaeé’ wartmas 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 (MEMm maén at. new“ ‘.,Lh;/r?ai)/(L{/jg°)= {0/le
71M: He]: (go/m. gem“. in ELEGPULLJL 1;: arrewe 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(m1)/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 34571:
_ ; .. 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 Lbdadw? \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) ...
View Full
Document
 Spring '09
 WARREN
 Physics

Click to edit the document details