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: 23.19 Here is why there will be a “circle” of light: Not aii the light rays from the spotlight that hit the
surface wili emerge. Rays with an angle greater than the critical angle ('3’,3 will be reﬂected internally and not
emerge. The angle QC is measured from the normal to the surface of the water. Therefore, the rays that will emerge are all within a cone whose vertex is the spotlight and whose. “base” is the circle that the problem
is asking about. Here’s the picture. .W7e’re given the depth of the pool 13 = 2.00 in. VVe’ll determine the critical angle ac and use this together
With E to ﬁnd at, the diameter of the circle. 876  CHAPTER 23. GEOMETRIC OPTICS From Equation 23.8 on page 1052 of the text, the critical angie satisﬁes . 71.2 1.00
88 z — t —— = 0
5111 m 1.33 22> 66 48.8 . From the above sketch, d
5 : Etan (9,; :> d : 2€tan 6,: : 2(200 In) tan 48.80 x 4.57 In. 23.33
a) Use the mirror equation (Equation 2323 on page 1060 of the text). l '1 __ 2 ,m Rs ,_ (10.0 cm)(—500ni) _
5 +8’ —R II} S — 2s—R :> 3 H 2(—50cm)—10.06m 04.5%].
_A b) Use the mirror magniﬁcation equation (Equation 23.24 on page 1061 of the text).
3’ 4.5 cm
2 —— = _—— = .090.
m s {—50 cm) 0 C) The image distance is positive, so the image is behind the mirror. Therefere, the image is virtual. d) The maghiﬁcatiori is positive, so the image is rightusideup. 23.36
a) The focal length is half the radius of curvature of the mirror (Equation 23.22 on page 1059 of the text), SO :—————=2.0m. E 4.0 m
2 2 Notice that this is a convex mirror, so according to our Cartesian sign conventions both R and f are positive.
b) ‘Use the mirror equation {Equation 23.23 on page 1060 of the text) to ﬁnd 5’. .C) i l 1 , f5 (2.0 m)(—3.0 m)
_ l __ = _ = : ——————— z 1.2 .
sTs’ 13:” s—f —3.0rn—2.0m m
Use the magniﬁcation equation (Equation 23.24 on page 1061 of the text) to ﬁnd '
3’ 1.2 m I
: —— : —~———~— 2 0.40.
m s (”3.0 m) Since the object is 2.2 In tall, the size of the image is 040(2. 2m)—0.88rn. d) The image distance is positive, so the image is in back of the mirror. Therefore the image is Virtual. B) The magniﬁcation is positive, so the image is rightside—up.
fl  You can USE Figure 23.49 on page 1003 of the text. Just replace the key with'a basketball player. 23.56
a) Much the way the bottom of a swimming pool does not look to he as deep as it really is, your arms Wlll appear shorter than they really are. Here is a more formal treatment:
We can ignore the glass, and concentrate on the wateruair interface. The only point of the glass is to allow light rays to come through and to keep the water out of your face. The glass will bend light rays, but
they will bend back when they leave the glass so Eight rays normal to the glass will be unaffected, those not
normal Wiil have a slight but negligible lateral shift. Let your ﬁnger tips be the object. Your eyes are in the air behind the glass, the object is in the water. So
now, in order to be in the “standard coordinate system," you are Swimming to the left with your arms out
in front of you, the origin is at the surface of your mask, and your eyes have a slightly positive coorciinate.
Suppose your arms are 2 long, and the index of refraction of the water is 1.33. Because the surface of your UDUJVJL'IJ. LLLU vr lit/'0' UHAJ" '1 'L'Iﬁ. 2.5. 900 t adius of curvature is infinite. So, the single surface refraction equation (Equation 23.35 on
1 s r ' mask is ﬂat, ' . .
page 1065 of the text) IS .. . _. . , ., .._.
. a 1.33 1.00 1.00—1.33 : _1 0012440 7523
*€i+§i:ngam==>“:?+?t com 5” 133 arms appear to be only 0 752 as long as they actually are.
70 cm)— — 53 cm long 'So your
752 but time is unchanged speed (Change of distance  b) . From part a) your arms will appear 0. 752{
a factor of 0. I' .' .. d'st 1103 are reduced by _
C) Since 1 a b d by afactor of 0.752 : .J ammo. me time) will he reduce w '_.. 23.63
3.) Use the lens niakers’ equation (Equation 23.51 on page 1070 of the text). 711 1.00 f= “—dn—T“ “W——T“‘—1‘"
(n2 “10(3—1 — 3—2) (150‘ 1'00) (20 cm H 25 cm) b) The diopter power is the reciprocal of the focai length expressed in meters. 1
diopter power = m = 0.50 dp. C) The focal length is positive, so the iens is converging.
Ci) Use the thin lens equation (Equation 23.52 on page 1071 of the text) to locate the image. 2:2.le02 cm :20111. _E+£=i m s’: sf :(w0.40m)(2i0m)=_0.50m.
s s’ f s+f —0.40n1+2.{lrn
Since the image distance is negative, the image is virtual. The magniﬁcation is
si _0_50 In
m .3 42.40 m The magniﬁcation is positive, so the image is upright. The ray diagram is the same as Figure 23.?0 on page
1075 of the text. 23.65 a) The focal lengths of the two lenses are found from 2.0dp :—1— 22> f120.501n :50 cm and 1 l
—5.0 tip 2 E => f2 : —O.20 In : ~20 cni Find the location 5’1 of the image produced by the ﬁrst iens relative to the ﬁrst lens with the thin lens
equation. 7 _i+1_1:> 1 +i_ l 2:55',_(50cm)(1'50cm)__75c'm
51 31—1”; —1500m s’IMSUcni 1T150c1n—45ficrn __ ' This image is 75 cm to the right of the ﬁrst lens, and thus 25 cm to the right of the second. it is the object
for the second lens, so 52 : +25 cm, a virtual object. Apply the thinlens equation to the second lens. 1 i 1 l 1 1 , (20 Cm)(25 cm) 2
m_. __—;_=>._ '—:~_—:;> :m~———u:— O .
 .92 ‘+ sg f2 25 cm T 3’1 ——20 cm 51 20 cm —25 cm 1X 1 cm The ﬁnal image is i x 102 cm to the ieft of the diverging lens and 50 cm to the left of the converging lens.
‘0) The magniﬁcation of the ﬁrst lens is The total magniﬁcation isithe product of the individual magniﬁcations. ' _. m = mlmg :.{—O.5G){—4) : 2. .3) _Since.thelﬁnal_ image distance is negativeJ the image is virtual. The total magniﬁcation is positive, so
the ﬁnai image is upright.  _ ' " 23.71 'There are three variables, 3, 3’, and f , and we are given three facts. So, with luck, we should be able to sort things out. Here are the facts: . . ‘ . . .
We are given that the image is inverted and three times the Size of the object, so tne magniﬁcation rs W300. Thus 5! tar—300. 1:3» 8 +3003=0m @ S Therefore, in the standard geometry, the object is to the left of the lens and the image to éhe right, so
8 — s is positive and is the distance between them. We are also told that distance is 5. 00 rn. ence, (2) s’—s=5.00m. 908 CHAPTER 23. GEOll/IETRIC OPTICS  Finally, we have the thin lens equation {Equation 23.52 on page 1071 of the text}, ' 1 1 1
(3) *g+;7=f. Now to sort things out:
Subtract Equation (1) from Equation (2) and find that H4005 = 5.00 In, so 8 2 —l.25 In. Use this value of s in either Equation (1) or (2) and find that
3’ n 3.75 In.
Now substitute these values fOr s and 3’ into Equation (3) and soive for f.
1 1 1
"1.25m +3.75m 2} =:> f20.938m. 2.3.95 b‘irst ﬁnd the focal lengths of the two lenses. From Equation 23.60 on page 1078 of the text, . 1 1
(11013“?r Power = ”*— :> fin meters 2 . .
in meters chopter power So, the two focal lengths are 1
=——=w0.10 2—1.0 ad 2 =. : . .
f1 —10.0dp ‘ .0m 0 cm 11 J”; 510%} 0200111 200cm
3.) Apply the thin lens equation (Equation 23.52 on page 1071 of the. text) to the objective lens and solve
for the image distance 3’. In the standard geometry, the bug—a—boo’s position is s : w25.0 cm.
3. . 1 .1 1 ‘ ,__ f8 __ (m10.0 cm)(——25.0 cm) H
SUI—f 2815+ _ 25.0cmm+10.0cm “17’1“” This image is the object for the second lens. Since
lenses are located 25.0 cm apart, it is Hence, to ﬁnd the location of the seco it is located 7.14 cm to the left of the ﬁrst tens, and the
located 7.14 cm + 25.0 cm 2 32.1 cm to the left of the second lens.
nd image, apply the lens equation with s = —32.1 cm. 1 l 1 , fs (20.0 CHIN—32.1 cm)
_... .L .__ = —.. :7; 2 L: m 2 . .
s ' s’ f 8 s é— 432.1 cm + 20.0 cm 53 1 cm The ﬁnal image is 53.1 cm to the right of the second lens. 255.37. 7 our) b) The total magniﬁcation is the product of the individual magniﬁcations. The magniﬁcation produced
by the ﬁrst lens is 5’ —7.14 cm
x — r —— 2 0.286.
ml 3 —25.0 cm
Themagniﬁcation produced the second lens is .
s’ 53.1 cm
 — z — : —l.55.
m2 __ s —32.1 cm So, the ﬁnal magniﬁcation is
m : mlmg :(O.286)(——1.65): —0.472. c) The ﬁnal image distance is positive, so the ﬁnal image is real. d) The total magniﬁcation is negative, so the ﬁnal image is inverted with reapect to the original object. ...
View
Full Document
 Spring '09
 HENRY
 Physics

Click to edit the document details