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Unformatted text preview: 0%. 29 / =t< 7. Interactive Solution 26.7 at www.wiley.com/college/cutnell of
fers one model for problems like this one. In a certain time, light
travels 6.20 km in a vacuum. During the same time, light travels
only 3.40 km in a liquid. What is the refractive index of the liquid? * 8. A ﬂat sheet of ice has a thickness of 2.0 cm. It is on top of a ﬂat
sheet of crystalline quartz that has a thickness of 1.1 cm. Light strikes
the ice perpendicularly and travels through it and then through the
quartz. In the time it takes the light to travel through the two sheets,
how far (in centimeters) would it have traveled in a vacuum? Section 26.2 Snell’s Law and the Refraction of Light 9. ssm A light ray in air is incident on a water surface at a 43° angle of
incidence. Find (a) the angle of reﬂection and (b) the angle of refraction. 10. A ray of light impinges from air onto a block of ice (n = 1.309) at a 600° angle of incidence. Assuming that this angle remains the same, ﬁnd the difference 62‘ m — (91mm in the angles of refraction
en the ice turns to water (n = 1.333). A spotlight on a
oat is 2.5 m above the
water, and the light
strikes the water at a
point that is 8.0 m hori
zontally displaced from
the spotlight (see the
drawing). The depth of the water is 4.0 m. Determine the distance d,
which locates the point where the light strikes the bottom. 12. Amber (n = 1.546) is a transparent brown—yellow fossil resin.
An insect, trapped and preserved within the amber, appears to be
2.5 cm beneath the surface when viewed directly from above. How
far below the surface is the insect actually located? 13. ssm A beam of light is traveling in air and strikes a material.
The angles of incidence and refraction are 630° and 47.00, respec—
tively. Obtain the speed of light in the material. 14. A scuba diver, submerged under water, looks up and sees sun—
light at an angle of 280° from the vertical. At what angle, measured
from the vertical, does this sunlight strike the surface of the water? 15. Light in a vacuum is incident on a transparent glass slab. The angle
of incidence is 350°. The slab is then immersed in a pool of liquid.
When the angle of incidence for the light striking the slab is 203°, the
angle of refraction for the light entering the slab is the same as when
the slab was in a vacuum. What is the index of refraction of the liquid? * 16. A silver medallion is sealed within a transparent block of plastic.
An observer in air, viewing the medallion from directly above, sees the
medallion at an apparent depth of 1.6 cm beneath the top surface of the
block. How far below the top surface would the medallion appear if the
observer (not wearing goggles) and the block were under water? * 17. Refer to Figure 26.4]; and assume the observer is nearly above
the submerged object. For this situation, derive the expression for
the apparent depth: d’ = d(n2/n1), Equation 26.3. (Hint: Use Snell ’s
law of refraction and the fact that the angles of incidence and re—
fraction are small, so tan 9 3 sin 0.) The drawing shows a
Ct angular block of glass
(n : 1.52) surrounded by
liquid carbon disulﬁde
(n = 1.63). A ray of light is
incident on the glass at
point A with a 300° angle
of incidence. At what angle
of refraction does the ray leave the glass at point B? PROBLEMS l 847 * 19. ssm www In Figure 26.6, suppose that the angle of incidence
is 61 = 30.0“, the thickness of the glass pane is 6.00 mm, and the re—
fractive index of the glass is n; = 1.52. Find the amount (in mm) by
which the emergent ray is displaced relative to the incident ray. * 20. Review Conceptual Example 4 as background for this problem.
A man in a boat is looking straight down at a ﬁsh in the water di—
rectly beneath him. The ﬁsh is looking straight up at the man. They
are equidistant from the air—water interface. To the man, the ﬁsh ap—
pears to be 2.0 m beneath his eyes. To the ﬁsh, how far above its
eyes does the man appear to be? ** 2‘1. ssm A small logo is embedded in a thick block of crown glass (71 = 1.52), 3.20 cm beneath the top surface of the glass. The block
is put under water, so there is 1.50 cm of water above the top surface
of the block. The logo is Viewed from directly above by an observer
in air. How far beneath die top surface of the water does the logo ap—
pear to be? ** 22. A beaker has a height of 30.0 cm. The lower half of the beaker is ﬁlled with water, and the upper half is ﬁlled with oil (n : 1.48).
To a person looking down into the beaker from above, what is the
apparent depth of the bottom? Section 26.3 Total Internal Reﬂection 23. ssm One method of determining the refractive index of a trans—
parent solid is to measure the critical angle when the solid is in air.
If 6C is found to be 405°, what is the index of refraction of the solid? 24. A point source of light is submerged 2.2 m below the surface of
a lake and emits rays in all directions. On the surface of the lake, di—
rectly above the source, the area illuminated is a circle. What is the
maximum radius that this circle could have? 25. Interactive Solution 26.25 at www.wiley.com/college/cutnell
provides one model for solving problems such as this. A glass block
(n = 1.56) is immersed in a liquid. A ray of light within the glass
hits a glass—liquid surface at a 750° angle of incidence. Some of
the light enters the liquid. What is the smallest possible refractive in
dex for the liquid? " 26. What is the critical angle for light emerging from carbon disul
ﬁde into air? Concept Simulation 26.1 at www.wiley.com/college/cutnell il«
lustrates the concepts that are pertinent to this problem. A ray of
light is traveling in glass and strikes a glass—liquid interface. The
angle of incidence is 580°, and the index of refraction of glass is
n z 1.50. (a) What must be the index of refraction of the liquid such
that the direction of the light entering the liquid is not changed?
(b) What is the largest index of refraction that the liquid can have,
such that none of the light is transmitted into the liquid and all of it
, reﬂected back into the glass? "' :’ The drawing shows a l
n2 2 1.63 1 ray of light traveling from
point A to point B, a dis—
tance of 4.60 m in a mater—
ial that has an index of re—
fraction 721. At point B, the
light encounters a different
substance whose index of
refraction is 712 = 1.63. The 
light strikes the interface at the critical angle of 9c = 48.1“. How
much time does it take for the light to travel from A to B? * 29. A layer of liquid B ﬂoats on liquid A. A ray of light begins in
liquid A and undergoes total internal reﬂection at the interface be
tween the liquids when the angle of incidence exceeds 365°. When Cﬂzé f 1328 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS Ice 62 M :Sin_1[(l.000)81n60.0 J: 4140
. I 1.309
_ . 1 (1.000)'sin 600° 0
Water 62’ water — Sln = The difference in the angles of refraction is 62 ,iee _62, water : 4140—4050 : REASONING AND SOLUTION The angle of incidence is found from the drawing to be 61 = tan“1 = 73°
25 m Snell's law gives the angle of refraction to be sin 62 I (ml/n2) sin (91 = (LOGO/1.333) sin 73° = 0.72 or (92 = 46°
The distance d is found from the drawing to be d= 8.0 in + (4.0 m) tan 02 : 12. REASONIN G AND SOLUTION Using Equation 26.3, we ﬁnd d=[—n—1—]d'=[ﬂ]2.50m= 712 1.000 l3. REASONING We begin by using Snell's law (Equation 26.2: ml sin 61 = r22 sin 62) to find the index of refraction of the material. Then we will use Equation 26.1, the
deﬁnition of the index of refraction (n = c/ v) to ﬁnd the speed of light in the material. SOLUTION From Snell's law, the index of refraction of the material is _ n1 sint91 _ (1.000) sin 63.0°
sin 492 Sin 470° 21.22 "2 Then, from Equation 26.1, we find that the speed of light v in the material is Chapter26 Problems 133] 18. ASONING Snell’s law will allow us to calculate the angle of refraction 192 B with which e ray leaves the glass at point B, provided that we have a value for the angle of incidence
61, B at this point (see the drawing). This angle of incidence is not given, but we can obtain it by considering what happens to the incident ray at point A. This ray is incident at an angle
61 A and refracted at an angle (92 A. Snell’s law can be used to obtain (92 A, the value for which can be combined with the geometry at points A and B to provide the needed value for
81 B. Since the light ray travels from a material (carbon disulfide) with a higher refractive
index toward a material (glass) with a lower refractive index, it is bent away from the normal
at point A, as the drawing shows. SOLUTION Using Snell’s law at point B, we have . . 1.52 .
(l.52)sin6’1B :(l.63)sm62,B or 51116sz = ﬁjsmﬂﬂ (1)
Glass J Carbon disulﬁde .
To ﬁnd (91 B we note from the drawing that
61 B +6?2 A = 90.00 or 617 B = 90.0°—62’ A (2) We can ﬁnd 62 A, which is the angle of refraction at point A, by again using Snell’s law: . . 1.63 .
(l.63)sint91 A =(1.52)sm62 A or smt92 A :[Ejsmﬁl A
7 a , _ 1
Carbon disulﬁde Glass Thus, we have A = sirr1 (0.536) = 32.40 a . 1.63 . 1.63 . o
sm62jA={TSE 511161’A:[—1'.5—2]31n30.0 20.536 or 62 Using Equation (2), we ﬁnd that 1332 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 91,3 = 9000—62 A = 90.00—32.40 = 57.60 With this value for 6’1 B in Equation (1) we obtain . 1.52 . 1.52 . .
sm 62,13 =£ﬁjsm6u3 =[Ejsm576o : 0.787 or 19.2, B = SID—1(0.787) = 51.90 19. The drawing at the right shows the geometry of the situation using
the same notation as that in Figure 26.6.
In addition to the text's notation, let t
represent the thickness of the pane, let L
represent the length of the ray in the pane,
let x (shown twice in the ﬁgure) equal the
displacement of the ray, and let the difference in angles (91 7 62 be given by q}. We wish to ﬁnd the amount x by which
the emergent ray is displaced relative to
the incident ray. This can be done by
applying Snell's law at each interface, and
then making use of the geometric and
trigonometric relations in the drawing. SOLUTION If we apply Snell's law (see Equation 26.2) to the bottom interface we obtain
n1 sin 19] = r22 sin 62‘ Similarly, if we apply Snell‘s law at the top interface where the ray emer es, we have n sin 6 = n sin 6 = n sin 19 . Com arin this with Snell's law at the
g 2 2 3 3 1 3 P g bottom face, we see that 711 sin 61 =711 sin 63, from which we can conclude that 63 = (91. Therefore, the emerging ray is parallel to the incident ray. From the geometry of the ray and thickness of the pane, we see that L cos 62 =t, from which it follows that L = t/cos 62. Furthermore, we see that x : L sin ¢ = L sin ((91 —192) .
Substituting for L, we ﬁnd I sin(6’1 —— 62) :L sing—19 =
x (I 2) cost92 Before we can use this expression to determine a numerical value for x, we must ﬁnd the
value of 492. Solving the expression for Snell's law at the bottom interface for 62, we have 1336 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS As 11L]. qui d decreases, the critical angle decreases. Therefore, nLiquid Cannot be less than the I
value calculated from this equation, in which BC = 75 .00 and nGIaSS = 1.56. SOLUTION Using Equation 264, we ﬁnd that H Li quid sin 60 = or nLiqwd = nGlaSS sin 19c : (1.56) sin 75.00 = 71 Glass 26. REASONING AND SOLUTION Using Equation 26.4 and taking the refractive index for
carbon disulﬁde from Table 26.1, we obtain ASONIN G AND SOL U TI ON
. The index of refraction 712 of the liquid must match that of the glass, or 722 = 1.50 . b. When none of the light is transmitted into the liquid, the angle of incidence must be equal
to or greater than the critical angle. According to Equation 26.4, the critical angle 6C is given by sin (9C = nz/nl, where 712 is the index of refraction of the liquid and n1 is that of the glass. Therefore, 1 _ 712 = 111 sin 60 = (1.50) sin 58.00 : 1.27 1f n2 were larger than 1.27, the critical angle would also be larger, and light would be
“ transmitted from the glass into the liquid. Thus, 722 = 1.27 represents the largest index of
1 refraction of the liquid such that none of the light is transmitted into the liquid. .' n = 1.63
28. EASONING The time it takes for the light to travel from A 2
o B is equal to the distance divided by the speed of light in the
substance. The distance is known, and the speed of light 12 in
the substance is equal to the speed of light c in a vacuum
divided by the index of refraction 11] (Equation 26.1). The index of refraction can be obtained by noting that the light is
incident at the critical angle QC (which is known). According to Equation 26.4, the index of refraction n1 is related to the critical angle and the index of
refraction 112 by 121 2 n2 / sin (90 . Chapter26 Problems 1337 SOLUTION The time t it takes for the light to travel from A to B is Distance . d '
(1) t = ______
Speed of light v
in the substance eed of light 6 in a vacuum and the The speed of light v in the substance is related to the sp
(Equation 26.1). Substituting this index of refraction 111 of the substance by v — c/n1 expression into Equation (1) gives d d d n
w: 1 (2)
v [i] c "1
nt at the critical angle 6C, we know that 771 sin 66 2:12 (Equation
and substituting the result into Equation (2) yields ‘ t: Since the light is incide
26.4). Solving this expression for M1 d "2 1.63
dn Sing m) P—J‘ O
t: 1: C : $111481 2 S c c 3.00><108 m/s each refractive index can be related to a critical angle for 26.4. By applying this expression to the A—B
ll obtain expressions for HB and 71C in terms of sions into the ratio, we will be able to 29. REASONING In the ratio nB/nC total internal reﬂection according to Equation
interface and again to the A—C interface, we wi the given critical angles. By substituting these expres
obtain a result from which the ratio can be calculated SOLUTION Applying Equation 26.4 to the A—B interface, we obtain n
. _ B _ .
srn6QAB —~—n or 118 —nA SinQC’AB
A Applying Equation 26.4 to the A—C interface gives n
s1n6c! AC — 7 or nC — n A Sin 60’ AC
A ...
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