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Speed of light in leaded glass v =1.66 x 108 mIs
Speed of light in vacuumc = 3M0 x 108 m/s
Required
Index of refraction of leaded glass ri =?
Analysis and Solution
The correct equation is n =
V
Substitute the values and their units, and solve the problem.
C
V
— 3.00 x 108 rn/s
l.66x108rn/s
=1.81
Paraphrase
The index of refraction of leaded glass is about 1.81.
2. Given
Speed of light in quartz v = 2.10 x io rn/s
Speed of light in vacuum c = 3.00 x iO rn/s
Required
Index of refraction of quartz n =?
Analysis and Solution
The correct equation is n =
L
Urnt D Ltght and Geometric Optics 95
Substitute the values and their units, and solve the problem.
C
V
— 3.00x108 rn/s2.10x108 rn/s
=1.43
Paraphrase
The index of refraction of quartz is about 1.43.
3. Given
Speed of light in the material v =1.24 xl 08 rn/s
Speed of light in vacuum c = 3.00 x 108 rn/s
Required
Index of refraction of the material n =?
Analysis and Solution
CThe correct equation is n = —
V
Substitute the values and their units, and solve the problem.
C
V
— 3.00 x 108m/s— 1.24x108m/s
= 2.42
Paraphrase
The index of refraction of the material is about 2.42. Thismatches the index of refraction of diamond. The material may bediamond.
96 Unit D: Light and Geometric Optics
Answers to Practice Problems 11.6
Given
Index of refraction of alcohol n 1 36
Speed of light in vacuum c = 300 xl 08 rn/s
Required
Speed of light in alcohol v =?
Analysis and Solution
The correct equation is n
Rearrange it to solve for the variable needed: v =
Substitute the values and their units, and solve the problem.
CV
n
— 3.OOxlO8m/s1.36
rz2.21x108m/s
Paraphrase
The speed of light in alcohol is about 2.21 x108m/s.
Given
Index of gallium phosphide n = 3.50
Speed of light in vacuum c = 3.00 xl o rn/s
Required
Speed of light in gallium phosphide v ?
Analysis and Solution
The correct equation is n =
Unit D Light and Geometnc Optics 97
Rearrange it to solve for the variable needed: v =n
Substitute the values and their units, and solve the problem.
C
‘.1
— 3.OOx 108 mIs
3.50
=8.57 iO rn/s
Paraphrase
The speed of light in gallium phosphide is about
8.57x107m/s.
3. Given
Index of refraction of sapphire n = 1.77
Speed of light in vacuum c = 3.00 xl 08 m/s
Required
Speed of light in sapphire v ?
Analysis and Solution
The correct equation is n =
Rearrange it to solve for the variable needed: v =
Substitute the values and their units, and solve the problem.
C
n
3.OOxlO8mIs1.77
=1.69x108m/S
Paraphrase
The speed of light in sapphire is about 1.69 x lO8mIs.
98 Unit D: Light and Geometric Optics
those used in Snell’s Iaw(01 and ). and discuss why each is the moreefTecilve of the t o in its context.Students can do Dl’) lnquiiy Acti itv after completing the problemsand D18 Inquiry Activity.
Answers to Practice Problems 11.7
1. Jdentifying air as medium 1 and water as medium 2:
Given
Index of refraction of air n1 = l.OO
Index of refraction of water n2 1.33
Angle of incidence 6 30°
Required
Angle of refraction2 =?
Analysis and Solution
The correct equation is n sin = n2 sin 02
Rearrange it to solve for the variable needed: sin 02
Substitute the values and their units, and solve the problem.
fl1sin6S1fl02r—____.i
1.00xSfl(3Oo)
1.33
0.3759
Take the inverse sine of both sides: 612 22°
Paraphrase
The angle of refraction is about 22°.
100 Jni D ight ard Ge’rntrjc Optics
2. Identifying water as medium 1 and diamond as medium 2:
Given
Index of refraction of water n1 =1.33
Index of refraction of diamond n2 = 2.42
Angle of incidence 01= 450
Required
Angle of refraction 02 =?
Analysis and Solution
The correct equation is n1 sin 01 n2 sin 02
Rearrange it to solve for the variable needed: sinG2 =
fl1 siii0
Substitute the values and their units, and solve the problem.
n sinGsinO2= 1 1
— 1.33 x sin(45°)2.42
= 0.3886
Take the inverse sine of both sides: 82 = 23°
Paraphrase
The angle of refraction is about 23°.
3. Identifying air as medium 1 and the lens as medium 2:Given
Index of refraction of air n1 =1.00
Index of refraction of lens n3 =1.41
Angle of incidence 8 55.0°
Unit D: Light and Geometric Optics 101
Required
Angle of refraction 2
Analysis and Solution
The correct equation is 111 Sin = n2 sin 62
n sin6511182=
1 1
Rearrange it to solve for the variable needed:
Substitute the values and their units, and solve the problem.
nsin8sin62= 1 1
n2
— 1.00 x sin(55.0°)1.41
0.5810
0 —355°Take the inverse sine of both sides: 2 —
Paraphrase
The angle of refraction is about 3550
Answers to Practice Problems 11.8
Identifying air as medium 1 and honey as medium 2:Given
Index of refraction of air n1 = 1.00
Angle of incidence 61 = 55.0°
Angle of refraction 6 = 19.5°
Required
Index of refraction of honey n2 =?
Analysis and Solution
The correct equation is n1 sin0 = n2 sin 62
102 Unit D Light and Geornetiic Optics
Rearrange it to solve for the variable needed: ,— 171 sin 01
s,n02
Substitute the values and their units, and solve the problem.
n sin6n2 sinO2
1.00 x sin(30°)— sin(19.5°)
1.4978
Paraphrase
The index of refraction of honey is about 1.50.
2. Identifying water as medium I and amber as medium 2:
GWen
Index of refraction of water n1 = 1.33
Angle of incidence 0, 350
Angle of refraction 0 = 350
Required
Index of refraction of amber n2 =?
Analysis and Solution
The correct equation is n1 sin 01 = n2 sin 62
Rearrange it to solve for the variable needed: n =n sin 01
2 sin62
Substitute the values and their units, and solve the problem.
n sinS2 sinS2
— 1.33xsin(35°)— sin(24°)
=1.8756
Unit D Light and Geometric Optics 103
Paraphrase
The index of refraction of amber is about 1.9.
3. Identifying flint glass as medium 1 and the lemon oil as medium2:
Given
Index of refraction of flint glass n1 1.61
Angle of incidence 0, = 40.00
Angle of refraction 6 = 4440
Required
Index of refraction of lemon oil n2 ?
Analysis and Solution
The correct equation is n1 sin 0, = n2 sin 02
Rearrange it to solve for the variable needed: n= i7 sin 61
2 sin62
Substitute the values and their units, and solve the problem.ri sin01 1
2 sin02
1.6lxsin(40.0°)sin(44.4°)
= 1.4791
Paraphrase
The index of refraction of lemon oil is about 1.48.
104 Unit D Light and Geometric Optics
10. Snell’s law relates the indices of refraction of two materials to theangles a light ray makes with the normal in the two materials. Thematerial with the larger index of refraction has the smaller angle, andthe material with the smaller index of refraction has the larger angle.When light passes from a material with a low index of refraction intoone with a high index of refraction, it bends toward the normal. Whenlight goes from a high to low index of refraction, it bends away fromthe nonnal.
Connect Your Understanding11. Given
Speed of light in the medium v = 1.2 x 108
Speed of light in vacuum c = 3.00 x l0 mis
Required
Index of refraction of the medium nAnalysis and Solution
The correct equation is ii =
Substitute the values and their units, and solve the problem.C
n=—V
— 3.OOxlO8mJs
— l.2xlO8m!s
= 2.500
Paraphrase
The index of refraction of leaded glass is about 2.5.12. Identifying air as medium 1 and jade as medium 2:
Given
Index of refraction of air n1 = 1.00
Index of refraction ofjade n, = 1.61
Angle of incidence = 80.00
Required
Angle of refraction,
=?
Analysis and Solution
The correct equation is n1 sin O = z, sin°2
Unit 0: Light and Geometric Optics 111
n sin0Rearrange it to solve br the variable needed: sin 0, =
127
Substitute the values and their units, and solve the problem.
n sin0sin0,=
117
= l.O0xsin(80.0°)
1.61=0.6117
Take the inverse sine of both sides: 07 = 37.71°
Paraphrase
The angle of refraction is about 37.7°
13. Identifying air as medium 1 and the material as medium 2:Given
Index of refraction of air = 1.00
Angle of incidence 01 = 25.0°
Angle of refraction 02 = 16.7°
Required
Index of refraction n, =?
Analysis and Solution
The correct equation is n1 sin 0 = n2 sin 02
n sin0Rearrange it to solve for the vanable needed: ii,
=- sm0,
Substitute the values and their units, and solve the problem.
n1 sin 0n=
2 sin0,
— 1.O0xsin(25.0°)
— sin(16.7°)
= 1.4707
Paraphrase
The index of refraction of the material is about 1.47; this matches theindex of refraction of Pyrex glass. The material may be Pyrex.
14. Identifying air as medium I and the material as medium 2:
Given
Index of refraction of air n1 = 1.00
Angle of incidence &[ = 60°
Angle of refraction 02 = 50°
112 Unit D: Light and Geometdc Optics
Required
Index of refraction n2 =?
Analysis and Solution
The correct equation is ii sin 01 = sin 02
nsin0Rearrange it to solve for the variable needed: n =2 sin0,
Substitute the values and their units, and solve the problem.
ii sin=
2 sin07
— l.O0xsin(60°)
sin(50°)
= 1.1305
Paraphrase
The index of refraction of the material is about 1.13.15. Given
Speed of light in the medium v = 2.26 x 108 mis
Speed of light in vacuumc 3.00 x 108 mIs
Required
Index of refraction of the medium ii =?
Analysis and Solution
The correct equation is ii = £
Substitute the values and their units, and solve the problem.
Cn=—
V
— 3.OOxlO8m!s— 2.26x 108 mis
= 1.3274
Paraphrase
The index of refraction of the material is about 1.33—it may bewater.
16. Given
Index of refraction of the condensate n = 1.76 xl 0
Speed of light in vacuum e = 3.00 x 108 mis
Required
Speed of light in the condensate: v
Analysis and Solution
Unit D: Light and Geometric Optics 113
The correct equation is n =
Rearrange it to solve for the variable needed: v = -
Substitute the values and their units, and solve the problem.
1‘1
— 3.OOxlO8mJs— l.76x10= l.7045x10’mls
Paraphrase
The speed of light in the Bose-Einstein condensate is about 17.1 mIs.17. Both reflection and refraction cause light rays to change direction.18. In reflection, light rays stay in the original material. In refraction,
light rays enter a different medium.19. An index of refraction less than 1.0 would mean that light travels
faster in the material than in a vacuum. Since light travels its fastestin a vacuum, this is impossible.
Reflection20. Students’ answers may vary but may include:
• I found it interesting that light rays can bend: this explains why Ikeep misjudging depths at the lake.
• I knew that fibre optics could bend light because I have had myknee “scoped,” but I didn’t know that this uses something called“total internal reflection.”
21. Students’ answers may vary but may include these terms: medium,refraction, index ofrefraction, dispersion, total internal reflection,fIbre optics, mirage.
114 Unit D: Light and Geometric Optics
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