required analysis and solution speed of light in...

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 =?



L

Urnt D Ltght and Geometric Optics 95


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 =?


CThe correct equation is n = —

V


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 =?


The correct equation is n

Rearrange it to solve for the variable needed: v =


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 ?



Unit D Light and Geometnc Optics 97

Rearrange it to solve for the variable needed: v =n


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 ?



Rearrange it to solve for the variable needed: v =


C

n

3.OOxlO8mIs1.77

=1.69x108m/S

Paraphrase

The speed of light in sapphire is about 1.69 x lO8mIs.


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.


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 =?


The correct equation is n sin = n2 sin 02

Rearrange it to solve for the variable needed: sin 02


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 =?


The correct equation is n1 sin 01 n2 sin 02

Rearrange it to solve for the variable needed: sinG2 =

fl1 siii0


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


The correct equation is 111 Sin = n2 sin 62

n sin6511182=

1 1

Rearrange it to solve for the variable needed:


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


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 =?


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


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 =?


The correct equation is n1 sin 01 = n2 sin 62

Rearrange it to solve for the variable needed: n =n sin 01

2 sin62


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 ?


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 ofjade n, = 1.61

Angle of incidence = 80.00

Required

Angle of refraction,

=?


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


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, =?


The correct equation is n1 sin 0 = n2 sin 02

n sin0Rearrange it to solve for the vanable needed: ii,

=- sm0,


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


Angle of incidence &[ = 60°

Angle of refraction 02 = 50°

112 Unit D: Light and Geometdc Optics

Required

Index of refraction n2 =?


The correct equation is ii sin 01 = sin 02

nsin0Rearrange it to solve for the variable needed: n =2 sin0,


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 =?


The correct equation is ii = £


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


Unit D: Light and Geometric Optics 113


Rearrange it to solve for the variable needed: v = -


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.


required analysis and solution speed of light in...

Documents