solucionario de ejercicios Capítulo 35 (5th Edition).pdf naturaleza de la luz y leyes de optica geometrica

Chapter 35 Solutons

The Moon's radius is 1.74 traveled by the light is:

35.1

d = 2(3.84

× 10 6

m and the Earth's radius is 6.37

× 10 8 m – 1.74 × 10 6 m – 6.37 × 10 6 m) = 7.52 × 10 8

× 10 6

m. The total distance

m

7.52 × 10 8 m This takes 2.51 s, so v = = 2.995 × 10 8 m/ s = 299.5 Mm/ s 2.51 s

∆ x

2(1.50 × 10 8 km)(10 km)(1000 00 m/ km) (22.0 (22.0 min)(60.0 min)(60.0 s/ min)

35.2

∆ x = ct ;

35.3

The experiment is most convincing if the wheel turns fast enough to pass outgoing light t  = 2l c through one notch notch and returning light light through the next:

c=

=

t 

 2l  θ  = ω t = ω    c  

ω  =

so

cθ 

2l

=

= 2.27 × 10 8 m / s = 227 M m / s

(2.998 × 10 8 )[2π  /  / (720)] 2(11.45 × 10 3 )

=

114 rad/ s

The returning light would be blocked by a tooth at one-half the angular speed, giving another data p oint. oint.

35.4

(a) (a)

For For the light light beam beam to make it through both slots, slots, the time time for for the light light to travel the d istanc istancee d  must equal the time for the disk to rotate through the angle θ , if  c is the speed of light, d  c

(b)

=

θ  , so ω 

ω  d ω  θ 

W e a re re g iv iv en en th th a t d = 2.50 2.50 m ,

c=

35.5

c=

ω  d ω  θ 

θ  =

1.00°  π ra d   60.0

  180°  

(2.50 m )(3.49 × 10 4 r a d = − 2.91 × 10 4 ra d

U s in g Sn e ll' s la w , sin θ 2

θ 2 = 25.5°

λ 2

=

λ 1 n1

= =

n1 n2

= 2 .91 91 × 10 − 4 s

ω  = 5555

rad,

) = 3.00 × 108

m s

=

rev re v   2 π  ra d s

300 Mm/ Mm/ s

sin θ 1

442 nm

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  = 3.49 × 10 4  1.00 rev  

r ad s

Chapter 35 Solutions

35.6

35.7

c

(a)

 f  =

(b)

λ glass

=

(c)

v glass

=

λ 

=

n1 sin θ1

3. 00 00 × 10 8 m / s − 6.328 × 10 7 m

λ air

=

n cair n

=

632.8 n m 1.50

=

323

4.74 × 10 14 H z

=

422 nm

3. 00 00 × 10 8 m / s 1.50

8 = 2. 00 00 × 10

m / s = 200 M m / s

= n2 sin θ 2

sin θ 1

= 1.333 .333 sin sin 45.0 45.0°

sin θ 1

= (1.33)(0.707) = 0.943 Figure for Goal Solutio n

θ 1 = 70.5° → 19.5° above the horizon

Goal Solution An u nd erwater scuba d iver sees sees the Sun Sun at an ap paren t angle of 45.0 45.0°° from from the vertical. vertical. What is the actual direction of the Sun? G:

The sunlight refracts as it enters the water from the air. Because the water has a higher index of  refraction, refraction, the light light slows slows dow n and bend s toward th e vertical vertical line line th at is nor ma l to the interface. Therefore, the elevation angle of the Sun above the water will be less than 45 ° as shown in th e diagram to the right, even though it appears to the diver that the sun is 45 ° above the horizon.

O:

We can use Snell’s law of refraction to find the precise angle of incidence.

A:

Snell’s law is:

n 1 s i n θ1

which gives

sin θ 1 sin θ 1

The sunlight is at θ 1

L:

= 70.5°

= n2 sin θ 2

= 1.333

sin 45.0 45.0 °

= (1.333)(0.707) = 0.943

to the vertical, so the Sun is 19.5 ° above the horizon.

The calc calculated ulated resu lt agrees with ou r pred iction. iction. When ap plying Snell’ Snell’ss law, it is is easy to mix up the index values and to confuse angles-with-the-normal angles-with-the-normal and angles-with-the-surfac angles-with-the-surface. e. Making a sketch sketch and a pred iction iction as we d id here helps avoid careless careless mistakes. mistakes.

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Chapter 35 Solutions

324

*35.8

(a)

n1 sin θ1

= n2 sin θ 2

1.00 sin 30.0° = n sin 19.24° n = 1.52

(c)

35.9

35.10

 f  =

=

(d)

v

(b)

λ  =

(a )

c

λ  c n v  f 

= = =

3. 00 00 × 10 8 m / s − 6.328 × 10 7 m 3. 00 00 × 10 8 m / s 1.52 1. 98 98 × 10 8 m / s 4. 74 74 × 10 14 /  s

Flin t G la la ss ss :

= =

(b)

W a ter :

(c)

C u bi bic Zi Zir co co ni n ia :

n1 sin θ1

v

n1 sin θ1

θ 2

n c n

= = v

= n2 sin θ 2 ;

n 2 = 1.90 =

35.11

v

c

c ; v

v=

= n2 sin θ 2 ;

= sin −1  

=

4.74 × 10 14 Hz

8 = 1.98 98 × 10

=

8

m s

1.66 3. 00 00 × 108 m s 1.333 c n

m / s = 198 M m/ m/ s

417 nm

3. 00 00 × 10

=

in air air and in in syrup. syrup.

=

= 1.81 81 × 10 8

m s

=

181 M m / s

= 2 .25 25 × 10 8

m s

=

225 M m / s

3. 00 00 × 108 m s 2.20

8 = 1.36 36 × 10

m s

=

136 M m / s

1.333 sin 37.0° = n 2 sin 25.0° c = 1.58 × 10 8 m / s = 158 M m / s 1.90

θ 2 = sin –1

(1.00)(sin 30°)  1.50

= 

 n 1 s in  n 2  

θ 1  

   

19.5°

θ 2 an d θ 3 are alternate interior angles formed by the ray cutting pa rallel norm als. So, θ 3 = θ 2 = 19.5 19.5°° . 1.50 sin θ 3 = (1.00) sin θ 4

θ 4 = 30.0°

Chapter 35 Solutions

35.12

(a )

W a t e r λ  =

λ 0

(b)

G la s s

λ  =

λ 0

*35.13

=

436 n m

= n w sin θ 2

sin θ 2

=

1 1.333

1.333

1.52

sin θ 1

=

=

327 nm

=

287 nm

1 1.333

sin (90.0° − 28.0°) = 0.662

= sin −1 0.66 662 = 41.5°

h=

(a )

n

436 n m

s in θ 1

θ 2

35.14

n

=

d 

ta n θ 2

=

3.00 m tan 41.5 1.5°

=

3.39 m

Fr o om m g e om om e tr tr y y,, 1. 1.25 m = d  sin 40.0° so d = 1.94 m

(b)

50 .0 .0° a b ov ov e h o r iz iz o n t al al

, o r p a r al alle l t o t h e

incident ray

*35.15

The incident light reaches the left-hand mirror at distance (1.00 m) tan 5.00° = 0.0875 m above its bottom edg e. The reflected reflected light first reaches the right-hand mirror at height 2(0.0875 m) = 0.175 m It bounces between the mirrors with this distance between points of contact with either. Since

1.00 m 0.175 m

= 5.72 5.72,, the light reflects

five five times from the right-hand mirror an d six times from from th e left. left.

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325

326

*35.16

Chapter 35 Solutions

At entry,

n1 s i n θ 1

= n2 sin θ 2

or

1.00 s in in 30 30.0° = 1. 1.50 s in in θ 2

θ 2 = 19.5° The distance h the light travels in the medium is given by (2.00 (2.00 cm ) h

cos θ 2 =

or

h=

( 2.00 cm) cm ) cos 19.5°

α

The angle of deviation upon entry is The offset offset distan ce comes from from

*35.17

sin α  =

d  : h

The d istance, istance, h, traveled traveled by the light light is h =

*35.18

t  =

h v

=

2. 12 12 × 10 − 2 m

2.00 cm cos 19.5° v

− = 1.06 06 × 10 10

8

2. 00 00 × 10 m / s

= θ1 − θ 2 = 30.0° − 19.5° = 10.5°

d  = (2.21 cm) sin 10.5° = 0.388 cm

The speed of light in the material is

Therefore,

= 2.12 cm

= 2.12 cm

c

3. 00 00 × 10 8 m / s

n

1.50

= =

= 2.00 00 × 10 8

m/ s

s = 106 106 ps

Applying Snell's law at the air-oil interface, n a ir sin θ  = n o il s in 20. 0°

y ie ld s

θ = 30.4°

App lying lying Snell's nell's law at th e oil-water oil-water interface interface n w sin θ ′ = n oil sin 20. 0°

*35.19

y ie ld s

θ ′ = 22.3°

time difference = (time for light to travel 6.20 m in ice) – (time to travel 6.20 m in air)

∆t  = 6.20 m − 6.20 m v ice

v

bu t

c

=

c n

(6.20 m )  1.309 − 1  = (6.20 − ∆t  = (6.20 (6.20 m ) (0.309) = 6.39 × 10 9     c

c

c

s = 6.39 6.39 ns

Chapter 35 Solutions

Consid er glass w ith an ind ex of refraction of 1.5, 1.5, w hich is 3 m m thick The sp eed of light i n the glass is

*35.20

3 × 108 m / s 1.5 3 × 10

The extra travel time is

−3 m

2 × 10 8 m / s

= 2 × 108 −

For light of wavelength 600 nm in vacuum and wavelength 3 × 10 − 3 m −7 4 × 10 m

the extra op tical tical path , in w avelengths, is

*35.21

−

m/ s

3 × 10

−3 m

3 × 108 m / s 600 n m 1.5

~ 10–11 s

= 400 n m

in glass,

3 × 10 − 3 m ~ 103 wavelengths 7 − × 6 10 m

(1.00) sin θ1 = (1.66) sin θ 2

Refraction proceeds according to (a) (a)

(1)

= v 2 cos θ 2

For the normal comp comp onent of veloc velocity ity to be be constant, constant,

v 1 cos θ 1

or

(c ) co s θ1 = (c 1.66) co s θ 2

We multiply Equations (1) and (2), obtaining:

sin θ1 co s θ1

= sin θ2 co s θ2

sin 2θ1

or

sin θ1

= sin 2θ 2

= 1.66 co s θ 1 ,

The p hysical

o r t a n θ 1

= 1.66

θ 1 = 58.9°

wh ich ich yields yields (b)

(2)

 

The solution θ1 = θ 2 = 0 does not satisfy satisfy Equat ion (2) (2) and m ust be rejected rejected . s o lu t io n is 2θ1 = 180° − 2θ 2 or θ 2 = 90.0° − θ 1 . Th en en Eq u at at io n (1) b e co m es es :

Light Light enteri entering ng the glass glass slows slows down and makes a smaller smaller angle with with the norm al. Both effe effect ctss reduce the velocity component parallel to the surface of the glass, so that component cannot remain constant, or will remain constant only in the trivial case

35.22

327

See the sketch sketch show ing the path of the light α an d γ  are angles of incidence at ray. mirrors 1 and 2. For triangle abca, 2α  + 2 γ  + β  = 180° or

β

= 180° − 2(α + γ )

(1)

Now for triangle triangle bcdb,

(90.0° − α ) + (90.0° − γ ) + θ  = 180° or

θ  = α  + γ 

(2)

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θ 1

= θ 2 = 0

Chapter 35 Solutions

328

Substituting Equation (2) into Equation (1) gives

β  = 180° − 2θ 

N ote: From Equat ion (2), (2), γ  = θ  − α . Thus, the ray will foll follow ow a path like that shown only if  α  < θ . For α  > θ , γ  is negative and multiple reflections from each mirror will occur before the incident an d reflect reflected ed rays intersect. intersect.

35.23

Let n ( x ) be the index of refraction at distance  x below the top of the atmosphere and its value at the planet surface. surface. Then, n ( x = h) = n be its n ( x ) = 1.000 +

(a) (a)

The total time required required to traverse traverse the atmosphere atmosphere is is t  =

t  =

(b)

 n − 1.000   x   h  

h dx

∫  0

v

h n x n − 1.000    h (n − 1.000)  h 2   h  n + 1.000   1 h  x  dx = + = ∫ 0 ( ) dx = ∫ 0 1.000 +   =  2     h    c c  c ch     c   2   3

20.0 × 10 m 3.00 × 10

8

m

 1.005 + 1.000   =   2 s  

66.8 µ s

The travel travel time time in the absenc absencee of an atmosphere atmosphere w ould ould be h /  c . presence of an atmosphere is

 n + 1.000   = 1.0025   2   .0025

Thus, the time in th e

times larger or 0.250% 0.250% long er .

Let n ( x ) be the index of refraction at distance  x below the top of the atmosphere and n ( x = h) = n be its its value at the planet surface. surface. Then,

35.24

n ( x ) = 1.000 +

(a) (a)

The total time required required to traverse traverse the atmosphere atmosphere is is t  =

(b)

 n − 1.000   x   h  

h dx

∫  0

v

hn x h n − 1.000)  h 2   = ∫ 0 ( ) dx = 1 ∫ 0 1.000 +  n − 1.000   x  dx = h + ( =  2     h    c c  c ch    

The travel travel time time in in the abs absenc encee of of an atmosphere would be h /  c . presence of an atmosphere is

 n + 1.000   times larger   2  

h  n + 1.000   c  

2

 

Thus, the time in th e

Chapter 35 Solutions

35.25

From Fig. 35.20

n v = 1.470 1.470 at 400 nm

Then

(1.00) sin θ

δ r  − δ v

n r  = 1.458 at 700 nm

and

(1.00) s in θ

sin θ 

sin sin 30.0°  

n1 sin θ 1

sin sin 30.0 30.0°  

− sin −1      

1.458

sin θ 

= n2 sin θ 2

1.470

θ 2

so

(1.00)( (1.00)(sin sin 30.0 30.0 ° )  

 

=

0.171°

 n sin θ 1   = sin −1  1   n 2    

θ 2

= sin −1     

θ 3

= ([(90.0° − 19.5° ) + 60.0°] − 180°) + 90.0° =

n 3 sin θ 3

=    

1.50

= n 4 sin θ 4

19.5°

θ 4

so

40.5°

 n sin θ 3   (1.50)(si (1.50)(sin n 40.5 40.5 ° )   = sin −1 3 = sin −1   =         1.00   n 4  

Taking Φ to be the apex angle and δ m in to be the angle angle of of minim um Equation 35.9, the index of refraction of the prism material is

35.27

sin n

Solving for δ mi n ,

35.28

= 1.458 sin θ r 

  − sin −1     = θ r  − θ v = sin −1    1.458    1.470  

∆δ  = sin −1    

35.26

= 1.470 sin θ v

an d

329

=

δ mi n

Φ     = 2 sin sin −1 n sin − Φ = 2 sin sin −1[( 2.20) sin (25. 0°)] − 50.0° =   2  

(a)

(1.0 (1.00) 0) sin 75.0 75.0°° = 1.45 1.458 8 sin sin θ 2;

(b)

Le t

θ 3

So

60.0° − θ 2

θ 2

deviation, deviation, from

 Φ + δ m in     2   sin ( Φ 2)

n (700 nm) = 1.458

+ β  = 90.0° ,

77.1°

θ 2 = 41.5°

+ α  = 90.0° ;

then α  + β  + 60.0° = 180°

− θ 3 = 0 ⇒ 60.0° − 41.5° = θ 3 =

(c) (c)

1.45 1.458 8 sin 18. 18.5° 5° = 1.0 1.00 0 sin sin θ 4

(d)

γ  = (θ 1 − θ 2 ) + [β  − (90.0° − θ 4 )]

18.5°

θ 4 = 27.6°

γ  = 75.0° − 41.5° + (90.0° − 18.5° ) − (90.0° − 27.6° ) = 42.6°

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86.8°

330

35.29

Chapter 35 Solutions

=

For the incoming ray,

sin θ 2

Using the figure to the right,

(θ 2 )violet

(θ 2 )re d

sin θ 1 n

sin 50.0 50.0°   = sin −1  sin = 27.48°   1.66   sin 50.0 50.0°  

= sin −1     θ 3′

For the outgoing ray,

= 60.0° – θ 2

(θ 4 )violet (θ 4 )re d

35.30

n

=

Φ,

35.31

sin θ 4

= n sin θ 3

= sin −1[1.66 sin 32.52° ] = 63.1 63.17 7°

= sin −1[1.62 sin 31.78° ]= 58.56 .56°

Φ + δ m in ≈ Φ so is also a small 2 ( sin θ  ≈ θ  when θ  18.2° . Since sin θ 1 = n sin θ 2 , this requirement becomes sin θ 1 > (1.50) s in in (18. 2°) = 0.468, or θ 1 > 27.9°

is in in rad ians.

Chapter 35 Solutions

= n sin θ 2 .

At the first refraction, (1.00 ) sin θ 1 the second surface is given by

35.32

n sin θ 3

= 1.00,

θ 3

or

The critical critical angle at

= sin −1(1.00 n )

But (90.0° − θ 2 ) + ( 90.0° − θ 3 ) + Φ = 180° , wh ich ich g ives θ 2

< sin −1(1.00 n ) and θ 2 > Φ − sin −1 (1.00 n ) .

Thus, to have θ 3 necessary necessary th at

 

− sin θ 1 > n sin Φ − sin 1

n

=

Si S in ce s in θ1

 1.00     n  

tan ta n

*35.34

sin (δ  + φ ) sin ( φ  /  / 2) sin 5° ( φ ) = 1.544sin − cos5° 1 2

= n s in θ 2 ,

   

2 θ 1 > sin −1     n

1

1

1

2

2

2

an d

φ  = 18.1°

so

= Φ − θ 2 α  = θ 1 − θ 2 β  = θ 4 − θ 3 θ 3

The total deviation is therefore δ  = α  + β  = θ 1 + θ 4

At exit:

− 1 s in in Φ − cos Φ   

2

Φ + (90.0° − θ 2 ) + (90.0° − θ 3 ) = 180°

Thus, θ 3

 1.00       n      

1

1.544sin

Note for use in every part:

A t e n tr tr y :

 

sin 5° + sin ( φ ) cos5° ( φ ) = sin (5° + φ ) = cos ( φ ) sin

so

At exit, the deviation is

n1 sin θ 1

= n 2 sin θ 2

or

= 60.0° − 30.0° = 30.0° 1.50 sin 30.0 30.0 ° = 1.00 sin θ 4

− θ 2 − θ 3 = θ 1 + θ 4 − Φ sin 48.6 48.6°    sin θ 2 = sin −1 = 30.0°   1.50  

or

so the path through the prism is symmetric symmetric when θ 1 (b)

δ  = 48.6° + 48.6° − 60.0° = 37.2°

(c)

A t e n tr tr y :

sin θ 2

=

A t e xit :

sin θ 4

= 1.50 sin (31.6°) ⇒

A t e n tr tr y: y:

sin θ 2

=

(d )

this requirement becomes

θ 1 > sin −1  n sin Φ − sin −1

or

At the first surface is

(a )

= Φ − θ 3 .

avoid total internal reflection at the second surface, it is

Through the application of trigonometric identities,

35.33

331

A t exit : sin θ 4

sin 45.6° 1.50

s in 51.6°

⇒

θ 2

= 28.4° θ 4

= 51.7°

θ 4

= sin −1[1.50 .50 sin sin (30.0°)] = 48.6°

= 48.6° .

θ 3

= 60.0° − 28.4° = 31.6°

δ  = 45.6° + 51.7° − 60.0° = 37.3°

⇒

θ 2

= 31.5°

θ 3

= 1.50 sin (28.5°) ⇒

θ 4

= 45.7°

δ  = 51.6° + 45.7° − 60.0° = 37.3°

1.50

= 60.0° − 31.5° = 28.5°

© 2000 2000 by Harcour t, Inc. Inc. All rights rights reserved .

Chapter 35 Solutions

332

n sin θ  = 1.

35.35

(a)

θ  = sin −1

  1   =  2.419  

(b)

θ  = sin −1

  1   =  1.66  

(c)

θ  = sin −1

  1   =  1.309  

s in θ c

35.36

35.37

From Table 35.1, 35.1,

=

n2 n1

θ c

;

24.4°

37.0°

49.8°

 n   = sin −1  2    n1  

(a )

D i a m on on d: d:

θ c

  = = sin −1    2.419  

33.4°

(b)

Flin t g la la ss ss :

θ c

  = = sin −1     1.66  

53.4°

(c)

Ice : Sin ce ce n 2 > n 1, there is is no critic critical al angle .

sin θ c

=

n2 n1

1.333

1.333

(Equation 35.10)

n 2 = n 1 sin 88.8° = (1.0003)(0.9998) = 1.000 08

*35.38

sin θ c

=

n air n pipe

=

1.00 1.36

= 0.735

θ c = 47.3°

Geometry show s that the angle of refracti refraction on at th e end is

θ r  = 90.0° – θ c = 90.0° – 47.3° = 42.7° Th en en , Sn Sn el ell' s la w a t th th e en en d , gives

35.39

For total internal reflection,

1.00 s in in θ  = 1.36 sin 42.7°

θ = 67.2°

n1 sin θ 1

= n2 sin 90.0°

(1.50) sin θ 1 = (1.33)(1.00)

or

θ 1 = 62.4°

Chapter 35 Solutions

35.40

333

To avoid internal reflection and come out through the vertical vertical face, face, light light inside the cube mu st have

θ 3

< sin −1 (1/ n) > 90.0° − sin −1 (1/

So

θ 2

Bu t

θ1 < 90.0°

n)

n sin θ 2

an d