solucionario capitulo 35 - paul e. tippens

Chapter 35. Refraction Physics, 6th Edition

Chapter 35. Refraction

The Index of Refraction (Refer to Table 35-1 for values of n.)

35-1. The speed of light through a certain medium is 1.6 x 108 m/s in a transparent medium,

what is the index of refraction in that medium?

8

8

3 x 10 m/s ;1.6 x 10 m/s

cnv

n = 1.88

35-2. If the speed of light is to be reduced by one-third, what must be the index of refraction for

the medium through which the light travels? The speed c is reduced by a third, so that:

23 2

3

3( ) ; ; ( ) 2x

x

c cv c n nv c

and n = 1.50

35-3. Compute the speed of light in (a) crown glass, (b) diamond, (c) water, and (d) ethyl

alcohol. (Since n = c/v, we find that v = c/n for each of these media.)

(a)

83 x 10 m/s ;1.50gv

vg = 1.97 x 108 m/s (b)

83 x 10 m/s ;2.42dv

vd = 1.24 x 108 m/s

(a)

83 x 10 m/s ;1.33wv

v = 2.26x 108 m/s (b)

83 x 10 m/s ;1.36av

va = 2.21 x 108 m/s

35-4 If light travels at 2.1 x 108 m/s in a transparent medium, what is the index of refraction?

8

8

(3 x 10 m/s) ;2.1 x 10 m/s

n n = 1.43

487


The Laws of Refraction

35-5. Light is incident at an angle of 370 from air to flint glass (n = 1.6). What is the angle of

refraction into the glass?

0(1.0)sin 37sin sin ; sin ;1.6g g a a gn n

g = 22.10

35-6. A beam of light makes an angle of 600 with the surface of water. What is the angle of

refraction into the water?

w

sinsin sin ; sin a aa a w w

w

nn nn

0

w(1)sin 60sin 0.651;

1.33

w = 40.60

35-7. Light passes from water (n = 1.33) to air. The beam emerges into air at an angle of 320

with the horizontal water surface? What is the angle of incidence inside the water?

= 900 – 320 = 580; w

sinsin sin ; sin a aa a w w

w

nn nn

0

w(1)sin 58sin 638;

1.33

w = 39.60

488

nw = 1.5air

600

water


35-8. Light in air is incident at 600 and is refracted into an unknown medium at an angle of 400.

What is the index of refraction for the unknown medium?

0

0x

sin (1)(sin 60 )sin sin ; ;sin sin 40a a

x x a a xnn n n

n = 1.35

35-9. Light strikes from medium A into medium B at an angle of 350 with the horizontal

boundary. If the angle of refraction is also 350, what is the relative index of refraction

between the two media? [ A = 900 – 350 = 550. ]

0

0

sin sin 55sin sin ; ;sin sin 35

B AA A B B

A B

nn nn

0

0

sin 55 1.43;sin 35

B

A

nn

nr = 1.43

35-10. Light incident from air at 450 is refracted into a transparent medium at an angle of 340.

What is the index of refraction for the material?

0

0

(1)sin 45sin sin ; ;sin 35A A m m mn n n

nm = 1.23

489

A 350350

B


*35-11. A ray of light originating in air (Fig. 35-20) is incident on water (n = 1.33) at an angle of

600. It then passes through the water entering glass (n = 1.50) and finally emerging back

into air again. Compute the angle of emergence.

The angle of refraction into one medium becomes the

angle of incidence for the next, and so on . . .

nair sin air = nw sin w = ng sin g = nair sin air

Thus it is seen that a ray emerging into the same medium

as that from which it originally entered has the same angle: e = i = 600

*35-12. Prove that, no matter how many parallel layers of different media are traversed by light,

the entrance angle and the final emergent angle will be equal as long as the initial and

final media are the same. The prove is the same as shown for Problem 35-11:

nair sin air = nw sin w = ng sin g = nair sin air; e = i = 600

Wavelength and Refraction

35-13. The wavelength of sodium light is 589 nm in air. Find its wavelength in glycerine.

From Table 28-1, the index for glycerin is: n = 1.47.

(589 nm)(1); ;1.47

g a a ag

a g g

n nn n

g = 401 nm

35-14. The wavelength decreases by 25 percent as it goes from air to an unknown medium.

What is the index of refraction for that medium?

490

air

nair = 1

nair = 1

nw = 1.33

ng = 1.55


A decrease of 25% means x is equal to ¾ of its air value:

0.750; 0.750; ;0.750

x air x airair

air x air

n nnn

nx = 1.33

35-15. A beam of light has a wavelength of 600 nm in air. What is the wavelength of this light

as it passes into glass (n = 1.50)?

g(1)(600 nm); ;

1.5g air air air

air g g

n nn n

g = 400 nm

35-16. Red light (620 nm) changes to blue light (478 nm) when it passes into a liquid. What is

the index of refraction for the liquid? What is the velocity of the light in the liquid?

b

(1)(620 nm); ;478 nm

air rL rL

air b

nn nn

nL = 1.30

*35-17. A ray of monochromatic light of wavelength 400 nm in medium A is incident at 300 at

the boundary of another medium B. If the ray is refracted at an angle of 500, what is its

wavelength in medium B?

0

0

sin sin (400 nm)sin50; ;sin sin sin30

A A A BB

B B A

B = 613 nm

491


Total Internal Reflection

35-18. What is the critical angle for light moving from quartz (n = 1.54) to water (n = 1.33).

2

1

1.33sin ;1.54c

nn

c = 59.70

35-19. The critical angle for a given medium relative to air is 400. What is the index of

refraction for the medium?

20 0

1

1sin ; ;sin 40 sin 40

airc x

nn nn

nx = 1.56

35-20. If the critical angle of incidence for a liquid to air surface is 460, what is the index of

refraction for the liquid?

20 0

1

1sin ; ;sin 46 sin 46

airc x

nn nn

nx = 1.39

35-21. What is the critical angle relative to air for (a) diamond, (b) water, and (c) ethyl alcohol.

Diamond:

2c

1

1.0sin ; sin ;2.42c

nn

c = 24.40

Water:

2c

1

1.0sin ; sin ;1.33c

nn

c = 48.80

492


Alcohol:

2c

1

1.0sin ; sin ;1.36c

nn

c = 47.30

35-22. What is the critical angle for flint glass immersed in ethyl alcohol?

2

1

1.36sin ;1.63c

nn

c = 56.50

*35-23. A right-angle prism like the one shown in Fig. 35-10a is submerged in water. What is the

minimum index of refraction for the material to achieve total internal reflection?

c < 450; 0

1.33 1.33sin ; ;sin 45

wc p

p p

n nn n

np = 1.88 (Minimum for total internal reflection.)

Challenge Problems

35-24. The angle of incidence is 300 and the angle of refraction is 26.30. If the incident medium

is water, what might the refractive medium be?

0

0x

sin (1.33)(sin 30 )sin sin ; ;sin sin 26.3w w

x x w w xnn n n

n = 1.50, glass

35-25. The speed of light in an unknown medium is 2.40 x 108 m/s. If the wavelength of light in

this unknown medium is 400 nm, what is the wavelength in air?

493

np nw = 1.33


8

8

(400 nm)(3 x 10 m/s); ;(2.40 x 10 m/s)

airair

x x

cv

x = 500 nm

35-26. A ray of light strikes a pane of glass at an angle of 300 with the glass surface. If the angle

of refraction is also 300, what is the index of refraction for the glass?

0

0

sin sin 60sin sin ; ;sin sin 30

g AA A g g

A g

nn n

n

nr = 1.73

35-27. A beam of light is incident on a plane surface separating two media of indexes 1.6 and

1.4. The angle of incidence is 300 in the medium of higher index. What is the angle of

refraction?

02 1 2 2

11 2 1

sin sin 1.6sin 30; sin ;sin 1.4

n nn n

1 = 34.80

35-28. In going from glass (n = 1.50) to water (n = 1.33), what is the critical angle for total

internal reflection?

2

1

1.33sin ;1.50c

nn

c = 62.50

35-29. Light of wavelength 650 nm in a particular glass has a speed of 1.7 x 108 m/s. What is the

index of refraction for this glass? What is the wavelength of this light in air?

494


8

8

3 x 10 m/s ;1.7 x 10 m/s

n n = 1.76

8

air 8

(3 x 10 m/s)(650 nm); 1.7 x 10 m/s

g g

air a

vv

; air = 1146 nm

35-30. The critical angle for a certain substance is 380 when it is surrounded by air. What is the

index of refraction of the substance?

21 0

1

1.0sin ; ;sin 38c

n nn

n1 = 1.62

35-31. The water in a swimming pool is 2 m deep. How deep does it appear to a person looking

vertically down?

1.00 2 m; ;1.33 1.33

air

w

nq qp n

q = 1.50 m

35-32. A plate of glass (n = 1.50) is placed over a coin on a table. The coin appears to be 3 cm

below the top of the glass plate. What is the thickness of the glass plate?

1.00 ; 1.50 (1.50)(3 cm);1.50

air

g

nq p qp n

p = 4.50 cm

495


Critical Thinking Problems

*35-33. Consider a horizontal ray of light striking one edge of an equilateral prism of glass (n =

1.50) as shown in Fig. 35-21. At what angle will the ray emerge from the other side?

0 00

1 11

sin 30 1.50 sin 30; sin ; 19.47sin 1.0 1.5

2 = 900 – 19.470 = 70.50;

3 = 49.470; 4 = 900 – 49.460; 4 = 40.530

00sin 40.5 1.00 ; sin 1.5sin 40.5 ;

sin 1.5

= 77.10

*35-34. What is the minimum angle of incidence at the first face of the prism in Fig. 35-21 such

that the beam is refracted into air at the second face? (Larger angles do not produce total

internal reflection at the second face.) [ First find critical angle for 4 ]

04 4c

1sin ; 41.8 ;1.5

airc

g

nn

Now find 1: 3 = 900 – 41.80 = 48.20

2 = 1800 – (48.20 + 600); 2 = 71.8 and 1 = 900 – 71.80 = 18.20

496

300

60 60

600n = 1.5

1

2

3

4


00

minmin

sin18.2 1.00 ; sin 1.5sin18.2 ;sin 1.50

min = 27.90

35-35. Light passing through a plate of transparent material of thickness t suffers a lateral

displacement d, as shown in Fig. 35-22. Compute the lateral displacement if the light

passes through glass surrounded by air. The angle of incidence 1 is 400 and the glass

(n = 1.50) is 2 cm thick.

00

22

sin 40 1.5 ; 25.4sin 1.0

3 = 900 – (25.40 + 500); 3 = 14.60

0 2 cmcos 25.4 ; R = 2.21 cm

R

;

03 3sin ; sin (2.21 cm)sin14.6 ;d d R

R

d = 5.59 mm

497

d2 cm R

32

500

400

t


*35-36. A rectangular shaped block of glass (n = 1.54) is submerged completely in water (n =

1.33). A beam of light traveling in the water strikes a vertical side of the glass block at

an angle of incidence 1 and is refracted into the glass where it continues to the top

surface of the block. What is the minimum angle 1 at the side such that the light does

not go out of the glass at the top? (First find c )

0

c1.33sin 0.864; 59.71.54c

2 = 900 – 59.70 = 30.30; 2 = 30.30

1 2 10

2 1

sin sin 1.54; ;sin sin30.3 1.33

nn

0

11.54sin 30.3sin ;

1.33

1 = 35.70

For angles smaller than 35.70, the light will leave the glass at the top surface.

*35-37. Prove that the lateral displacement in Fig. 35-22 can be calculated from

1 11

2 2

cossin 1cos

nd tn

Use this relationship to verify the answer to Critical Thinking Problem 35-35.

1

1

cos ; cos

d dpp

498

ng = 1.54

c

2

1

nw = 1.33

1d1

ap

2

dt

2

1

t1


11 2

1

sintan ; tan ;cos

p a at t

22

2 1

sintan ; ; cos cos

da t a t p

2

1 1 2

1

sinsin cos coscos

d tp a

t t

Simplifying this expression and solving for d, we obtain:

2 11

2

sin cossincos

td t

Now, n2 sin2 = n1 sin1 or

12 1

2

sin sinnn

Substitution and simplifying, we finally obtain the expression below:

1 11

2 2

cossin 1cos

nd tn

Substitution of values from Problem 35-35, gives the following result for d:

499


d = 5.59 mm

500

solucionario capitulo 35 - paul e. tippens

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