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Chapter 35

By Larry Ramirez,2014-07-09 07:24
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Chapter 35

    thChapter 35. Refraction Physics, 6 Edition

    Chapter 35. Refraction

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

    835-1. The speed of light through a certain medium is 1.6 x 10 m/s in a transparent medium,

    what is the index of refraction in that medium?

    8c3 x 10m/s n = 1.88 n??;8v1.6 x 10m/s

    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:

    cc32 and n = 1.50 vcnn????(); ; 3x2vc()23x

    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.)

    883 x 10m/s3 x 10m/s88(a) v = 1.97 x 10 m/s (b) v = 1.24 x 10 m/s v?v?;;gdgd1.502.42

    883 x 10m/s3 x 10m/s88(a) v = 2.26x 10 m/s (b) v = 2.21 x 10 m/s v?v?;;awa1.331.36

    8 35-4 If light travels at 2.1 x 10 m/s in a transparent medium, what is the index of refraction?

    8(3 x 10m/s) n = 1.43 n?;82.1 x 10m/s

The Laws of Refraction

    035-5. Light is incident at an angle of 37 from air to flint glass (n = 1.6). What is the angle of

    refraction into the glass?

    0(1.0)sin370 ? = 22.1 nn?????sinsin; sin;gggaag1.6

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    035-6. A beam of light makes an angle of 60 with the surface of water. What is the angle of

    refraction into the water?

    nsin?0n = 1.5 waa 60 ??nnsinsin; sin???waawwair nw

    water 0(1)sin600 ? = 40.6 ???sin0.651;ww1.33

     035-7. Light passes from water (n = 1.33) to air. The beam emerges into air at an angle of 32

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

    nsin?000aa? = 90 32 = 58; ??nnsinsin; sin???waawwnw

    0(1)sin580 ? = 39.6???sin638;ww1.33

     0035-8. Light in air is incident at 60 and is refracted into an unknown medium at an angle of 40.

    What is the index of refraction for the unknown medium?

    0nsin?(1)(sin60)aa n = 1.35 ???nnnsinsin; ;??xxaax0sinsin40?x

     035-9. Light strikes from medium A into medium B at an angle of 35 with the horizontal

    0boundary. If the angle of refraction is also 35, what is the relative index of refraction

    000between the two media? [ ? = 90 35 = 55. ] A

    00A 35 ? nsinsin55?0BA35 ???nnsinsin; ;??AABB0nsinsin35?ABB

    0nsin55B n = 1.43 ??1.43;r0nsin35A

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    0035-10. Light incident from air at 45 is refracted into a transparent medium at an angle of 34.

    What is the index of refraction for the material?

    0(1)sin45 n = 1.23 nnn????sinsin; ;mAAmmm0sin35

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

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

    into air again. Compute the angle of emergence. ?air = 1 nair

     The angle of refraction into one medium becomes the n = 1.33 w

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

    n = 1.55 g

     n sin ?= n sin ? = n sin ? = n sin ? airair wwggairairn = 1 air

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

    0 as that from which it originally entered has the same angle: ? = ? = 60 ei

    *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:

    0n sin ?= n sin ? = n sin ? = n sin ?; ? = ? = 60 airair wwggairairei

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.

    ?nn(589 nm)(1)?gaaa ? = 401 nm ???; ;?ggnn1.47?agg

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    thChapter 35. Refraction Physics, 6 Edition

    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?

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

    ??nnxairxair n = 1.33 ????0.750; 0.750; ;nxair0.750n??airxair

    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)?

    n??(1)(600 nm)ngairairair ? = 400 nm ???; ;?gg1.5nn?airgg

    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?

    n?n?(1)(620 nm)airrLr n = 1.30 ???; ;nLLn478 nm??bairb

     0*35-17. A ray of monochromatic light of wavelength 400 nm in medium A is incident at 30 at

    0the boundary of another medium B. If the ray is refracted at an angle of 50, what is its

    wavelength in medium B?

    0sinsin(400 nm)sin50????AAAB ? = 613 nm ???; ;?BB0sinsinsin30???BBA

Total Internal Reflection

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

    n1.3302 ? = 59.7 ???sin;ccn1.541

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    035-19. The critical angle for a given medium relative to air is 40. What is the index of

    refraction for the medium?

    nn1air2 n = 1.56 ????sin; ;nxcx00sin40sin40n1

     035-20. If the critical angle of incidence for a liquid to air surface is 46, what is the index of

    refraction for the liquid?

    nn1air2 n = 1.39 ????sin; ;nxcx00nsin46sin461

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

    n1.002 Diamond: ? = 24.4 ????sin; sin;cccn2.421

    n1.002 Water: ? = 48.8 ????sin; sin;cccn1.331

    n1.002 Alcohol: ? = 47.3 ????sin; sin;cccn1.361

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

    n1.3602 ? = 56.5 ???sin;ccn1.631

    *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?

     npn = 1.33 wn1.331.330w ? < 45; ????sin; ;nccp0nnsin45pp

     n = 1.88 (Minimum for total internal reflection.) p

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    thChapter 35. Refraction Physics, 6 Edition

    Challenge Problems

    0035-24. The angle of incidence is 30 and the angle of refraction is 26.3. If the incident medium

    is water, what might the refractive medium be?

    0nsin?(1.33)(sin30)ww n = 1.50, glass ???nnnsinsin; ;??xxwwx0sinsin26.3?x

     835-25. The speed of light in an unknown medium is 2.40 x 10 m/s. If the wavelength of light in

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

    8?c(400 nm)(3 x 10m/s)air ? = 500 nm ??; ;?xair8v(2.40 x 10m/s)?xx

     035-26. A ray of light strikes a pane of glass at an angle of 30 with the glass surface. If the angle

    0of refraction is also 30, what is the index of refraction for the glass?

    0nsinsin60?gA n = 1.73 ???sinsin; ;nn??rAAgg0sinsin30n?Ag

35-27. A beam of light is incident on a plane surface separating two media of indexes 1.6 and 1.4.

    0The angle of incidence is 30 in the medium of higher index. What is the angle of

    refraction?

    0sinsin1.6sin30??nn02122 ? = 34.8 ???; sin;?11sin1.4nn?121

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

    internal reflection?

    n1.3302 ? = 62.5 ???sin;ccn1.501

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    835-29. Light of wavelength 650 nm in a particular glass has a speed of 1.7 x 10 m/s. What is the

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

    83 x 10m/s n = 1.76 n?;81.7 x 10m/s

    8v?(3 x 10m/s)(650 nm)gg ; ? = 1146 nm ??; ?airair81.7 x 10m/sv?aira

     035-30. The critical angle for a certain substance is 38 when it is surrounded by air. What is the

    index of refraction of the substance?

    n1.02 n = 1.62 ???sin; ;n1c10nsin381

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

    vertically down?

    nq1.002 mair q = 1.50 m ???; ;q1.331.33pnw

    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?

    qn1.00air p = 4.50 cm ????; 1.50(1.50)(3 cm);pqpn1.50g

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    thChapter 35. Refraction Physics, 6 Edition

    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?

    00sin301.50sin3000 ?????; sin; 19.47 6011n = 1.5 ?sin1.01.510 30 ?2 000 ? = 90 19.47 = 70.5; ?;?;(~,;,;;:,?;;,!, 2 ?3 ?1 ? ?4 0000 ? = 49.47; ? = 90 49.46; ? = 40.53 344

    6060000sin40.51.00 00 ? = 77.1 ???; sin1.5sin40.5;sin1.5?

*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

    n10000air Now find ?: ? = 90 41.8 = 48.2 ?????sin; 41.8;1344cc1.5ng

    000000? = 180 (48.2 + 60); ? = 71.8 and ? = 90 71.8 = 18.2 221

    0sin18.21.000 0 ? = 27.9???; sin1.5sin18.2;minmin?sin1.50min

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    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

    0passes through glass surrounded by air. The angle of incidence ? is 40 and the glass 1

    (n = 1.50) is 2 cm thick. 0 40

    00sin401.50 50 ???; 25.42?sin1.02?? 3 2t 2 cm 0000R ? = 90 (25.4 + 50); ? = 14.6 33d

    2 cm0 ; cos25.4; R = 2.21 cm?R

    d0 d = 5.59 mm sin; sin(2.21 cm)sin14.6;?????dR33R

    *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 ? and is refracted into the glass where it continues to the top 1

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

    not go out of the glass at the top? (First find ? ) c = 1.33 nw

    1.330 ?c ?????sin0.864; 59.7cc1.54

    ? 20000 ? = 90 59.7 = 30.3; ? = 30.322? 1

    n = 1.54 gsinsin1.54??n121 ??; ;0sinsin30.31.33n?21

    01.54sin30.30 ? = 35.7??sin;111.33

    0For angles smaller than 35.7, the light will leave the glass at the top surface.

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    thChapter 35. Refraction Physics, 6 Edition

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

    ??ncos?11 dtsin1?,?(,1ncos?22)?

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

    dd cos; ???p1pcos?1

    sin?paa1? 1 tan; tan;?????12costt?1

    ?1 sin?d ?22t ???atatptan; ; ?t 2 ?2coscos??21d p

    dsin???a 2?1 t(,?d 1 sinpacoscos???112(,?? costt(,?1(,)?

    tsincos??21 Simplifying this expression and solving for d, we obtain: ?,dtsin?1cos?2

    n1Now, n sin? = n sin?or sinsin???2211 21n2

    Substitution and simplifying, we finally obtain the expression below:

    ??ncos?11 dtsin1?,?(,1ncos?22)?

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

    d = 5.59 mm

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