 (Parte 1 de 6) 3.1. IDENTIFY: For reflection, raθθ=. SET UP: The desired path of the ray is sketched in Figure 3.1.

EXECUTE: 14.0 cmtan

EVALUATE: The angle of incidence is measured from the normal to the surface. Figure 3.1

3.2 IDENTIFY: For reflection, raθθ=.

SET UP: The angles of incidence and reflection at each reflection are shown in Figure 3.2. For the rays to be perpendicular when they cross, 90α=°. same direction. As 90θ→°, 0α→°. This corresponds to the incident and reflected rays traveling in nearly opposite directions.

Figure 3.2

3-2 Chapter 3 3.3. IDENTIFY and SET UP: Use Eqs.(3.1) and (3.5) to calculate v and .λ

1.47 ccnv vn

EVALUATE: Light is slower in the liquid than in vacuum. By ,vfλ= when v is smaller, λ is smaller.

3.4. IDENTIFY: In air, 0cfλ=. In glass, 0n λλ=.

EVALUATE: In glass the light travels slower than in vacuum and the wavelength is smaller.

3.5. IDENTIFY: cnv =. 0n λλ=, where 0λis the wavelength in vacuum.

3.6. IDENTIFY: 0n λλ=.

SET UP: From Table 3.1, water1.33n= and benzene1.501n=.

EXECUTE: (a) waterwaterbenzenebenzene0nnλλλ==. water benzene water

benzene

3.7. IDENTIFY: Apply Eqs.(3.2) and (3.4) to calculate and .rbθθ The angles in these equations are measured with respect to the normal, not the surface. (a) SET UP: The incident, reflected and refracted rays are shown in Figure 3.7.

EXECUTE: 42.5raθθ==° The reflected ray makes an

Figure 3.7 (b) sinsin,abnθθ= where the angles are measured from the normal to the interface.

bn n

The refracted ray makes an angle of 90.066.0bθ°−=° with the surface of the glass. EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index.

3.8. IDENTIFY: Use the distance and time to find the speed of light in the plastic. cnv =.

EVALUATE: In air light travels this same distance in 8 2.50 m8.3 ns

The Nature and Propagation of Light 3-3

3.9. IDENTIFY and SET UP: Use Snell s law to find the index of refraction of the plastic and then use Eq.(3.1) to calculate the speed v of light in the plastic.

EXECUTE: sinsinaabbnnθθ=

EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent toward the normal. 3.10. IDENTIFY: Apply Snell s law at both interfaces. SET UP: The path of the ray is sketched in Figure 3.10. Table 3.1 gives 1.329n=for the methanol.

EXECUTE: (a) At the air-glass interface glass(1.0)sin41.3sinnα°=. At the glass-methanol interface

EVALUATE: The angle αis 25.2°. The index of refraction of methanol is less than that of the glass and the ray is bent away from the normal at the glass→methanol interface. The unknown liquid has an index of refraction greater than that of the glass, so the ray is bent toward the normal at the glass→liquid interface.

Figure 3.10 3.1. IDENTIFY: Apply Snell s law to each refraction.

SET UP: Let the light initially be in the material with refractive index anand let the final slab have refractive index bn. In part (a) let the middle slab have refractive index 1n. EXECUTE: (a) 1st interface: 11sinsinaannθθ=. 2nd interface: 11sinsinbbnnθθ=. Combining the two equations gives sinsinaabbnnθθ=. This is the equation that would apply if the middle slab were absent. (b) For N slabs, 11sinsinaannθθ=, 1122sinsinnnθθ=, , 22sinsinNNbbnnθθ−−=. Combining all these equations gives sinsinaabbnnθθ=.

EVALUATE: The final direction of travel depends on the angle of incidence in the first slab and the refractive indices of the first and last slabs. 3.12. IDENTIFY: Apply Snell’s law to the refraction at each interface.

EXECUTE: (a) air water air

water

1.00arcsin sin arcsin sin35.0 25.5 .

EVALUATE: (b) This calculation has no dependence on the glass because we can omit that step in the air air glass glass wate waterchain: si n sin sin .rnn nθ θθ==

3.13. IDENTIFY: When a wave passes from one material into another, the number of waves per second that cross the boundary is the same on both sides of the boundary, so the frequency does not change. The wavelength and speed of the wave, however, do change.

SET UP: In a material having index of refraction n, the wavelength is 0,n λλ= where λ0 is the wavelength in vacuum, and the speed is .cn

EXECUTE: (a) The frequency is the same, so it is still f. The wavelength becomes 0,n λλ= so λ0 = nλ. The speed is ,cvn = so c = nv.

(b) The frequency is still f. The wavelength becomes 0nn and the speed becomes

cn v nvv

EVALUATE: These results give the speed and wavelength in a new medium in terms of the original medium without referring them to the values in vacuum (or air).

3-4 Chapter 3

3.14. IDENTIFY: Apply the law of reflection.

SET UP: The mirror in its original position and after being rotated by an angle θ are shown in Figure 3.14. α is the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled φ are equal and the two angles labeled φ′are equal because of the law of reflection. The two angles labeled θ are equal because the

EVALUATE: This result is independent of the initial angle of incidence.

Figure 3.14 3.15. IDENTIFY: Apply sinsinaabbnnθθ=.

EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal. 3.16. IDENTIFY: Apply Snell’s law.

EXECUTE: 1.33arcsin sin arcsin sin45.0 38.21.52a ba

°. But this is the angle from the normal to the surface,

so the angle from the vertical is an additional 15° because of the tilt of the surface. Therefore, the angle is 53.2°. EVALUATE: Compared to Example 3.1, aθ is shifted by 15° but the shift in bθ is only 53.249.3.9−=°. 3.17. IDENTIFY: The critical angle for total internal reflection is aθthat gives 90bθ=°in Snell’s law.

SET UP: In Figure 3.17 the angle of incidence aθ is related to angle θ by 90aθθ+=°. EXECUTE: (a) Calculate aθ that gives 90bθ=°. 1.60an=, 1.00bn= so sinsinaabbnnθθ= gives

EVALUATE: The critical angle increases when the ratio abnn increases.

Figure 3.17

The Nature and Propagation of Light 3-5

3.18. IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will occur at the cube surface if the angle of incidence is greater than or equal to the critical angle.

SET UP: At the critical angle θc, Snell s law gives nglass sin θc = nair sin 90° and likewise for water. EXECUTE: (a) At the critical angle θc , nglass sin θc = nair sin 90°. 1.53 sin θc = (1.0)(1) andθc = 40.8°.

(b) Using the same procedure as in part (a), we have 1.53 sin θc = 1.3 sin 90° and θc = 60.6°. EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air. 3.19. IDENTIFY: Use the critical angle to find the index of refraction of the liquid. SET UP: Total internal reflection requires that the light be incident on the material with the larger n, in this case the liquid. Apply sinsinaabbnnθθ= with a = liquid and b = air, so liqann= and 1.0.bn=

liq crit

(a) sinsinaabbnnθθ= (a = liquid, b = air)

bn n

(b) Now sinsinaabbnnθθ= with a = air, b = liquid

1.48aa b bn n

EVALUATE: For light traveling liquid → air the light is bent away from the normal. For light traveling air → liquid the light is bent toward the normal. 3.20. IDENTIFY: The largest angle of incidence for which any light refracts into the air is the critical angle for waterair→.

SET UP: Figure 3.20 shows a ray incident at the critical angle and therefore at the edge of the ring of light. The radius of this circle is r and 10.0 md=is the distance from the ring to the surface of the water.

EXECUTE: From the figure, crittanrdθ=. critθis calculated from sinsinaabbnnθθ= with 1.333an=, critaθθ=,

Figure 3.20

3.21. IDENTIFY and SET UP: For glass → water, crit48.7.θ=° Apply Snell s law with critaθθ= to calculate the index of refraction an of the glass.

EXECUTE: crit crit

EVALUATE: For total internal reflection to occur the light must be incident in the material of larger refractive index. Our results give glasswater,n> in agreement with this.

3-6 Chapter 3

3.2. IDENTIFY: If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface is crit.θ SET UP: The ray has an angle of incidence of 0°at the first surface of the glass, so enters the glass without being bent, as shown in Figure 3.2. The figure shows that crit90αθ+=°. EXECUTE: (a) For the glass-air interface crit,aθθ= 1.52,an= 1.00bn=and 90.bθ=° sinsinaabbnnθθ= gives

(b) Now the second interface is glasswater→and 1.333bn=. sinsinaabbnnθθ= gives crit (1.3)(sin90 )sin 1.52

EVALUATE: The critical angle increases when the air is replaced by water and rays are bent as they refract out of the glass.

Figure 3.2

3.23. IDENTIFY: Apply sinsinaabbnnθθ=.

SET UP: The light is in diamond and encounters an interface with air, so 2.42an=and 1.00bn=. The largest aθis when 90bθ=°.

3.24. IDENTIFY: Snell’s law is sinsinaabbnnθθ=. cvn =.

SET UP: aira=, glassb=.

sin sin38.1aa b

EVALUATE: n is larger for the violet light and therefore this light is bent more toward the normal, and the violet light has a smaller speed in the glass than the red light.

3.25. IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12and the transmitted light is polarized along the axis of the polarizer. When polarized light of intensity maxIis incident on a polarizer, the transmitted intensity is 2 maxcosIIφ=, where φis the angle between the polarization direction of the incident light and the axis of the filter.

EVALUATE: Adding the middle filter increases the transmitted intensity. 3.26. IDENTIFY: Apply Snell’s law. SET UP: The incident, reflected and refracted rays are shown in Figure 3.26.

sin sin 37a ba b n θθ

The Nature and Propagation of Light 3-7

EVALUATE: The refractive index of b is greater than that of a, and the ray is bent toward the normal when it refracts.

Figure 3.26

3.27. IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals the polarizing angle, so p54.5.θ=° Use Eq.(3.8) to calculate the refractive index of the glass. Then use Snell s law to calculate the angle of refraction.

EXECUTE: (a) pglassairptan gives tan(1.0)tan54.51.40.ba n n

(b) sinsinaabbnnθθ=

bn n

reflected ray and the refracted ray are perpendicular to each other. This agrees with Fig.3.28.

Figure 3.27

3.28. IDENTIFY: Set 0/10II=, where I is the intensity of light passed by the second polarizer.

SET UP: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12and the transmitted light is polarized along the axis of the polarizer. When polarized light of intensity maxIis incident on a polarizer, the transmitted intensity is 2 maxcosIIφ=, where φis the angle between the polarization direction of the incident light and the axis of the filter.

EXECUTE: (a) After the first filter 02 I=and the light is polarized along the vertical direction. After the second

cos1/10φ=and 71.6φ=°. EVALUATE: When the incident light is polarized along the axis of the first filter, φmust be larger to achieve the same overall reduction in intensity than when the incident light is unpolarized. 3.29. IDENTIFY: From Malus s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters.

SET UP: If the angle between the two axes is θ, the intensity of the emerging light is I = Imax cos2θ.

EXECUTE: At angle θ, I = Imax cos2θ, and at the new angle α, 12 I = Imax cos2α. Taking the ratio of the intensities gives maxcos cosIIII α θ

=, which gives us coscos2 θα=. Solving for α yields cosarccos2

EVALUATE: Careful! This result is not cos2θ.

3-8 Chapter 3 3.30. IDENTIFY: The reflected light is completely polarized when the angle of incidence equals the polarizing angle pθ, where ptanbann θ=.

EVALUATE: The polarizing angle depends on the refractive indicies of both materials at the interface.

3.31. IDENTIFY: When unpolarized light of intensity 0Iis incident on a polarizing filter, the transmitted light has intensity 1 02Iand is polarized along the filter axis. When polarized light of intensity 0Iis incident on a polarizing

EVALUATE: The transmitted intensity depends on the angle between the axes of the two filters. 3.32. IDENTIFY: After passing through the first filter the light is linearly polarized along the filter axis. After the second filter, 2 max(cos)Iφ=, where φ is the angle between the axes of the two filters.

SET UP: The maximum amount of light is transmitted when 0φ=.

and the transmitted intensity decreases. 3.3. IDENTIFY and SET UP: Apply Eq.(3.7) to polarizers #2 and #3. The light incident on the first polarizer is unpolarized, so the transmitted light has half the intensity of the incident light, and the transmitted light is polarized. (a) EXECUTE: The axes of the three filters are shown in Figure 3.3a.

2 maxcosIIφ=

Figure 3.3a

After the first filter the intensity is 1 102II= and the light is linearly polarized along the axis of the first polarizer.

0.125I and the light is linearly polarized along the axis of the third polarizer. (b) The axes of the remaining two filters are shown in Figure 3.3b.

After the first filter the intensity is 1 102II= and the

light is linearly polarized along the axis of the first polarizer.

Figure 3.3b

EVALUATE: Light is transmitted through all three filters, but no light is transmitted if the middle polarizer is removed.

The Nature and Propagation of Light 3-9

3.34. IDENTIFY: Use the transmitted intensity when all three polairzers are present to solve for the incident intensity

0I. Then repeat the calculation with only the first and third polarizers.

SET UP: For unpolarized light incident on a filter, 1 02II= and the light is linearly polarized along the filter axis.

For polarized light incident on a filter, 2 max(cos)Iφ=, where maxIis the intensity of the incident light, and the emerging light is linearly polarized along the filter axis.

EXECUTE: With all three polarizers, if the incident intensity is 0Ithe transmitted intensity is

 002( )(cos23.0 ) (cos[62.0 23.0 ]) 0.256II I=− =° ° 2

EVALUATE: The transmitted intensity is greater when all three filters are present. 3.35. IDENTIFY: The shorter the wavelength of light, the more it is scattered. The intensity is inversely proportional to the fourth power of the wavelength.

SET UP: The intensity of the scattered light is proportional to 1/λ4, we can write it as I = (constant)/ λ4. EXECUTE: (a) Since I is proportional to 1/λ4, we have I = (constant)/ λ4. Taking the ratio of the intensity of the red light to that of the green light gives

= 0.374, so IR = 0.374I.

(b) Following the same procedure as in part (a) gives 4 4

520 nm

= 2.35, so IV = 2.35I.

EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about 7 times as great as that of the red light, so this scattered light will have a blue-violet color. 3.36. IDENTIFY: As the wave front reaches the sharp object, every point on the front will act as a source of secondary wavelets. SET UP: Consider a wave front that is just about to go past the corner. Follow it along and draw the successive wave fronts. EXECUTE: The path of the wavefront is drawn in Figure 3.36. EVALUATE: The wave fronts clearly bend around the sharp point, just as water waves bend around a rock and light waves bend around the edge of a slit.

(Parte 1 de 6)