Solucionario Capitulo 33 - Paul E. Tippens

Physics, 6th Edition

Chapter 33. Light and Illumination

Chapter 33. Light and Illumination Light and the Electromagnetic Spectrum 33-1. An infrared spectrophotometer scans the wavelengths from 1 to 16 µm. Express these range in terms of the frequencies of the infrared rays. f1 =

c 3 x 108 m/s c 3 x 108 m/s 13 = = 30.0 x 10 Hz; f = = = 1.88 x 1013 Hz 2 λ1 1 x 10-6 m λ2 16 x 10-6 m Range of frequencies: 1.88 x 1013 Hz to 30 x 1013 Hz

33-2. What is the frequency of violet light of wavelength 410 nm? f =

c 3 x 108 m/s = ; λ 410 x 10-9 m

f = 7.32 x 1014 Hz

33-3. A microwave radiator used in measuring automobile speeds emits radiation of frequency 1.2 x 109 Hz. What is the wavelength?

λ=

c 3 x 108 m/s = ; f 1.2 x 109 Hz

λ = 250 mm

33-4. What is the range of frequencies for visible light? ( Range of λ: 700 nm to 400 nm ) f1 =

c 3 x 108 m/s c 3 x 108 m/s 14 = = 4.29 x 10 Hz; f = = = 7.50 x 1014 Hz 2 -9 -9 λ1 700 x 10 m λ2 400 x 10 m Range of frequencies: 4.29 x 1014 Hz to 7.50 x 1014 Hz

33-5. If Plank’s constant h is equal to 6.626 x 10-34 J s, what is the energy of light of wavelength 600 nm? E = hf =

hc (6.625 x 10-34 J ⋅ s)(3 x 108 m/s) = ; λ 600 x 10-9 m 205

E = 3.31 x 10-19 J


Chapter 33. Light and Illumination 33-6. What is the frequency of light whose energy is 5 x 10-19 J? E 5.00 x 10-19 J f = = ; h (6.625 x 10-34 J ⋅ s)

f = 7.55 x 1014 Hz

33-7. The frequency of yellow-green light is 5.41 x 1014 Hz. Express the wavelength of this light in nanometers and in angstroms?

λ=

c 3 x 108 m/s = ; f 5.41 x 1014 Hz

λ = 555 x 10-9 m(1 x 1010 A/m);

λ = 555 nm λ = 5550 A

33-8. What is the wavelength of light whose energy is 7 x 10-19 J? E=

hc ; λ

λ=

hc (6.625 x 10-34 J ⋅ s)(3 x 108 m/s) = ; E 7 x 10-19 J

λ = 284 nm

The Velocity of Light 33-9. The sun is approximately 93 million miles from the earth. How much time is required for the light emitted by the sun to reach us on earth? 93 x 106 mi t= = 500 s ; 186,000 mi/s

t = 8.33 min

33-10. If two experimenters in Galileo’s experiment were separated by a distance of 5 km, how much time would have passed from the instant the lantern was opened until the light was observed? t=

s 5000 m = ; c 3 x 108 m/s

t = 17.0 x 10-6 s or 17.0 µs

206



33-11. The light reaching us from the nearest star, Alpha Centauri, requires 4.3 years to reach us. How far is this in miles? In kilometers? s = ct = (186,000 mi/s)(3.154 x 107 s/yr)(4.30 yr); s = ct = (3 x 108 m/s)(3.154 x 107 s/yr)(4.30 yr);

s = 2.53 x 1013 mi s =4.07 x 1013 km

33-12. A spacecraft circling the moon at a distance of 384,000 km from the earth communicates by radio with a se on earth. How much time elapses between the sending and receiving of a signal? s = (384,000 km)(1000 m/km) = 3.10 x 108 m;

t=

4 x 1010 m ; 3 x 108 m/s

3.84 x 108 m = 1.28 s t= 3 x 108 m/s

t = 133 s

33-13. An spacecraft sends a signal that requires 20 min to reach the earth. How far away is the spacecraft from earth? s = ct = (3 x 108 m/s)(20 min)(60 s/min);

s = 3.60 x 1011 m

Light Rays and Shadows 33-14. The shadow formed on a screen 4 m away from a point source of light is 60 cm tall. What is the height of the object located 1 m from the source and 3 m from the shadow? Similar trianges: 60 cm h = 100 cm 400 cm (400 cm)(60 cm) h= ; 100 cm

60 cm

1m h = 2.40 m

207

3m

h



33-15. A point source of light is placed 15 cm from an upright 6-cm ruler. Calculate the length of the shadow formed by the ruler on a wall 40 cm from the ruler. h 6 cm = ; 55 cm 15 cm

h

6 cm

h = 22.0 cm 15 cm

40 cm

33-16. How far must an 80-mm-diameter plate be placed in front of a point source of light if it is to form a shadow 400 mm in diameter at a distance of 2 m from the light source? x 2000 mm = ; 80 mm 400 mm

x = 400 mm

80 mm

400 mm

x 2000 mm

33-17. A source of light 40 mm in diameter shines through a pinhole in the tip of a cardboard box 2 m from the source. What is the diameter of the image formed on the bottom of the box if 60 mm

the height of the box is 60 mm? y 40 mm = ; 60 mm 2000 mm

40 mm

y

y = 1.20 mm

2000 mm

*33-18. A lamp is covered with a box, and a 20-mm-long narrow slit is cut in the box so that light shines through. An object 30 mm tall blocks the light from the slit at a distance of 500 mm. Calculated the length of the umbra and penumbra formed on a screen located 1.50 m from 500 mm

the slit. (Similar ∆’s) a 500 + a = ; 20 mm 30 mm u 20 = ; 1500 + a a

1500 mm

a

a = 1000 mm

30 mm

20 mm

u 20 = ; u = 50 mm 1500 + 1000 1000

b

c

b 500 − b p 20 = ; b = 200 mm; Now, = from which p = 130 mm 20 30 1500 − b b 208

u

p



Illumination of Surfaces 33-19. What is the solid angle subtended at the center of a 3.20-m-diameter sphere by a 0.5-m2 area on its surface? A 0.5 m 2 Ω= 2 = ; R (1.6 m) 2

Ω = 0.195 sr

33-20. A solid angle of 0.080 sr is subtended at the center of a 9.00 cm diameter sphere by surface area A on the sphere. What is this area? A = ΩR2 = (0.08 sr)(0.09 m)2;

A = 6.48 x 10-4 m2

33-21. An 8½ x 11 cm sheet of metal is illuminated by a source of light located 1.3 m directly above it. What is the luminous flux falling on the metal if the source has an intensity of 200 cd. What is the total luminous flux emitted by the light source? A = (0.085 m)(0.11 m); Ω=

A = 9.35 x 10-3 m2 1.3 m

A 9.35 x 10−3 m 2 = = 5.53 x 10-3sr 2 2 R (1.3 m)

F = IΩ = (200 cd)(5.53 x 10-3 sr); Total Flux = 4πI = 4π(200 cd);

F = 1.11 lm Total Flux = 2510 lm

33-22. A 40-W monochromatic source of yellow-green light (555 nm) illuminates a 0.5-m2 surface at a distance of 1.0 m. What is the luminous intensity of the source and how many lumens fall on the surface? F = (680 lm/W)(40 W) = 27,200 lm;

I=

FR 2 (27, 200 lm)(1 m) 2 = ; A 0.5 m 2

209

I=

F A ; Ω= 2 Ω R

I = 54,400 cd



33-23. What is the illumination produced by a 200-cd source on a small surface 4.0 m away? E=

I 200 cd = ; 2 R (4 m) 2

E = 12.5 lx

33-24. A lamp 2 m from a small surface produces an illumination of 100 lx on the surface. What is the intensity of the source? I = ER2 = (100 lx)(2 m)2;

I = 400 cd

33-25. A table top 1 m wide and 2 m long is located 4.0 m from a lamp. If 40 lm of flux fall on this surface, what is the illumination E of the surface? E=

F 40 lm = ; A (1 m)(2 m)

E = 20 lx

33-26. What should be the location of the lamp in Problem 33-25 in order to produce twice the illumination? ( Illumination varies inversely with square of distance.) E1R12 = E2R22 = (2E1)R22;

R2 =

R12 (4m) 2 = ; 2 2

R2 = 2.83 m

*33-27. A point source of light is placed at the center of a sphere 70 mm in diameter. A hole is cut in the surface of the sphere, allowing the flux to pass through a solid angle of 0.12 sr. What is the diameter of the opening? [ R = D/2 = 35 mm ] A = ΩR2 = (0.12 sr)(35 mm)2; A = 147 mm2 A=

π D2 ; 4

D=

4A 4(147 mm 2 ) = ; π π

210

D = 13.7 mm



Challenge Problems 33-28. When light of wavelength 550 nm passes from air into a thin glass plate and out again into the air, the frequency remains constant, but the speed through the glass is reduced to 2 x 208 m/s. What is the wavelength inside the glass? (f is same for both) v v f = air = glass ; λair λglass

λglass =

vglass λair vair

(2 x 108 m/s)(550 nm) = ; 3 x 108 m/s

λglass= 367 nm

33-29. A 30-cd standard light source is compared with a lamp of unknown intensity using a grease-spot photometer (refer to Fig. 33-21). The two light sources are placed 1 m apart, and the grease spot is moved toward the standard light. When the grease spot is 25 cm from the standard light source, the illumination is equal on both sides. Compute the unknown intensity? [ The illumination E is the same for each. ] Ix Is = ; rx2 rs2

Ix =

I s rx2 (30 cd)(75 cm) 2 = ; rs2 (25 cm) 2

Ix

30 cd 75 cm

25 cm

Ix = 270 cd

33-30. Where should the grease spot in Problem 33-29 be placed for the illumination by the unknown light source to be exactly twice the illumination of the standard source? E x = 2 Es ;

I x 2I s = ; rx2 rs2

4.5 1 = 2; 2 (1- x ) x

270 cd 2(30 cd) = ; (1- x) 2 x2

4.5 x 2 = (1 − x) 2 ;

3.12 x = 1;

x=

1m 30 cd

2.12 x = 1 − x;

1 = 0.320 m; 3.12

211

x

270 cd 1m–x

x = 32.0 cm from standard

Ix



33-31. The illumination of a given surface is 80 lx when it is 3 m away from the light source. At what distance will the illumination be 20 lx? ( Recall that I = ER is constant ) E1 R12 = E2 R22 ;

R2 =

E1 R12 (80 lx)(3 m) 2 = ; E2 (20 lx)

R2 = 600 m

33-32. A light is suspended 9 m above a street and provides an illumination of 35 lx at a point directly below it. Determine the luminous intensity of the light. E=

I ; I = ER 2 = (36 lx)(9 m) 2 ; 2 R

I = 2920 cd

*33-33. A 60-W monochromatic source of yellow-green light (555 nm) illuminates a 0.6 m2 surface at a distance of 1.0 m. What is the solid angle subtended at the source? What is the luminous intensity of the source? F = (680 lm/W)(60 W) = 40,800 lm;

I=

Ω=

F (40,800 lm) = ; Ω 0.60 sr

A 0.6 m 2 = = 0.600 sr R 2 (1 m) 2

I = 68,000 cd

*33-34. At what distance from a wall will a 35-cd lamp provide the same illumination as an 80-cd lamp located 4.0 m from the wall? [ E1 = E2 ] I1 I 2 = ; r12 r22

r2 =

I 2 r12 (35 cd)(4 m) 2 = ; I1 80 cd

r2 = 2.65 m

*33-35. How much must a small lamp be lowered to double the illumination on an object that is 80 cm directly under it? R2 =

E2 = 2E1 and E1R12 = E2R22 so that: E1R12 = (2E1)R22

R12 (80 cm) 2 = ; 2 2

R2 = 56.6 cm; y = 80 cm – 56.6 cm = 23.4 cm

212



*33-36. Compute the illumination of a given surface 140 cm from a 74-cd light source if the normal to the surface makes an angle of 380 with the flux. I cos θ (74 cd)(cos 380 ) E= = ; R2 (1.40 m) 2

E = 29.8 lx

*33-37. A circular table top is located 4 m below and 3 m to the left of a lamp that emits 1800 lm. What illumination is provided on the surface of the table? What is the area of the table top if 3 lm of flux falls on its surface? tan θ =

3m 4m

θ = 36.90 I=

F = 4πI

F 1800 lm = ; 4π 4π

I cos θ (143 lm/sr) cos 36.90 E= = ; R2 (5 m) 2

R=5 m

4m

3m

I = 143 lm/sr

E = 4.58 lx;

θ

A=

F = 0.655 m2 E

*33-38. What angle θ between the flux and a line drawn normal to a surface will cause the illumination of that surface to be reduced by one-half when the distance to the surface has not changed? E1 =

I1 ; R12

E2 =

I cosθ ; R22

E1 = 2 E2 ;

I1 = I 2 and R1 = R2

Substitution yields: 2I cosθ = I and cos θ = 0.5

213

or

θ = 600



*33-39. In Michelson’s measurements of the speed of light, as shown in Fig. 33-11, he obtained a value of 2.997 x 108 m/s. If the total light path was 35 km, what was the rotational frequency of the eight-sided mirror? The time for light to reappear from edge 1 to edge 2 is: t=

s 35, 000 m = = 1.167 x 10-4s c 3 x 108 m/s

The time for one revolution is 8t: T = 8(1.167 x 10-4 s) f =

1 1 = ; T 8(1.167 x 10-4s)

f = 1071 Hz

*33-40. All of the light from a spotlight is collected and focused on a screen of area 0.30 m2. What must be the luminous intensity of the light in order that an illumination of 500 lx be achieved? E=

I ; A

I = EA = (500 lx)(0.30 m 2 ) = 150 cd

I = 150 cd

*33-41. A 300-cd light is suspended 5 m above the left edge of a table. Find the illumination of a small piece of paper located a horizontal distance of 2.5 m from the edge of the table? R = (2.5 m) 2 + (5 m) 2 = 5.59 m 2.5 m tan θ = ; 5m E=

R θ

θ = 26.6

0

2.5 m

I cos θ (300 cd)(cos 26.60 ) = ; R2 (5.59 m) 2

214

E = 8.59 lx

5m



Critical Thinking Problems 33-42. A certain radio station broadcasts at a frequency of 1150 kHz; A red beam of light has a frequency of 4.70 x 1014 Hz ; and an ultraviolet ray has a frequency of 2.4 x 1016 Hz. Which has the highest wavelength? Which has the greatest energy? What are the wavelengths of each electromagnetic wave? [ Recall that h = 6.625 x 10-34 J.]

λ=

c 3 x 108 m/s = ; f 1.150 x 106 Hz

λ = 261 m

E = hf = 7.62 x 10-28 J

λ=

c 3 x 108 m/s = ; f 4.70 x 1014 Hz

λ = 639 nm

E = hf = 3.11 x 10-19 J

λ=

c 3 x 108 m/s = ; f 2.40 x 1016 Hz

λ = 12.5 nm

E = hf = 1.59 x 10-17 J

The radio wave has highest wavelength; The ultraviolet ray has highest energy. *33-43. An unknown light source A located 80 cm from a screen produces the same illumination as a standard 30-cd light source at point B located 30 cm from the screen. What is the luminous intensity of the unknown light source? Ix Is = ; rx2 rs2

Ix =

I s rx2 (30 cd)(80 cm) 2 = ; rs2 (30 cm) 2

Ix = 213 cd

*33-44. The illumination of a surface 3.40 m directly below a light source is 20 lx. Find the intensity of the light source? At what distance below the light source will the illumination be doubled? Is the luminous flux also doubled at this location? I = ER2 = (20 lx)(3.40 m)2 ; I = 231 cd E2 = 2E1 and E1R12 = E2R22 = (2E1)R22 R2 =

R12 (3.40 m) 2 = ; 2 2

R2 = 2.40 m ∆F =0

215

R2 R1



*33-45. The illumination of an isotropic source is EA at a point A on a table 30 cm directly below the source. At what horizontal distance from A on the table top will the illumination be reduced by one half? EA =

I ; RA2

EB =

I 2 I cos θ = ; 2 RA RB2

I cos θ ; RB2

E A = 2 EB ;

1 ; 2 cos θ

tan θ =

RB

θ

RA 30 cm

R 2A 1 = ; 2 R B 2 cosθ

2 So that: (cos θ ) =

I A = IB

but

RA = cos θ RB

1 (cos θ )3 = ; 2

x ; 30 cm

x

cos θ = 3 0.5 = 0.794; and θ = 37.50

x = (30 cm) tan 37.50 ;

x = 23.0 cm

*33-46. In Fizeau’s experiment to calculate the speed of light, the plane mirror was located at a distance of 8630 m. He used a wheel containing 720 teeth (and voids). Every time the rotational speed of the wheel was increased by 24.2 rev/s, the light came through to his eye. What value did he obtain for the speed of light? An increase of f = 24.2 Hz is needed for light to reappear from Edge 1 to Edge 2. Thus, the time for one revolution of entire wheel is: T=

1 1 = = 0.0413 s f 24.2 Hz

Now, the time for one tooth and one void between 1 and 2 is: t=

0.0413 s = 5.74 x10−5 s; 720 teeth c=

2(8630 m) ; 5.74 x 10-5s

Time for one round trip.

c = 3.01 x 108 m/s

216

Solucionario Capitulo 33 - Paul E. Tippens

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