Tài liệu Physics exercises_solution: Chapter 34 pptx

f If the object is at infinity, then the image is at the focal point.. g The image is larger than the object magnification greater than one for... So for an object on the left side of th

34.1: If up is the  -direction and right is the x y  -direction, then the object is at

),,(atis),

0.350

m0.350m

28.0m040.0mirror

tree mirror tree

mirror

tree mirror

d

d h h d

d h

h

34.3: A plane mirror does not change the height of the object in the image, nor does the

distance from the mirror change So, the image is 39.2cmto the right of the mirror, and its height is 4.85cm

2

cm0.34

1cm

0.22

21

11

mirror

cm,20.1cm16.5

cm33.0cm)600.0

34.6: a)

cm5.16

1cm

0.22

21

111

mirror

cm,240.0cm16.5

cm).606(cm)600.0

virtual

m1058.5

1m

75.1

11

11

s

m

1013.2)m106794)(

1014.3(

1014.31058.5

75.1

4 3

11

11 10

1cm

00.3

21

111,cm00

ball) The magnification is 0.0667

cm0.21

cm40

34.9: a) 1 1 1 1 1 1 Also

s f

f s

s m f

s

sf s fs

f s s f s f s

s which means the image is always smaller and inverted since the magnification is negative

For  2 0     1

f

f m f f s f

s f

d) Concave mirror: 0s f s0,and we have a virtual image to the right

of the mirror  1,

f

f

m so the image is upright and larger than the object

f s f

s

sf s

f s f

f s

f

f

,since

m   so the image is smaller

f) If the object is at infinity, then the image is at the focal point.

g) If the object is next to the mirror, then the image is also at the mirror

h)

i) The image is erect if s

j) The image is inverted if s

k) The image is larger if 0s2f

l) The image is smaller if s2f ors0

m) As the object is moved closer and closer to the focal point, the magnification INCREASES to infinite values

e) The image is erect (magnification greater than zero) for s  f

f) The image is inverted (magnification less than zero) for s f

g) The image is larger than the object (magnification greater than one) for

34.13: a)

cm0.12

1cm

0.20

21

11

mirror

cm,409.0cm12.0

cm5.45cm)9.0

0.12

0

1cm

0.32

21

111

309.1

n

s

n a b

34.16: a) 1.00 0 5.26cm,

cm00.7

33.1

n s

33.1

n s

b a b

n

n n

R

h s

h s

R

n n s

n s





Also, the magnification calculation yields:

.tan

andtan

s n

s n y

y m s

y n s

y n s

y s

y

b

a b

a b

and,,

n s

n R

h s

h s

h n

n s

tan

s n

s n y

y m s

y n s

y n n

n

a

a b

a b b a a

60.060.1

n n s

n s

cm00.3

60.060.1cm0.12

n n s

n s

cm00.3

60.060.1cm00.2

n n s

n s

34.19:

)(

)()

()

R s s

s n R s

R s R s

sR n

R

n n s

n

s

n

b a

a b b a

cm)3.00cm601(cm160

cm.0

60.060.1cm0.24

n n s

n

s

mm,0.578mm

50.1)cm0.24)(

60.1(

cm8.14

s n y

60.060

.1cm0.24

n n s

n

s

mm,0.326mm

50.1)cm0.24)(

60.1(

cm)35.8

s n y

33.000

.1cm0.14

33

n n s

n s

so the fish appears to be at the center of the bowl

.33.1cm)

0.17)(

00.1(

cm)0.17)(

33.1

s n m

b a

cm0.14

33.033.100.1

n n s

n s

which is outside the bowl

34.23: For s18cm:

cm0.18

1cm

0.14

11

111

0.18

0.63

c) and d) From the magnification, we see that the image is real and inverted

For s7 0cm:

cm00.7

1cm

0.14

11

11

00.7

0

c) and d) From the magnification, we see that the image is virtual and erect

34.24: a) 48.0cm,

cm0.12

1cm

0.16

11

11

diverging

16.0

12.0)(cm)850.0

c)

400.0

30.1

s

s y

.31cm00.7

1cm00.7

11

11

f s

s

(to the left)

cm,85.4cm

75.15

1cm

0.7

50.4

s s

s

s y

y

cm2170

.90

11711.0

11

s

cm,154cm)217(711

1cm

00.5

1)48.0(

1cm0.18

11

1)1(1

s

cm3.10

11

44

4

cm3.13

;cm3.13

;cm3.13

;cm44.4

;cm44.4

;cm

3

13

8 7

6 5

4 3

2 1

f f

f f

f f

34.30: We have a converging lens if the focal length is positive, which requires:

.011011)

1

(

1

2 1 2

R

n

(i) {R1 R2}{R1,R2  }(ii)R1 0,R2 0

(iii) {|R1||R2 |}{R1,R2 0} Hence the three lenses in Fig (35.29a)

We have a diverging lens if the focal length is negative, which requires:

.011011)1(

1

2 1 2

0)iii(0)

ii(},

{}{

34.31: a) The lens equation is the same for both thin lenses and spherical mirrors, so the

derivation of the equations in Ex (34.9) is identical and one gets:

.also

and,1

111

1

1

s f

f s

s m f

s

sf s fs

f s s f s f

interpretation of the results For a lens, the outgoing side is not that on which the object

lies, unlike for a mirror So for an object on the left side of the lens, a positive image distance means that the image is on the right of the lens, and a negative image distance means that the image is on the left side of the lens

c) Again, for Ex (34.10) and (34.12), the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above

cm0.17

1cm

0.12

11

11

s

,cm34.04

.2

cm800.04

.22

.7

)0.17

1cm

0.48

11

111

cm24.1646.0

cm800.0646

.03.26

0

s

34.34: a) 11.1cm,

cm0.36

1cm

0.16

11

11



f f s

16

36cm)80.0

c)

m240

m024.0m

s y

y m

.mm60mm

60

1mm100.6

11

11

85-mm lens

m9.6

m0.036m

0

mm

149mm

150

1mm

1040

11

11

135-mm lens

m085.0

11

m90.3

11

s

s

mm,39.0mm175090.3

0869.0

1cm

4.20

11

11



s f s

s

m109.50

m00.5

34.39: a) 1.4 10

mm200,000

s m

mm200,000

s m

mm200,000

s m

11

cm8

11

11

2 2

s

cm

6cm

12

11

cm4

11

1

1

2 2

D

f f

34.42: The square of the aperture diameter (~ the area) is proportional to the length of the

250

1mm

23.1

mm830

sf s sf

f s s f s s

For s45cm450mm,s56mm

For s,s f 50mm

The range of distances between the lens and film is 50mmto56mm

34.44: a) 0.153m 15.3cm.

m150.0

1m

00.9

1111



s f s s

b)    58.8

153.0

00.9

1power

4.36

11

cm25

11

s

m30.1

1power

11

111

s

cm60.2

40.1cm0.40

n n s

n s

m600.0

1m

25.0

11

s

m600.0

11

111

s

34.48: a) Angular magnification 4.17

cm00.6

cm0.25cm0



f M

cm00.6

1cm

0.25

11

11

cm00.8

1m

0.25

11

11

b) 4.13 (1.00mm)(4.13) 4.13mm

cm06.6

cm0

34.50: 80.0mm 8.00cm.

rad025.0

mm00

150

.6

111150.650

s s s s

s

s m

.cm0.2250

.6,

cm38.300

.4

150

.6

11

00.5(

mm)0.5mm160(mm)250(mm)250(

2 1



f f

s M

317

mm10

y m

34.53: a) The image from the objective is at the focal point of the eyepiece, so

cm9.17cm80.1cm7.192

.cm837.0cm

800.0

1cm

9.17

11

11

cm837.0

cm9.17

s

s m

cm80.1

cm0.25)4.21(cm0.252



f m M

34.54: Using the approximation s1  f, and then ,

1

1 1

mm122

mm9.1

;mm122mm9.1mm120:

mm124mm

4

;mm124mm4mm120:

mm

4

.5.8mm16

mm136

mm16

;mm136mm16mm120:

mm

16

1

1 1

f

s

s m s

f

The eyepiece magnifies by either 5 or 10, so:

a) The maximum magnification occurs for the 1.9-mm objective and 10x

eyepiece:

.640)10)(

64(

5.8(

cm0.952

m950.0

11

m3000

11

c) M 6.33arctan(60.0 3000)6.33(60.0) (3000)0.127rad

34.56: f1 f2 d s  f1 d s  f2 1.80m0.0900m1.71m

.0.1900.9

1712

f y s

s y

y



f

f M

s

1m12.0m75.0

1m

3.1m75.0

11

34.62: a) There are three images formed.

34.63: The minimum length mirror for a woman to see her full height h, is h , as 2shown in the figure below

34.64: 2.25   4.00m 1.25s4.00ms3.2m.

s

s s

s

7.20 m from the wall Also:

.m43.4m

20.7

1m2.3

1221



R R s s

0.60

m00.80

.6000.6

1m133.0

12

21



R R s s

m180.0

21

m0.13

12

s

s

0.13

0894.0m

50

y

b) The height of the image is less then 1% of the true height of the car, and is less than the image would appear in a plane mirror at the same location This gives the

illusion that the car is further away then “expected.”

34.67: a) R0and s0,soa realimage(s0) is produced for virtual object

positions between the focal point and vertex of the mirror So for a 24.0 cm radius mirror, the virtual object positions must be between the vertex and 12.0 cm to the right of the mirror b) The image orientation is erect, since 0

s m

c)

34.68: The derivations of Eqs (34.6) and (34.7) are identical for convex mirrors, as long

as one recalls that R and s are negative Consider the diagram below:

2

211

s

s y

y m f

R s s

R f s R s

21

cm0.8

12

11

s

0

8.5()

m s

s s

5,0,0since5

22

5

.4

3and

10

3

R s

R



34.71: a)

f s s

11

s

s ds

s d ds

s d s s f

s ds

d

But.1

11

1s

1

2

2 2

2

2

m

 Images are always inverted longitudinally

cm000.150

21

cm000.200

12

s s

cm000.150

21

cm100.200

12

s s

(ii) Front face:    0.600000,  2 0.6000002 

000.200

000.120

m m s

s m

964.119

m m s

s m

2 2

2 2

2

2 2

a

b a

b b

a b

a

b a a

b b a

n

n m n

n n

n s

s n

n s

s ds

s d

ds

s d s

n s

n R

n n s

n s

n ds d

s

d    

s d dt

s d v R s

sR s

R s

211

2

sR R

m50.22

3 2

2 2

R v v

b)

m25.1s

m50.2

2 2

R v v

Note: The signs are somewhat confusing If a real object is moving with v > 0, this implies it is moving away from the mirror However, if a virtual image is moving with v > 0, this implies it is moving from “behind” the mirror toward the vertex.

34.74: In this context, the microscope just takes an image and makes it visible The real

optics are at the glass surfaces

.31.1mm

50.2

mm780.0mm50.20

s

n s

n

s

n a b

Note that the object and image are measured from the front surface of the second plate, making the image virtual

34.75: a) Reflection from the front face of the glass means that the image is just h

below the glass surface, like a normal mirror

b) The reflection from the mirrored surface behind the glass will not be affected

because of the intervening glass The light travels through a distance 2d of glass, so the

path through the glass appears to be 2 ,

c) The distance between the two images is just 2

n d

34.76: a) The image from the left end acts as the object for the right end of the rod.

cm0.6

60.060.1cm0.23

n n s

n s

So the second object distance is s2 40.0cm28.3cm11.7cm

Also: 1.6023.0 0.769.

3.28

s n

s n m

b a

c) The object is real and inverted

cm0.12

60.01

cm7.11

60.1

2 2

n n s

n s

Also:    1.57  0.7691.57 1.21.

7.11

5.1160.1

2 1

s n

s n m

b a

e) So the final image is virtual, and inverted

f) y1.50mm1.211.82mm

cm0.6

60.060.1cm0.23

n n s

n s

So the second object distance is s2 25.0cm28.3cm3.3cm

Also: 1.6023.0 0.769.

3.28

s n

s n m

b a

b) The object is virtual

cm0.12

60.01

cm3.3

60.1

2 2

2 2

n n s

n s

Also:

3.3

87.160.1

2 1 2

s n m

b a

d) So the final image is real and inverted

e) y  ym 1.50mm0.6931.04mm

34.78: For the water-benzene interface to get the apparent water depth:

.cm33.70

50.1cm50.6

33.1

1)cm60.2cm33.7(

50.1

n

s

n a b

34.79: 1 2.00.

2

12

12

,but

n R

s s R

n n s

60.060.1cm0.12

n n s

n s

So the object distance for the far end of the rod is 50.0cm(36.9cm)86.9cm

cm

3.540

1cm9.86

60

n n s

n s

b) The magnification is the product of the two magnifications:

.92.100

.1,

92.1)0.12)(

60.1(

9.36

2 1 2

s n

s n m

b a

cm00.4

80.080.1

n n s

00.4

80.01

cm00.1

80

n n s

n s

which is 4.50 cm from the center of the sphere

50.9

0.150

cm50.9

1cm

0.15

0.10

58.01

cm0.15

58.1R

s

n R

n n

34.83: a) From the diagram:

)50.1(

190.0sin

But.sin50.1190.0sin

R R

r R

cm190.0





 r

So the diameter of the light hitting the surface is 2r 0.254cm

b) There is no dependence on the radius of the glass sphere in the calculation above

34.84: a) Treating each of the goblet surfaces as spherical surfaces, we have to pass,

from left to right, through four interfaces For the empty goblet:

cm12cm

4.00

0.501.50

1

1 1

n n s

n

s

cm6.64cm

40.3

50.01

cm4.11

50.1cm

4.11cm12cm60

31.9cm

40.3

50.050

.1cm4.71

1cm

4.71cm80.6cm6

cm.9.37cm

00.4

50.01

cm91.9

50.1cm91.9cm60.0cm31

So the image is 37.9cm2(4.0cm)29.9cmto the left of the goblet

b) For the wine-filled goblet:

cm12cm

00.4

50.050.11

1 1

n n s

n

s

cm7.14cm

40.3

13.037

.1cm4.11

50.1cm

4.11cm12cm60

1.11cm

40.3

13.050

.1cm9.7

37.1cm

9.7cm7.14cm80

cm,73.3cm

00.4

50.01

cm5.10

50.1cm

5.10cm1.11cm60

to the right of the goblet

3

13

43

n n s

n s

.33

1133.1

R s R s R

n n s

n s

So the final image is a distance 3R from the right-hand side of the sphere, or 4R to

the right of the center of the globe

34.86: a) and

R

n n f

n n R

n n n f

n R

n n s

n s

f

f n n f

n f

n R

n n f

n R

n n f

n

b a b a a b b a

b a

f s

f R

f f n s

n f s

f n R

n n s

n s

canweso

f n n

n R

b a

b

a

34.87: Below, x is the distance from object to the screen’s original position.

f x

f x

f s

s

1cm22

1cm26

1and

1cm30

1cm30

11

1

1

1 1

cm450

left of the lens, so s46.3cm30cm16.3cm The corresponding focal length is 10.56 cm

34.88: We have images formed from both ends From the first:

cm

0.30cm

00.6

55.055.1cm0.25

n n s

55.0cm

0.65

1cm

0.30

55.1

n n s

n s

cm

3.50cm

3.20cm0



34.89: a) For the first lens: 30.0cm.

cm0.12

11

cm0.20

11

s s

0

20

0

0.28cm,

0.28cm

0.12

11

cm0.21

11

s

s

So the image is 28.0 cm to the left of the second lens, and is therefore 19.0 cm to the left of the first lens

b) The final image is virtual

c) Since the magnification is mm1m2 (1.50)(1.33)2.00, the final image

is erect and has a height y(2.00)(2.50mm)5.00mm

cm0.28

1cm

0.12

1)60.0(11)1(1

2 1

n f

cm,158cm

35

11cm45

11

s

cm

76.145

158cm)

50.0

11cm157

11

s s

cm,5.0157

45cm)1.76(cm,0

and the image is erect

c) Putting an identical lens just 45 cm from the first means that the first lens’s image becomes an object for the second, a distance of 113 cm to the right of the secondlens

113

7.26)cm76.1(andcm,7.26cm

35

11cm113

11

s

s

and the image is inverted

34.91: 80.0cm.

cm0.40

11

cm0.80

11

0.40

11

cm0.28

11

s s

So the object distance for the third lens in 52.0 cm – (16.47 cm) = 35.53 cm

cm,318cm

0.40

11

cm53.35

11

s

virtual and 318 cm to the left of the third mirror, or equivalently 214 cm to the left of the first mirror

34.92: a)

cm00.3

11

cm0.18

11

11andcm0

s s s

s

s.

s s

( )2 (18.0cm) 54.0cm2 0 14.2cm,3.80cm So the screen must either be 3.80 cm or 14.2 cm from the object

2.14

80.3:

cm80



s

s m s

.74.380.3

2.14:

cm2



s

s m s

34.93: a) Bouncing first off the convex mirror, then the concave mirror:

.33.4m5.56

m600.0

m600.0

1m

56.5

1m360.0

21

m600.0

12

x s

s x R

s

s

But the object distance for the concave mirror is just

.33.4m5.56

m20.34.33m

s

So for the concave mirror:

360.0

21m20.333.4

33.4m56.521

x R

s s

m

24.0,m72.00

20.38.175

1m56.5

1360.0

2112

x s

s x R

s

s

But the object distance for the convex mirror is just

.1m56.5

m600.02.33m

600

s

So for the convex mirror:

360.0

21

m600.033.2

1m56.521

x R

s s

0.24m

m,13.00

600.000.25

34.94: Light passing straight through the lens:

a)

cm0.32

11

cm0.85

11

s

c) The image is real

d) The image is inverted

For light reflecting off the mirror, and then passing through the lens:

a)

cm,0.20cm

0.10

11

cm0.20

11

s

mirror, which becomes the new object for the lens, is at the same location as the object

So the final image position is 51.3 cm to the right of the lens, as in the first case above

c) The image is real

d) The image is erect

34.95: Parallel light coming in from the left is focused 12.0 cm from the left lens, which

is 8.00 cm to right of the second lens Therefore:

cm,80.4cm

0.12

11

cm00.8

11

s

and this is where the first focal point of the eyepiece is located The second focal point is obtained by sending in parallel light from the right, and the symmetry of the lens set-up enables us to immediately state that the second focal point is 4.80 cm to the left of the first lens

34.96: a) With two lenses of different focal length in contact, the image distance from

the first lens becomes exactly minus the object distance for the second lens So we have:

.111111

11and11111

1

2 2 1 1 2 1

2 2 1

1 1 1 1

1

f s f s s s

s s s

f s f s



b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact All we need in order to calculate the system’s focal length is calculate the individual focal lengths, and then use the formula from part (a)

11)(

1

R R n n

.061.0cm00.9

1cm

1)46.0(11)(

1:CCl

2 1

cm

93.8112

.011

1

1 2

f

34.97: At the first surface, ( 14.4cm) 23.04cm

00.1

60.1

n s

n

a

b b

a

At the second surface,

cm

80.048

.060.0

04.2360.152.23)cm0.23(60.1

00.1cm

7.14

t t

t s

n

n t s

a b

(Note, as many significant figures as possible should be kept during the calculation, since numbers comparable in size are subtracted.)