Solucionario Capitulo 34 - Paul E. Tippens

Chapter 34. Reflection and Mirrors Physics, 6th Edition

Chapter 34. Reflection and Mirrors

Reflection from Plane Mirrors

34-1. A man 1.80 m tall stands 1.2 m from a large plane mirror. How tall is his image? How far

is he from his image? ( Image symmetric, reversed.)

Image distance q = object distance p:

q = p = 1.2 m; y’ = y = 1.8 m

33-2. What is the shortest mirror length required to enable a 1.68 m woman to see her entire

image? ( Virtual image is behind the mirror. Rays show that only half length is needed.)

Mirror height = (1/2) object height

5 ft, 8 in.2

h h = 2 ft, 10 in.

It doesn’t matter where she stands.

*33-3. A plane mirror moves at a speed of 30 km/h away from a stationary person. How fast

does this persons image appear to be moving in the opposite direction?

Each time mirror moves, image also moves, so that speed is doubled: vi = 60 km/h

*33-4. The optical lever is a sensitive measuring device that utilizes minute rotations of a plane

mirror to measure small deflections. The device is illustrated in Fig. 34-19. When the

mirror is in position 1, the light ray follows the path IVR1. If the mirror is rotated through

an angle to position 2, the ray will follow the path IVR2. Show that the reflected beam

turns through an angle 2, which is twice the angle through which the mirror itself turns.

473

1.8 m

1.2 m

1.68 m


Remember the basic principles of ray tracing and that the angle of incidence always

equals the angle of reflection. Refer to the figure drawn on the following page.

*34-4. (Cont.) Show that the deviated ray turns through twice the angle turned by mirror.

In horizontal initial position, i = r. Now from figure:

i 1 = i 1 + r 2 = r 1 +

i 2 = r 2 = ( i 1 + ) + ( r 1 + )

R1R2 = 2

Images Formed by Spherical Mirrors

4-5. A light bulb 3 cm high is placed 20 cm in front of a concave mirror with a radius of

curvature of 15 cm. Determine the nature, size, and location of the image formed. Sketch

the ray-tracing diagram. [ y = 3 cm, p = 20 cm ]

(20 cm)(7.5 cm) ;20 cm - 7.5 cm

pfqp f

q = 12 cm, real

' (12 cm)(3 cm); ' ;(20 cm)

y q qyM yy p p

y’ = -1.80 cm

q = 12.0 cm, y’ = - 1.80 cm; Real, inverted, and diminished.

34-6. A spherical concave mirror has a focal length of 20 cm. What are the nature size and

location of the image formed when a 6 cm tall object is located 15 cm from this mirror?

(15 cm)(20 cm) ;15 cm - 20 cm

pfqp f

474

2

1

I

R2

R1N1

2

N2

virtual image

C

F


q = -60 cm, virtual and enlarged

' ( 60 cm)(6 cm); ' ;(20 cm)

y q qyM yy p p

y’ = +18 cm

q = -60 cm, y’ = 18.0 cm; virtual, erect, and enlarged.

34-7. A 8-cm pencil is placed 10 cm from a diverging mirror of radius 30 cm. Determine the

nature, size, and location of the image formed. Sketch the ray-tracing diagram.

For diverging mirror: f = (R/2) = -15 cm

(10 cm)(-15 cm) ;10 cm - (-15 cm)

pfqp f

q = -6.00 cm, virtual

' ( 6 cm)(8 cm); ' ;(10 cm)

y q qyM yy p p

y’ = +16 cm

q = -6.00 cm, y’ = 4.80 cm; virtual, erect, and diminished.

34-8. A spherical convex mirror has a focal length 25 cm. What are the nature, size, and

location of the image formed of a 5-cm tall object located 30 cm from the mirror?

For diverging mirror: f = -25 cm

(30 cm)(-25 cm) ;30 cm - (-25 cm)

pfqp f

q = -13.6 cm, virtual

475

virtual image

CF

virtual image

CF


' ( 13.6 cm)(5 cm); ' ;(30 cm)

y q qyM yy p p

y’ = +2.27 cm

q = -13.6 cm, y’ = 2.27 cm; virtual, erect, and diminished.

34-9. An object 5 cm tall is place halfway between the focal point and the center of curvature

of a concave spherical mirror of radius 30 cm. Determine the location and magnification

the of the image? f = (R/2) = 15 cm. p = 22.5 cm

(22.5 cm)(15 cm) ;22.5 cm - 15 cm

pfqp f

q = 45 cm, real

' (45 cm); ;(22.5 cm)

y qM My p

M = -2.00

q = 45.0 cm, M = - 2.00; Real, inverted, and enlarged.

34-10. A 4-cm high source of light is placed in front of a spherical concave mirror whose radius

is 40 cm. Determine the nature size and location of the images formed for the following

object distances: (a) 60 cm, (b) 40 cm, (c) 30 cm, (d) 20 cm, and (e) 10 cm. Draw the

appropriate ray-tracing diagrams. (In the interest of space, diagrams are not drawn)

Givens: f = (40 cm/2) = 20 cm; y = 4 cm, p = 60, 40, 30, 20, and 10 cm.

;pfqp f

' ; 'y q qyM yy p p

476

C

F


(a)

(60 cm)(20 cm)60 cm - 20 cm

q = 30 cm, real

(30 cm)(4 cm)'(60 cm)

y

= -2 cm, inverted

(b)

(40 cm)(20 cm)40 cm - 20 cm

q = 40 cm, real

(40 cm)(4 cm)'(40 cm)

y

= -4 cm, inverted

(c)

(30 cm)(20 cm)30 cm - 20 cm

q = 60 cm, real

(60 cm)(4 cm)'(30 cm)

y

= -8 cm, inverted

(d)

(20 cm)(20 cm)20 cm - 20 cm

q = , no image

( )(4 cm)'(20 cm)

y

= , no image

(e)

(10 cm)(20 cm)10 cm - 20 cm

q = -20 cm, virtual

( 20 cm)(4 cm)'(10 cm)

y

= 8 cm, erect

34-11. At what distance from a concave spherical mirror of radius 30 cm must an object be

placed to form an enlarged, inverted image located 60 cm from the mirror?

Enlarged and inverted means object between F and C.

and the image is beyond the radius.

f = (R/2) = (30 cm/2); f = +15 cm

q = +60 cm , positive since image is real.

(60 cm)(15 cm) ;60 cm - 15 cm

qfpq f

p = 20 cm

477

C

F


Magnification

34-12. What is the magnification of an object if it is located 10 cm from a mirror and its image is

erect and seems to be located 40 cm behind the mirror? Is this mirror diverging or

converging? ( Erect image means diverging mirror.)

q = -40 cm , p = 10 cm

( 40 cm) ;10 cm

qMp

M = -4.00

34-13. A Christmas tree ornament has a silvered surface and a diameter of 3 in. What is the

magnification of an object placed 6 in. from the surface of this ornament? [ R= D/2 ]

1.5 in. ;2 2Rf

f = -0.75 in. (–) since converging

(6 in.)(-0.75 in.) ;6 in. - (-0.75 in.

q q = -0.667 in., virtual

( 0.667 in.) ;6.00 in.

qMp

M = +0.111

34-14. What type of mirror is required to form an image on a screen 2 m away from the mirror

when an object is placed 12 cm in front of the mirror? What is the magnification?

It must be a converging mirror, since image is real.

Also, from position and size, it must be enlarged and

478

R = 3 in.

virtual image

CF

virtual image

C

F

C

F


inverted. Thus, q = +2.00 m, p = 12 cm.

(2 m) ;0.12 m

qMp

M = -16.7

(0.12 m)(2 m)0.12 m + 2 m

pqfp q

; f = +11.3 cm Positive means it is converging.

*34-15. A concave shaving mirror has a focal length of 520 mm. How far away from it should an

object be placed for the image to be erect and twice its actual size?

2; 2 ; ;q pfM q p qp p f

2 ; 2( ) ;pfp p f fp f

520 mm2 2 ; ;2 2fp f f p

p = 260 mm

*34-16. If a magnification of +3 is desired, how far should the mirror of Problem 34-15 be

placed from the face?

3; 3 ; ;q pfM q p qp p f

3 ; 3( ) ;pfp p f fp f

479

virtual image

C

F


2 2(520 mm)3 3 ; ;3 3fp f f p

p = 357mm

*34-17. An object is placed 12 cm from the surface of a spherical mirror. If an erect image is

formed that is one-third the size of the object, what is the radius of the mirror. Is it

converging or diverging? (It is diverging since image is erect and diminished.)

1 (12 cm); ; 4 cm3 3 3

q pM q qp

(12 cm)(-4 cm) ;12 cm + (-4 cm)

pqfp q

f = -6.00 cm, diverging

*34-18. A concave spherical mirror has a radius of 30 cm and forms an inverted image on a wall

90 cm away. What is the magnification? [ f = (30 cm/2) = 15 cm; q = +90 cm ]

(90 cm)(15 cm) 90 cm18 cm; ;90 cm - 15 cm 18 cm

p M

M = -5.00

Challenge Problems

34-19. What are the nature, size, and location of the image formed when a 6-cm tall object is

located 15 cm from a spherical concave mirror of focal length 20 cm?

(15 cm)(20 cm)15 cm - 20 cm

q = -60 cm, virtual

480

virtual image

C

F


( 60 cm)(6 cm)'(15 cm)

y

= 24 cm, erect

q = -60.0 cm, y’ = 24.0 cm; virtual, erect, and enlarged.

34-20. An erect image has a magnification of +0.6. Is the mirror diverging or converging?

What is the object distance if the image distance is –12 cm?

The mirror is diverging since image is diminished, erect, and virtual.

( 12 cm)0.6; ;0.6 0.6

q qM pp

p = 20 cm

34-21. An object is located 50 cm from a converging mirror whose radius is 40 cm. What is the

image distance and the magnification?

(50 cm)(20 cm) ;50 cm - 20 cm

pfqp f

q = 33.3 cm, real

' (33.3 cm) ;(50 cm)

y qMy p

M = -0.667

q = 33.3 cm, M = - 0.667; Real, inverted, and diminished.

34-22. What is the focal length of a diverging mirror if the image of an object located 200 mm

from the mirror appears to be a distance of 120 mm behind the mirror?

481


(200 mm)( 120 mm) ;200 mm - (-120 mm)

pqfp q

f = -75.0 mm

34-23. A silver ball is 4.0 cm in diameter. Locate the image of a 6-cm object located 9 cm from

the surface of the ball. What is the magnification? [ R = 4 cm/2 = 2 cm. ]

f = (R/2)=(-2 cm/2) = -1 cm; p = 9 cm;

(9 cm)(-1 cm) ;9 cm - (-1 cm)

pfqp f

q = -0.9 cm

( 0.9 cm) ;9.00 cm

qMp

M = +0.100

34-24. An object 80 mm tall is placed 400 mm in front of a diverging mirror of radius –600

mm. Determine the nature size and location of the image

f = (R/2)=(-600 mm/2) = -300 mm

(400 mm)(-300 mm) ;400 mm - (-300 mm)

pfqp f

q = -171 cm

' ( 171 mm)(80 mm); y' ;400 mm

y qMy p

y’ = +34.3 mm

*34-25. An object 10 cm tall is located 20 cm from a spherical mirror. If an erect image 5-cm

tall is formed, what is the focal length of the mirror?

482

R = 4 cm

virtual image

CF

R = 3 in.

virtual image

CF


' 5 cm 0.5; ; (0.5)(20 cm); 10 cm10 cm

y qM M q Mp qy p

(20 cm)(-10 cm) ;20 cm + (-10 cm)

pqfp q

f = -20 cm

*34-26. What is the magnification if the image of an object is located 15 cm from a diverging

mirror of focal length –20 cm? [ q = -15 cm; f = -20 cm ]

( 15 cm)(-20 cm) 60 cm;15 cm ( 20 cm)

qfpq f

( 15 cm) ;60 cm

qMp

M = +0.250

*34-27. An object is placed 200 mm from the vertex of a convex spherical mirror whose radius

is 400 mm. What is the magnification of the mirror?

f = R/2 = -400 mm/2 = -200 mm; p = 200 mm

(200 mm)(-200 mm) ;200 mm - (-200 mm)

pqqp q

q = - 100 mm

Magnification:

( 100 mm) ;200 mm

qMp

M = +½

*34-28. A convex spherical mirror has a radius of –60 cm. How far away should an object be

held if the image is to be one-third the size of the object?

f = R/2 = -60m/2 = -30 mm; M = +1/3

483

p

R = 400 mm.

virtual image

C F

R = -60 cm

virtual image

CF


1 ; ; 3 3

q p pfM q qp p f

; ( 1)( ) 3 ; 3 0

3p pf p f f p f f

p f

;

p = -2f = -2(-30 cm); p = 60 cm

*34-29. What should be the radius of curvature of a convex spherical mirror to produce an image

one-fourth as large as the object which is located 40 in. from the mirror?

1 -1; ; ;4 4 4 40 in.

q p pf fM qp p f f

-40 in. + f = 4f

40 in.3

f

; f = -26.7 in.

*34-30. A convex mirror has a focal length of –500 mm. If an object is placed 400 mm from the

vertex, what is the magnification?

(400 mm)(-500 mm) ( 222 mm)222 mm; ;400 mm - (-500 mm) 400 mm

qq Mp

M = +0.556

*34-31. A spherical mirror forms a real image 18 cm from the surface. The image is twice as

large as the object. Find the location of the object and the focal length of the mirror.

Since the image is real and enlarged, we draw as shown:

Note: q = -18 cm, M = -2 (inverted)

484

C

F


( 18 cm)2; ;2 2

q qM pp

p = 9.00 cm

(9 cm)(-18 cm) ;9 cm +(-18 cm)

pqfp q

f = 6.00 cm

*34-32. A certain mirror placed 2 m from an object produces an erect image enlarged three

times. Is the mirror diverging or converging? What is the radius of the mirror?

An erect, enlarged image is consistent

only for converging mirrors.

3; 3 3(2 m);qM q pp

q = -6.00 m

(2 cm)(-6 cm) ;2 cm +(-6 cm)

pqfp q

f = +3.00 cm, converging

*34-33. The magnification of a mirror is –0.333. Where is the object located if its image is

formed on a card 540 mm from the mirror? What is the focal length?

540 mm0.333; ;0.333 0.333

q qM pp

p = 1.62 m

(1.62 m)(0.540 m) ;1.62 m + 0.540 m

pqfp q

f = +405 mm

485

virtual image

C

F


*34-34. What should be the radius of curvature of a concave mirror to produce an image one-

fourth as large as an object 50 cm away from the mirror?

The magnification must be negative if image is diminished by concave mirror:

1 (50 cm); ;4 4 4

q pM qp

q = 12.5 cm

(50 cm)(12.5 cm) ;50 cm + 12.5 cm

pqfp q

f = +10 cm

*34-35. A spherical shaving mirror as a magnification of +2.5 when an object is located 15 cm

from the surface. What is the focal length of the mirror?

2.5; 2.5(15 cm);qM qp

q = -37.5 cm

(15 cm)(-37.5 cm) ;15 cm +(-37.5 cm)

pqfp q

f = +25.0 cm, converging

Critical Thinking Problems

34-36. A baseball player is 6 ft tall and stands 30 ft in front of a plane mirror. The distance from

the top of his cap to his eyes is 8 in. Draw a diagram showing location of the images

formed of his feet and of the top of his cap? What is the minimum length of mirror

486

virtual image

C

F


required for him to see his entire image. If he walks 10 ft closer to the mirror what is the

new separation of object and image?

From the drawing of reflected rays, you see that:

y = ½(8 in.) + ½(72 in. – 8 in.) ; ymin = 36 in.

Since the image moves closer by the same amount that the ballplayer does a new object

distance of 20 ft means a separation of 2(20 ft) = 40 ft

*34-37. The diameter of the moon is 3480 km and it is 3.84 x 108 m away from the earth. A

telescope on the earth utilizes a spherical mirror, whose radius is 8.00 m to form an image

of the moon. What is the diameter of the image formed. What is the magnification of the

mirror? ( y = 3.48 x 10-6 m, p = 3.84 x 108 m. )

6

8

(3.84 x 10 m)(4 m) 4.00 m3.84 x 10 m - 4 m

pfqp f

(at F)

8

' 4 m; ;3.84 x 10 m

y qM My p

M = -1.04 x 10-8

-8 6 ' ( 1.04 x 10 )(3.48 x 10 m);y My y = 3.62 cm

*34-38. An image 60 mm long is formed on a wall located 2.3 m away from a source of light 20

mm high. What is the focal length of this mirror? Is it diverging or converging? What is

the magnification? ( The image in inverted so that M is negative, image is inverted.)

q = p + 2.3 m;

( 2.3 m) 3p

q pMp

487

Moon

p

2.3 m

C

F


p + 2.3 m = 3p; p = 1.15 m; q = p + 2.3 m = 3.45 m

(1.15 m)(3.45 m) ;1.15 m + 3.45 m

pqfp q

f = +0.862 m

The positive focal length indicates that the mirror is converging.

(3.45 m) ;1.15 m

qMp

M = -3.00

*34-39. Derive an expression for calculating the focal length of a mirror in terms of the object

distance p and the magnification M. Apply it to Problem 34-35. Derive a similar relation

for calculating the image distance q in terms of M and p. Apply it to Problem 34-33.

; ; ; q pf pfM q Mp q Mpp p f p f

-M(p – f) = f; -Mp + Mf = f; Mf – f = Mp ( 1)Mpf

M

M = 2.5, p = 15 cm;

(2.5)(15 cm) ;( 1) (2.5 1)

MpfM

f = +25 cm

488


; ; ; q q qf q qfM p pp M q f M q f

-(q – f) =M f; -q + f = Mf Mf – f = -q ( 1)qf

M

M = -0.333, q = 540 mm;

540 mm( 1) 0.333 1

qfM

f = +405 mm

*34-40. A concave mirror of radius 800 mm is placed 600 mm from a plane mirror that faces it.

A source of light placed midway between the mirrors is shielded so that the light is first

reflected from the concave surface. What are the position and magnification of the image

489

p

q

600 mmvirtual image

C

F


formed after reflection from the plane mirror? (Hint: treat the image formed by the first

mirror as the object for the second mirror.)

0.600 m 0.300 m; 0.400 m2 2

Rp f

First find q formed by spherical mirror.

(0.300 m)(0.400 m) ; 1.20 m(0.300 m - 0.400 m)

pfq qp f

; (behind mirror)

Since plane mirror is 0.600 m in left of spherical mirror, the final image q’= p’ is:

q’ = 1.20 m + 0.600 m = 1.8 m; q’ = 1.8 m to left or behind the plane mirror.

1 2 1 2( 1.2 m) 4; 1; x ( 4)( 1);0.300 m

qM M M Mp

M = +4

490

Solucionario Capitulo 34 - Paul E. Tippens

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