Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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;
1.2 m
1.8 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
h
5 ft, 8 in.
2
1.68 m
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.
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.
473
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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
 i 1 =  i 1 + r 2 =  r 1 + 
N1
N2
R1

R2
2
 i 2 =  r 2 = ( i 1 + ) + ( r 1 + )
2
R1R2 = 2 
1

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 ]
q
pf
(20 cm)(7.5 cm)

;
p f
20 cm - 7.5 cm
M
q = 12 cm, real
y ' q
qy (12 cm)(3 cm)

; y' 

;
y
p
p
(20 cm)
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?
q
pf
(15 cm)(20 cm)

;
p f
15 cm - 20 cm
C
F
q = -60 cm, virtual and enlarged
M
y ' q
qy (60 cm)(6 cm)

; y' 

;
y
p
p
(20 cm)
y’ = +18 cm
q = -60 cm, y’ = 18.0 cm; virtual, erect, and enlarged.
474
virtual image
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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.
virtual image
For diverging mirror: f = (R/2) = -15 cm
q
pf
(10 cm)(-15 cm)

; q = -6.00 cm, virtual
p  f 10 cm - (-15 cm)
M
y ' q
qy (6 cm)(8 cm)

; y' 

;
y
p
p
(10 cm)
C
F
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?
virtual image
For diverging mirror: f = -25 cm
q
pf
(30 cm)(-25 cm)

; q = -13.6 cm, virtual
p  f 30 cm - (-25 cm)
M
y ' q
qy (13.6 cm)(5 cm)

; y' 

;
y
p
p
(30 cm)
C
F
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
q
M
C
pf
(22.5 cm)(15 cm)

;
p f
22.5 cm - 15 cm
q = 45 cm, real
y ' q
(45 cm)

; M
;
y
p
(22.5 cm)
M = -2.00
F
q = 45.0 cm, M = - 2.00; Real, inverted, and enlarged.
475
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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.
q
pf
;
p f
M
y ' q

;
y
p
y' 
qy
p
(a) q 
(60 cm)(20 cm)
= 30 cm, real
60 cm - 20 cm
y' 
(30 cm)(4 cm)
= -2 cm, inverted
(60 cm)
(b) q 
(40 cm)(20 cm)
= 40 cm, real
40 cm - 20 cm
y' 
(40 cm)(4 cm)
= -4 cm, inverted
(40 cm)
(c) q 
(30 cm)(20 cm)
= 60 cm, real
30 cm - 20 cm
y' 
(60 cm)(4 cm)
= -8 cm, inverted
(30 cm)
(d) q 
(20 cm)(20 cm)
= , no image
20 cm - 20 cm
y' 
()(4 cm)
= , no image
(20 cm)
(e) q 
(10 cm)(20 cm)
= -20 cm, virtual
10 cm - 20 cm
y' 
(20 cm)(4 cm)
= 8 cm, erect
(10 cm)
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.
C
and the image is beyond the radius.
F
f = (R/2) = (30 cm/2);
f = +15 cm
q = +60 cm , positive since image is real.
p
qf
(60 cm)(15 cm)

;
q f
60 cm - 15 cm
476
p = 20 cm
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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.)
C
q = -40 cm , p = 10 cm
q (40 cm)
M

;
p
10 cm
virtual image
F
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 ]
f 
R 1.5 in.

;
2
2
q
virtual image
f = -0.75 in. (–) since converging
(6 in.)(-0.75 in.)
; q = -0.667 in., virtual
6 in. - (-0.75 in.
M
q (0.667 in.)

;
p
6.00 in.
C
F
R = 3 in.
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
inverted. Thus, q = +2.00 m, p = 12 cm.
M
f 
q (2 m)

;
p 0.12 m
pq
(0.12 m)(2 m)

;
p  q 0.12 m + 2 m
C
F
M = -16.7
f = +11.3 cm
477
Positive means it is
converging.
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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?
M
q
 2;
p
q  2 p;
pf
2 p 
;
p f
q
pf
;
p f
C
 2( p  f )  f ;
2 p  2 f  f ;
virtual image
F
p
f 520 mm

;
2
2
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?
M
q
 3;
p
q  3 p;
3 p  3 f  f ;
p
q
pf
pf
; 3 p 
;
p f
p f
2 f 2(520 mm)

;
3
3
 3( p  f )  f ;
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.)
M
f 
q
1
 p (12 cm)
 ; q

;
p
3
3
3
pq
(12 cm)(-4 cm)

;
p  q 12 cm + (-4 cm)
q  4 cm
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;
p
(90 cm)(15 cm)
 18 cm;
90 cm - 15 cm
M
478
90 cm
;
18 cm
q = +90 cm ]
M = -5.00
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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?
q
(15 cm)(20 cm)
= -60 cm, virtual
15 cm - 20 cm
y' 
C
F
(60 cm)(6 cm)
= 24 cm, erect
(15 cm)
virtual image
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.
M
q
q (12 cm)
 0.6; p 

;
p
0.6
0.6
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?
q
pf
(50 cm)(20 cm)

;
p f
50 cm - 20 cm
M
y ' q (33.3 cm)


;
y
p
(50 cm)
q = 33.3 cm, real
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?
f 
pq
(200 mm)(120 mm)

;
p  q 200 mm - (-120 mm)
479
f = -75.0 mm
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
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;
q
pf
(9 cm)(-1 cm)

;
p  f 9 cm - (-1 cm)
M
virtual image
q = -0.9 cm
q (0.9 cm)

;
p
9.00 cm
C
F
R = 4 cm
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
virtual image
f = (R/2)=(-600 mm/2) = -300 mm
q
pf
(400 mm)(-300 mm)

;
p  f 400 mm - (-300 mm)
M
C
F
R = 3 in.
q = -171 cm
y ' q
(171 mm)(80 mm)

; y' 
;
y
p
400 mm
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?
M
y ' 5 cm

 0.5;
y 10 cm
f 
M
q
;
p
q   Mp  (0.5)(20 cm); q  10 cm
pq
(20 cm)(-10 cm)

;
p  q 20 cm + (-10 cm)
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 ]
p
qf
(15 cm)(-20 cm)

 60 cm;
q  f 15 cm  (20 cm)
480
M
q (15 cm)

; M = +0.250
p
60 cm
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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?
virtual image
f = R/2 = -400 mm/2 = -200 mm;
q
p = 200 mm
pq
(200 mm)(-200 mm)

;
p  q 200 mm - (-200 mm)
M
Magnification:
p
C
F
R = 400 mm.
q = - 100 mm
q (100 mm)

;
p
200 mm
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?
virtual image
f = R/2 = -60m/2 = -30 mm;
M
q 1
 ;
p 3
p
pf

;
3
p f
q
p
;
3
M = +1/3
q
(  1)( p  f )  3 f ;
pf
p f
C
F
R = -60 cm
 p  f 3f  0;
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?
M
q 1
 ;
p 4
q
-40 in. + f = 4f
p
pf

;
4
p f
f 
40 in.
;
3
-1
f

;
4 40 in.  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?
q
(400 mm)(-500 mm)
q (222 mm)
 222 mm; M 

;
400 mm - (-500 mm)
p
400 mm
481
M = +0.556
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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)
M
f 
q
 2;
p
p
q (18 cm)

;
2
2
pq
(9 cm)(-18 cm)

;
p  q 9 cm +(-18 cm)
C
F
p = 9.00 cm
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
C
only for converging mirrors.
M
q
 3;
p
f 
virtual image
F
q  3 p  3(2 m); q = -6.00 m
pq
(2 cm)(-6 cm)

;
p  q 2 cm +(-6 cm)
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?
M
q
 0.333;
p
f 
p
q
540 mm

;
0.333
0.333
pq
(1.62 m)(0.540 m)

;
p  q 1.62 m + 0.540 m
482
p = 1.62 m
f = +405 mm
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*34-34. What should be the radius of curvature of a concave mirror to produce an image onefourth as large as an object 50 cm away from the mirror?
The magnification must be negative if image is diminished by concave mirror:
M
q
1
 ;
p
4
f 
q
p (50 cm)

;
4
4
pq
(50 cm)(12.5 cm)

;
p  q 50 cm + 12.5 cm
q = 12.5 cm
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?
M
q
 2.5;
p
q  2.5(15 cm);
C
F
virtual image
q = -37.5 cm
f 
pq
(15 cm)(-37.5 cm)

;
p  q 15 cm +(-37.5 cm)
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
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
483
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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. )
Moon
pf
(3.84 x 106 m)(4 m)
q

 4.00 m (at F)
p f
3.84 x 108m - 4 m
M
y ' q
4 m

; M
;
y
p
3.84 x 108 m
M = -1.04 x 10-8
y '  My  (1.04 x 10-8 )(3.48 x 106 m);
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;
p + 2.3 m = 3p;
f 
M
q ( p  2.3 m)

 3
p
p
p
C
p = 1.15 m; q = p + 2.3 m = 3.45 m
pq
(1.15 m)(3.45 m)

;
p  q 1.15 m + 3.45 m
2.3 m
f = +0.862 m
The positive focal length indicates that the mirror is converging.
M
F
q (3.45 m)

;
p
1.15 m
484
M = -3.00
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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.
M
q
;
p
-M(p – f) = f;
q   Mp;
-Mp + Mf = f;
M = 2.5, p = 15 cm; f 
M
q
q
;
p
-(q – f) =M f;
p
 Mp 
Mf – f = Mp
pf
p f
f 
Mp
( M  1)
Mp
(2.5)(15 cm)

; f = +25 cm
( M  1)
(2.5  1)
q
;
M
p
f 
qf
;
q f
q
qf

M q f
Mf – f = -q
-q + f = Mf
M = -0.333, q = 540 mm;
pf
;
p f
q
540 mm

( M  1) 0.333  1
485
f 
q
( M  1)
f = +405 mm
Physics, 6th Edition
Chapter 34. Reflection and Mirrors
*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
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
p
 0.300 m;
2
p
R
f   0.400 m
2
C
q
F
600 mm
virtual image
First find q formed by spherical mirror.
q
pf
(0.300 m)(0.400 m)

;
p  f (0.300 m - 0.400 m)
q  1.20 m ;
(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;
M1 
q (1.2 m)

 4;
p
0.300 m
q’ = 1.8 m to left or behind the plane mirror.
M 2  1; M1 x M 2  (4)(1);
486
M = +4