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310 Chapter 23
12. (a) No. The screen is needed to reflect the light toward your eye.
(b) Yes. The light is traveling toward your eye and diverging away from the position of the image, the same as if the object was located at that position.
14. (d). The entire image would appear because any portion of the lens can form the image. The image would be dimmer because the card reduces the light intensity on the screen by 50%.
PROBLEM SOLUTIONS
23.1 If you stand 40 cm in front of the mirror, the time required for light scattered from your face to travel to the mirror and back to your eye is
Δ t d c
( ) ×
= × −
8 2 7^109
m m s s
Thus, the image you observe shows you ~10 −^9 s younger than your current age.
23.2 (a) With the palm located 1.0 m in front of the nearest mirror, that mirror forms an image, I (^) P 1 , of the palm located 1.0 m behind the nearest mirror.
(b) The farthest mirror forms an image, I (^) B 1 , of the back of the hand located 2.0 m behind this mirror and 5.0 m in front of the nearest mirror. This image serves and an object for the nearest mirror, which then forms an image, I (^) B 2 , of the back of the hand located 5.0 m behind the nearest mirror.
(c) The image I (^) P 1 (see part a) serves as an object located 4.0 m in front of the farthest mirror, which forms an image I (^) P 2 of the palm, located 4.0 m behind that mirror and 7.0 m in front of the nearest mirror. This image then serves as an object for the nearest mirror, which forms an image I (^) P 3 of the palm , located 7.0 m behind the nearest mirror.
(d) Since all images are located behind the mirror, all are virtual images.
23.3 (1) The first image in the left-hand mirror is 5.00 ft behind the mirror, or 10.0 ft from the person.
(2) The first image in the right-hand mirror serves as an object for the left-hand mirror. It is located 10.0 ft behind the right-hand mirror, which is 25.0 ft from the left-hand mirror. Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror, or 30.0 ft from the person.
(3) The fi rst image in the left-hand mirror serves as an object for the right-hand mirror. It is located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that mirror. This image then serves as an object for the left-hand mirror. The distance from this object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is 35.0 ft behind the mirror, or 40.0 ft from the person.
Mirrors and Lenses 311
23.4 The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror.
The image of the choir is
0 800. m + 5 30. m =6 10. m
from the organist. Using similar triangles gives ′ = h 0 600
m.
m m
or h ′ = ( 0 600 )⎛⎝⎜ ⎞⎠⎟ =
m. m m m
23.5 In the figure at the right, θ ′ =θ since they are vertical angles formed by two intersecting straight lines. Their complementary angles are also equal or α ′ =α. The right triangles PQR and P QR ′ have the common side QR and are then congruent by the angle-side-angle theorem. Thus, the corresponding sides PQ and P Q ′ are equal, or the image is as far behind the mirror as the object is in front of it.
23.6 (a) Since the object is in front of the mirror, p > 0. With the image behind the mirror, q < 0. The mirror equation gives the radius of curvature as
2 1 1 1 1 00
R p q 10 0 10 0
. cm. cm.
cm
or R = 2 ⎛⎝⎜ 10 0.^ cm⎞⎠⎟ = +2 22. 9
cm
(b) The magnification is M q p
= − = −^ (−^. = +
cm) 10 0 cm
23.7 (a) Since the mirror is concave, R > 0. Because the object is located in front of the mirror, p > 0. The mirror equation, 1 p + 1 q = 2 R , then gives the image distance as
q pR p R
( )( ) (^2) ( ) −
cm cm cm ccm = +13 3. cm
Since q > 0 , the image is located 13 3. cm in front of the mirror.
continued on next page
Mirrors and Lenses 313
23.11 The magnifi ed, virtual images formed by a concave mirror are upright, so M > 0.
Thus, M q p
h h
= − = ′ = 5 00.^ cm= +2 50. 2.00 cm
, giving
q = − 2 50. p = − 2 50. ( + 3 00. cm ) = −7 50. cm
The mirror equation then gives 1 2 1 1 1 3 00
f R p q 7 7
. cm .50 cm
.50 cm
or (^) f = 7 50.^ cm=5 00.
cm
23.12 Realize that the magnitude of the radius of curvature, R , is the same for both sides of the hubcap. For the convex side, R = − R ; and for the concave side, R = + R. The object distance p is positive (real object) and has the same value in both cases. Also, we write the virtual image distance as q = − q in each case. The mirror equation then gives:
For the convex side,
q R p or^ q^
R p R p
[1]
For the concave side,
q R p or^ q^
R p R p
[2]
Comparing Equations [1] and [2], we observe that the smaller magnitude image distance, q = 10 0. cm, occurs with the convex side of the mirror. Hence, we have 1 10 0
. cm R p
[3]
and for the concave side, q = 30 0. cm gives 1 30 0
. cm R p
[4]
(a) Adding Equations [3] and [4] yields 2 3 1 30 0
p
= +^ p = + .
cm
or cm
(b) Subtracting [3] from [4] gives 4 3 1 30 0
R
= −^ R =
cm
or cm
23.13 The image is upright, so M > 0 , and we have
M q p
= − = + 2 0. , or q = − 2 0. p = − 2 0 2. ( 5 cm ) = − 50 cm
The radius of curvature is then found to be 2 1 1 1 2
R p q 5 5
5 cm 0 cm
0 cm
, or R =
2 0 .50 m = 1 0
. m
314 Chapter 23
23.14 (a) Your ray diagram should be carefully drawn to scale and look like the diagram given below:
(b) From the mirror equation with p = +10 0. cm and f = −15 0. cm, the image distance is
q pf p f
( ) −( ) − −
cm cm cm (^) (( cm)
cm cm cm .
and the magnification is M = − q p = − −( 6 00. cm) ( 10 0. cm ) = + 0 600. > 0 .Thus, you should find that the image is upright, located 6 00. cm behind the mirror,six-tenths the size of the object.
23.15 The focal length of the mirror may be found from the given object and image distances as 1 f = 1 p + 1 q , or
f p q p q
( )( )
cm cm cm cm
..1 cm
For an upright image twice the size of the object, the magnification is
M = − q p = +2 00. , giving q = − 2 00. p
Then, using the mirror equation again, 1 p + 1 q = 1 f becomes 1 1 1 1 2 00 2 00
p q p p p f
or p f = = = 2 00
cm cm
23.16 (a) The mirror is convex, so f < 0, and we have f = − f = −8 0. cm. The image is virtual, so q < 0, or q = − q. Since we also know that q = p 3 , the mirror equation gives
1 1 1 3 1 p q p p f
p 8 0. cm
and p = +16 cm
This means that we have a real object located 16 cm in front of the mirror.
(b) The magnification is M = − q p = + q p = +1 3. Thus, the image is upright and one-third the size of the object.
316 Chapter 23
23.20 (a) From 1 1 2 p q R
R p p R
p p
( (^) ) 2 −
m m
The table gives the image position at a few critical points in the motion. Between p = 3 00. m and p = 0 500. m, the real image moves from 0.600 m to positive infinity. From
Object Distance, p
Image Distance, q 3.00 m 0.600 m
0.500 m ±^ ∞
p = 0 500. cm to p = 0 , the virtual image moves from negative infinity to 0.
Note the “jump” in the image position as the ball passes through the focal point of the mirror.
(b) The ball and its image coincide when p = 0 and when
1 p + 1 p = 2 p = 2 R , or p = R = 1 00. m
From Δ y = v 0 yt a ty^2
- , with v 0 y = 0 , the times for the ball to fall from p = 3 00. m to these positions are found to be
t y a (^) y
( )
(− ) −
m m s 2 s and
t = ( − ) −
m m s 2 s
23.21 From n p
n q
n n R
1 + 2 = 2 −^1 , with R → ∞, the image position is found to be
q n n
= − p = − ⎛ ⎝⎜^
(^2) ) = − 1
. cm. cm
or the virtual image is 38 2. cm below the upper surface of the ice.
23.22 The center of curvature of a convex surface is located behind the surface, and the sign convention for refracting surfaces (Table 23.2 in the textbook) states that R > 0 , giving R = + 8 00. cm. The object is in front of the surface ( p > 0 ) and in air ( n 1 = 1 00. ), while the second medium is glass ( n 2 = 1 50. ). Thus, n (^) 1 p + n (^) 2 q = ( n (^) 2 − n 1 ) R becomes
1 00 1 50 1 50 1 00 8 00
p q.
cm and reduces to q p p
( ) −
cm cm
(a) If p = 20 0. cm, q = ( )( ) −
cm cm cm cm cm
(b) If p = 8 00. cm, q = ( )( ) −
cm cm cm cm cmm
continued on next page
Mirrors and Lenses 317
(c) If p = 4 00. cm, q = ( (^) )( (^) ) −
cm cm cm cm
cmm
(d) If p = 2 00. cm, q = ( (^) )( (^) ) −
cm cm cm cm
cmm
23.23 Since the center of curvature of the surface is on the side the light comes from, R < 0 , giving
R = − 4 0. cm. Then, n p
n q
n n R
(^1) + 2 = 2 −^1 becomes
q.
cm cm
, or q = − 4 0. cm
Thus, the magnifi cation M h h
n n
q p
1 2
gives
⎠⎟^
(− (^) ) (
h n q n p
(^1) h 2
cm cm)) ( 2 5.^ mm) =^ 3 8. mm
23.24 For a plane refracting surface (^) ( R →∞)
n p
n q
n n R
(^1) + 2 = 2 −^1 becomes q n n
= − 2 p 1
(a) When the pool is full, p = 2 00. m and
q = − ⎛⎝⎜ 1 00⎞⎠⎟ ( ) = − 1 333
. m. m
or the pool appears to be 1 50. m deep.
(b) If the pool is half filled, then p = 1 00. m and q = − 0 750. m. Thus, the bottom of the pool appears to be 0.75 m below the water surface or 1 75. m below ground level.
23.25 As parallel rays from the Sun (^) (object distance, p → ∞) enter the transparent sphere from air (^ n 1 =1 00.^ ), the center of curvature of the surface is on the side the light is going toward (back side). Thus, R > 0. It is observed that a real image is formed on the surface opposite the Sun, giving the image distance as q = + 2 R.
Then n p
n q
n n R
(^1) + 2 = 2 −^1 becomes 0 2
n R
which reduces to n = 2 n −2 00. and gives n = 2 00.
Mirrors and Lenses 319
(a) If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00 cm thick pane of glass, then p = + 100 cm and t =6 00. cmand the position of the final image relative to the glass-air interface is
q = − ⎛⎝⎜ ⎞⎠⎟ ( ) + ⎛⎝⎜ ⎞⎠⎟
⎡ cm (. 000 cm ) 79 0 cm 0 790m ⎣⎢^
(b) If the thickness of the glass is negligible ( t → 0 ), the distance of the final image from the glass-air interface is
q n n
n n p n n
a p g
g w
a w
⎠⎟^
⎠⎟^
⎛⎝⎜ ⎞⎠⎟ ( cm ) = −. cm = −. m
so we see that the 6.00 cm thickness of the glass in part (a) made a 4.00 cm difference in the apparent position of the jellyfish.
(c) The thicker the glass, the greater the distance between the final image and the outer surface of the glass.
23.28 The wall of the aquarium (assumed to be of negligible thickness) is a plane ( R → ∞) refract-
ing surface separating water ( n 1 = 1 333. )and air ( n 2 = 1 00. ). Thus, n p
n q
n n R
(^1) + 2 = 2 −^1 gives the
image position as q n n p p = −
⎠⎟^
1 1 333.^
. When the object position changes by Δ p , the change in
the image position is Δ
q p = − 1 333.
. The apparent speed of the fish is then given by
v image cm s = = ( ) = =
q Δ Δ t
p t 1 333
. ccm s
23.29 With R (^) 1 = + 2 00. cm and R 2 = +2 50. cm, the lens maker’s equation gives the focal length as
1 1
f (^) 1 2 2 5 n R R = ( − ) −
= (. − ) −
. cm. 00
cm cm 1
or f = (^) − =
cm 1 cm
23.30 The lens maker’s equation is used to compute the focal length in each case.
(a)
f (^) 1 2 n R R = ( − ) −
f = ( − ) − f (− )
. (^). cm. cm ⎥ =. cm
(b)
f 12 0 = ( − ) − ( − )
. (^). cm. cm ⎥ f = 16 4. cm
320 Chapter 23
23.31 The focal length of a converging lens is positive, so f = +10 0. cm. The thin lens equation then yields a focal length of
q pf p f
p p
( (^) ) −
cm cm
(a) When p = +20 0. cm ,
q = ( (^) )( (^) ) −
cm cm cm cm
cmm and M q p
cm. cm
so the image is located 20.0 cm beyond the lens , is real ( q > 0) , is inverted ( M < 0) ,
and is the same size as the object (^) ( M =1 00. (^) ).
(b) When p = f = +10 0. cm , the object is at the focal point and no image is formed. Instead, parallel rays emerge from the lens.
(c) When p = 5 00. cm,
q = ( (^) )( (^) ) −
cm cm cm cm
cmm and M q p
= − = − −^ 10 0 = +
cm. cm
so the image is located 10.0 cm in front of the lens , is virtual ( q < 0) , is upright ( M^ >^ 0)^ , and is twice the size of the object (^) ( M =2 00. (^) ).
23.32 (a) and (b) Your scale drawings should look similar to those given below:
Figure (a) (^) Figure (b)
A carefully drawn-to-scale version of Figure (a) should yield a real, inverted image that is located 20 cm in back of the lens and the same size as the object. Similarly, a carefully drawn-to-scale version of Figure (b) should yield an upright, virtual image located 10 cm in front of the lens and twice the size of the object.
(c) The accuracy of the graph depends on how accurately the ray diagrams are drawn. Sources of uncertainty: a parallel line from the tip of the object may not be exactly parallel; the focal points may not be exactly located; lines through the focal points may not be exactly the correct slope; the location of the intersection of two lines cannot be determined with complete accuracy.
322 Chapter 23
23.35 (a) The real image case is shown in the ray diagram. Notice that p + q = 12 9. cm, or q = 12 9. cm − p. The thin lens equation, with f = 2 44. cm , then gives
1 1 12 9
p p 2 44
. cm. cm
or p^2 − (12 9. cm) p + 31 5. cm 2 = 0
Using the quadratic formula to solve gives
p = 9 63. cm or p =3 27. cm
Both are valid solutions for the real image case.
(b) The virtual image case is shown in the second diagram. Note that in this case, q = − ( 12 9. cm + p ), so the thin lens equation gives
1 1 12 9
p p 2 44
. cm. cm
or p^2 + (12 9. cm) p − 31 5. cm 2 = 0
The quadratic formula then gives p = 2 10. cm or p = −15 0. cm.
Since the object is real, the negative solution must be rejected, leaving p = 2 10. cm.
23.36 We must first realize that we are looking at an upright, magnified, virtual image. Thus, we have a real object located between a converging lens and its front-side focal point, so q < 0 , p > 0 , and f > 0.
The magnification is M q p = − = + 2 , giving q = − 2 p. Then, from the thin lens equation,
1 1 2
p p p f
− = + = or^ f^ =^2 p =^ 2 2 84(^.^ cm) =^ 5 68. cm
23.37 It is desired to form a magnifi ed, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magnification then gives
M
h h
q p
×
1 80. m 24.0 10 3 m , or q = 75 0. p
Also, we know that p + q = 3 00. m. Therefore, p + 75 0. p =3 00. m, giving
(b) p = = × − =
m m mm
continued on next page
Mirrors and Lenses 323
(a) The thin lens equation then gives 1 1 75 0
p p p f
or f = ⎛⎝⎜ 75 0⎞⎠⎟ p = ⎛⎝⎜ ⎞⎠⎟ ( ) = 76 0
. mm 9 9 0. mm
23.38 To have a magnification of M = − q p = +3 00. , it is necessary that q = −3 00. p. The thin lens equation, with f = +18 0. cm for the convergent convex lens, gives the required object distance as
1 1 3 00
p p p 18 0
... cm or p = ( (^) ) =
cm cm
23.39 Since the light rays incident to the first lens are parallel, (^) p 1 = ∞ and the thin lens equation gives q (^) 1 = f 1 = − 10 0. cm.
The virtual image formed by the first lens serves as the object for the second lens, so p (^) 2 = 30 0. cm + q 1 =40 0. cm. If the light rays leaving the second lens are parallel, then q 2 = ∞ and the thin lens equation gives f (^) 2 = p 2 = 40 0. cm.
23.40 (a) Solving the thin lens equation for the image distance q gives
1 1 1 q f p
p f p f
or q pf p f
(b) For a real object, p > 0 and p = p. Also, for a diverging lens, f < 0 and f = − f. The result of part (a) then becomes
q p f p f
p f p f
( − ) − −( )
Thus, we see that q < 0 for all numeric values of p and f. Since negative image distances mean virtual images, we conclude that a diverging lens will always form virtual images of real objects.
(c) For a real object, p > 0 and p = p. Also, for a converging lens, f > 0 and f = f. The result of part (a) then becomes
q p f p f
= p f −
> 0 if − > 0
Since q must be positive for a real image, we see that a converging lens will form real images of real objects only when p > f (or p > f since both p and f are positive in this situation).
Mirrors and Lenses 325
23.43 From the thin lens equation, q p f (^1) p f 1 1 1 1
( (^) )( (^) ) −
cm cm cm cm
= − 8 00. cm.
The magnification by the first lens is M q (^1) p 1 1
( − ) = +
cm cm
The virtual image formed by the first lens is the object for the second lens, so p (^) 2 = 6 00. cm + q 1 = +14 0. cm and the thin lens equation gives
q p f (^2) p f 2 2 2 2
( ) −( ) − −
4.0 cm cm 4.0 cm 16
..0 cm
cm ( )
The magnification by the second lens is M q (^2) p 2 2
( − ) = +
cm cm
, so the overall
magnifi cation is M = M M 1 2 = (^) ( +2 00. (^) ) +( 0 533. (^) ) = + 1 07..
The position of the final image is 7 47. cm in front of the second lens , and its height is h ′ = M h = (^) ( +1 07. (^) )( 1 00. cm ) = 1 07. cm.
Since M > 0 , the final image is upright ; and since q 2 < 0 , this image is virtual.
23.44 (a) We start with the final image and work backward. From Figure P23.44, observe that q 2 = − (50 0. cm −31 0. cm ) = −19 0. cm. The thin lens equation
then gives p q f (^2) q f 2 2 2 2
(− )( ) − −
.0 cm cm 19.0 cm
cm
= + cm
The image formed by the first lens serves as the object for the second lens and is located 9.74 cm in front of the second lens.
Thus, q 1 = 50 0. cm − 9 74. cm =40 3. cm and the thin lens equation gives
p q f (^1) q f 1 1 1 1
( (^) )( (^) ) −
0.3 cm cm 0.3 cm
. cm
= +13 3. cm
The original object should be located 13 3. cm in front of the first lens.
(b) The overall magnification is
M M M q p
q p
⎠⎟^
1 2 1 =^ −
1
2 2
40 3. cm 13.3 ccm
cm cm
⎝⎜^
⎠⎟^ −
⎛ (− ) ⎝⎜
(c) Since M < 0 , the final image is inverted ; and since q 2 < 0 , it is virtual.
326 Chapter 23
23.45 Note: Final answers to this problem are highly sensitive to round-off error. To avoid this, we retain extra digits in intermediate answers and round only the final answers to the correct number of signifi cant fi gures.
Since the fi nal image is to be real and in the film plane, q (^) 2 = + d.
Then, the thin lens equation gives p q f q f
d (^2) d 2 2 2 2
( ) −
cm cm
From Figure P23.45, observe that d < 12 0. cm. The above result then shows that p 2 < 0 , so the object for the second lens will be a virtual object.
The object of the second lens ( L 2 ) is the image formed by the first lens ( L 1 ) , so
q d p d (^1 2) d
= ( − ) − = − + −
cm cm cm 13.0 ccm cm 13.0 cm
⎝⎜^
⎠⎟ =^ 12 0 −^ −
2 . d d
If d = 5 00. cm, then q 1 = + 15 125. cm; and when d = 10 0. cm, q 1 = + 45 333. cm.
From the thin lens equation, p q f q f
q (^1) q 1 1 1 1
1 1
( ) −
cm cm
When q 1 = + 15 125. cm ( d =5 00. cm) , then p 1 = 1 82. × 103 cm =18 2. m.
When q 1 = + 45 333. cm ( d =10 0. cm ), then p 1 = 22 4. cm =0 224. m.
Thus, the range of focal distances for this camera is 0 .224^ m to 18.2 m.
23.46 (a) From the thin lens equation, the image distance for the first lens is
q p f (^1) p f 1 1 1 1
( )( ) −
cm cm cm cm = +30 0. cm
(b) With q 1 = +30 0. cm, the image of the first lens is located 30.0 cm in back of that lens. Since the second lens is only 10.0 cm beyond the first lens, this means that the first lens is trying to form its image at a location 20.0 cm beyond the second lens.
(c) The image the first lens forms (or would form if allowed to do so) serves as the object for the second lens. Considering the answer to part (b) above, we see that this will be a virtual object, with object distance p 2 = −20 0. cm.
(d) From the thin lens equation, the image distance for the second lens is
q p f (^2) p f 2 2 2 2
( − )( ) − −
cm cm cm 000
cm = +. cm
(e) M q (^1) p 1 1
cm 15.0 cm
(f) M^
q (^2) p 2 2
(− )
cm 20.0 cm
(g) M = M M 1 2 = ( −2 00. ) +( 0 200. ) = − 0 400.
(h) Since q 2 > 0 , the fi nal image is real , and since M < 0 , that image is inverted.
328 Chapter 23
(c) If p = 3 f = +36 0. cm, q = − 9 00. cm.
(d) If p = f = +12 0. cm, q = − 6 00. cm.
(e) If p = f 2 = +6 00. cm, q = − 4 00. cm.
23.51 As light passes left to right through the lens, the image position is given by
q p f (^1) p f 1 1 1 1
( (^) )( (^) ) −
0 cm 0.0 cm 00 cm 0.0 cmm
= + 400 cm
This image serves as an object for the mirror with an object distance of p (^) 2 = 100 cm − q 1 = − 300 cm (virtual object). From the mirror equation, the position of the image formed by the mirror is
q p f (^2) p f 2 2 2 2
( − ) −( ) − − −
3 cm cm 3 cm 5
cm
cm ( )
This image is the object for the lens as light now passes through it going right to left. The object distance for the lens is p (^) 3 = 100 cm − q 2 = 100 cm − −( 60 0. cm), or p 3 = 160 cm. From the thin lens equation,
q p f (^3) p f 3 3 3 3
( (^) )( (^) ) −
cm cm cm c
. mm
= + 160 cm
Thus, the final image is located 160 cm to the left of the lens.
The overall magnification is M M M M q p
q p
q p
⎠⎟^
⎠⎟^
1 1
2 2
3 3
, or
M = ⎛⎝⎜ − ⎞⎠⎟ −
(− ) ( − )
cm cm
cm cm
⎝⎜^
⎠⎟ =^ −
cm 0 800 cm
Since M < 0 , the final image is inverted.
23.52 Since the object is midway between the lens and mirror, the object distance for the mirror is p 1 = + 12 5. cm. The mirror equation gives the image position as
1 2 1 2 20 0
q 1 (^) R p 1 50
= − = − = −^ =
. cm. cm. cm ..0 cm
, or q 1 = + 50 0. cm
This image serves as the object for the lens, so p (^) 2 = 25 0. cm − q 1 = −25 0. cm. Note that since p 2 < 0 , this is a virtual object. The thin lens equation gives the image position for the lens as
q p f (^2) p f 2 2 2 2
(− ) −( ) − − −
25.0 cm cm 25.0 cm
1 16.7 cm
cm ( )
Since q 2 < 0 , this is a virtual image that is located 50.3 cm in front of the lens or 2 5 3. cm behind the mirror. The overall magnification is
M M M q p
q p
⎠⎟^
1 2 1 =^ −
1
2 2
50 0. cm 12 .5 ccm
cm 25.0 cm
⎠⎟^
(− ) (− )
Since M > 0 , the final image is upright.
Mirrors and Lenses 329
23.53 A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the fl at face. We next consider the light’s exit from the curved surface, for which R = − 6 00. cm.
The incident rays are parallel, so p = ∞.
Then, n p
n q
n n R
(^1) + 2 = 2 −^1 becomes 0 1 00^ 1 00^ 1 56 6 00
q. cm
from which q = 10 7. cm.
23.54 (a) The thin lens equation gives the image distance for the first lens as
q p f (^1) p f 1 1 1 1
( )( ) −
cm cm cm cm = + 40 0. cm
The magnification by this lens is then M q (^1) p 1 1
cm cm
The real image formed by the first lens is the object for the second lens. Thus, p (^) 2 = 50 0. cm − q 1 = +10 0. cm and the thin lens equation gives
q p f (^2) p f 2 2 2 2
( )( ) −
0.0 cm cm 0.0 cm
. cm = +10 0. cm
The final image is 10 0. cm in back of the second lens.
(b) The magnifi cation by the second lens is M q (^2) p 2 2
0 cm 0 cm
. , so the overall magnifi cation is M = M M 1 2 = ( −1 00. ) −( 1 00. ) = +1 00.. Since this magnification has a value of unity, the final image is the same size as the original object, or h ′ = M h 1 = (+ 1 00. )( 2 00. cm) = 2 00. cm.
The image distance for the second lens is positive, so the final image is real.
(c) When the two lenses are in contact, the focal length of the combination is 1 1 1 1 20 0
f f 1 (^) f 2 5 00
. cm. cm , or f = 4 00. cm
The image position is then
q pf p f
( )( ) −
.00 cm cm .00 cm cm
2 20 0. cm
