Unformatted text preview: _______________________________ 50. REASONING
a. Given the focal length (f = −0.300 m) of the lens and the image distance di, we will
111
employ the thinlens equation
+=
(Equation 26.6) to determine how far from the
d o di f
van the person is actually standing, which is the object distance do. The image and the
person are both behind the van, so the image is virtual, and the image distance is negative:
di = −0.240 m. b. Once we have determined the object distance do, we will use the magnification equation
h
d
m = i = − i (Equation 26.7) to calculate the true height ho of the person from the height
ho
do
hi = 0.34 m of the image.
SOLUTION
a. Solving Equation 26.6 for do, we obtain
1
11
=−
do f di or do = 1
1
=
= 1.2 m
11
1
1
−
−
f di −0.300 m ( −0.240 m ) b. Taking the reciprocal of both sides of hi
d
= − i (Equation 26.7) and solving for ho
ho
do yields ho
d
=− o
hi
di or d 1.2 m ho = −hi o = − ( 0.34 m ) = 1.7 m
d −0.240 m i 1380 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________
51. REASONING Since the object distance and the focal length of the lens are given, the
thinlens equation (Equation 26.6) can be used to find the image distance. The height of the
image can be determined by using the magnification equation (Equation 26.7).
SOLUTION
a. The object distance do, the image distance di, and the focal length f of the lens are related
by the thinlens equation:
1
11
+=
(26.6)
d o di f Solving for the image distance gives
111
1
1
=−
=
−
di f do 12.0 cm 8.00 cm or di = −24 cm b. The image height hi (the height of the magnified print) is related to the object height ho,
the image distance di, and the object distance do by the magnification equation:
d −24 cm (26.7)
hi = − ho i = − ( 2.00 mm ) = 6.0 mm
do 8.00 cm ______________________________________________________________________________
52. REASONING The height of the mountain’s image is given by the magnification equation
as hi = –hodi/do. To use this expression, however, we will need to know the image distance
di, which can be determined using the thinlens equation. Knowing the image distance, we
can apply the expression for the image height directly to calculate the desired ratio.
SOLUTION According to the thinlens equation, we have 1
1
1
+
=
di do
f (1) For both pictures, the object distance do is very large compared to the focal length f.
Therefore, 1/do is negligible compared to 1/f, and the thinlens equation indicates that di = f.
As a result, the magnification equation indicates that the image height is given by
hi = – ho di
do =– ho f
do (2) Chapter 26 Problems 1381 Applying Equation (2) for the two pictures and noting that in each case the object height ho
and the focal length f are the same, we find ( hi )5 km
( hi )14 km ho f –
d
( do )14 km 14 km
o 5 km =
=
=
= 2.8 ho f ( do )5 km 5.0 km
–
d o 14 km ______________________________________________________________________________
53. REASONING The distance from the lens to the screen, the image distance, can be
obtained directly from the thinlens equation, Equation 26.6, since the object distance and
focal length are known. The width and height of the image on the screen can be
determined by using Equation 26.7, the magnification equation.
SOLUTION
a. The distance di to the screen is 11
1
1
1
=−
=
−
= 2.646 × 10−4 mm −1
di f do 105.00 mm 108.00 mm
so that di = 3.78 × 103 mm = 3.78 m .
b. According to the magnification equation, the width and height of the image on the screen
are Width d 3.78 × 103 mm 2
hi = ho − i = ( 24.0 mm ) − = − 8.40 × 10 mm d
108 mm o
The width is 8.40 × 10 2 mm . Height d
hi = ho − i
d
o 3.78 × 103 mm 3 = ( 36.0 mm ) − = − 1.26 × 10 mm 108 mm The height is 1.26 × 103 mm .
______________________________________________________________________________ 1382 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 54. REASONING In order for a sharply focused image to appear on the image sensor, the
distance between the sensor and the lens must be equal to the image distance di, which can
111
be found from the thinlens equation
+=
(Equation 26.6), where
d o di f
f = 200.0 mm = 0.2000 m is the focal length of the lens and do is the distance between the
object and the lens. Since there are two object distances (do1, do2), there will be two image
distances (di1, di2). When the converging lens forms a real image, increasing the object
distance decreases the image distance. Therefore, the smaller object distance of do1 = 3.5 m
will yield the greater image distance di1, and the greater object distance of do2 = 50.0 m will
yield the smaller image distance di2. Thus, the distance x that the lens must move is equal to
the difference between the image distances:
x = d i1 − d i2 (1) 111
1
+=
(Equation 26.6) for di yields di =
. Using this
11
d o di f
−
f do
result with the first and second object distances, we have that SOLUTION Solving di1 = 1
1
1
−
f do1 and di2 = 1
1
1
−
f...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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