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Unformatted text preview: We consider two cases: in case 1, the object is placed 18 cm in front of a
diverging lens. The magnification for this case is given by m1. In case 2, the object is
moved so that the magnification m2 is reduced by a factor of 2 compared to that in case 1.
1 In other words, we have m 2 = m 1. Using Equation 26.7, we can write this as
2 – d i2
d o2 1 d i1 =– 2 d o1 (1) 1386 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS This expression can be solved for do2. First, however, we must find a numerical value for
di1, and we must eliminate the variable di2.
SOLUTION
The image distance for case 1 can be found from the thinlens equation [Equation 26.6:
(1 / d o ) + (1 / d i ) = (1 / f ) ]. The problem statement gives the focal length as f = –12 cm .
Since the object is 18 cm in front of the diverging lens, d o1 = 18 cm . Solving for di1, we
find
1
1
1
1
1
=–
=
–
or d i1 = –7.2 cm
d i1 f d o1 –12 cm 18 cm where the minus sign indicates that the image is virtual. Solving Equation (1) for do2, we
have
d d o2 = 2 d i2 o1 (2) d i1 To eliminate di2 from this result, we note that the thinlens equation applied to case 2 gives
d –f
1
1
1
=–
= o2
d i2
f d o2
f d o2 or d i2 = f d o2
d o2 − f Substituting this expression for di2 into Equation (2), we have 2 f d o2 d o1 d o2 = d o2 − f d i1 or d o1 d o2 − f = 2 f d i1 Solving for do2, we find d 18 cm + 1 = 48 cm
do2 = f 2 o1 + 1 = (–12cm) 2 –7.2 cm d i1 ______________________________________________________________________________ 62. REASONING To find the distance through which the object must be moved, we must
obtain the object distances for the two situations described in the problem. To do this, we
combine the thinlens equation and the magnification equation, since data for the
magnification is given. SOLUTION
a. Since the magnification is positive, the image is upright, and the object must be located
within the focal point of the lens, as in Figure 26.28. When the magnification is negative
and has a magnitude greater than one, the object must be located between the focal point Chapter 26 Problems 1387 and the point that is at a distance of twice the focal length from the lens, as in Figure 26.27.
Therefore, the object should be moved away from the lens .
b. According to the thinlens equation, we have
1
1
1
+
=
di do
f di = or do f (1) do – f According to the magnification equation, with di expressed as in Equation (2), we have m=– di
do =– 1 do f f =
do do – f f – do or do = f ( m – 1)
m (2) Applying Equation (2) to the two cases described in the problem, we have ( d o )+ m = f ( m –1) f ( +4.0 –1) 3.0 f
=
=
m
+4.0
4.0 (3) ( do ) – m = f ( m –1) f ( –4.0 –1) 5.0 f
=
=
m
–4.0
4.0 (4) Subtracting Equation (3) from Equation (4), we find that the object must be moved away
from the lens by an additional distance of ( do ) – m – ( do )+ m = 5.0 f
4.0 3.0 f 2.0 f 0.30 m
=
=
= 0.15 m
4.0
4.0
2.0
______________________________________________________________________________
– 63. REASONING AND SOLUTION Let d represent the distance between the object and the
screen. Then, do + di = d. Using this expression in the thinlens equation gives 1
1
1
+
=
do d − do f or do2 – ddo + df = 0 With d = 125 cm and f = 25.0 cm, the quadratic formula yields solutions of
do = +35 cm and do = +90.5 cm
______________________________________________________________________________ 1388 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 64. REASONING According to Equation 2.1, the average speed s of a moving object is the
distance D traveled, divided by the elapsed time ∆t:
s= D
∆t (2.1) In order to determine the average speed of the image of the horse over each interval of
∆t = 2.0 seconds, we will need to find the image distance di,0 at the beginning of each
interval and the image distance di,f at the end of each interval. The horse is closer to the lens
than the focal point F of the lens, so the image of the horse is virtual, and the image
distances are negative. Therefore, the distance D traveled by the image of the horse will be
equal to the magnitude of the difference between the final and initial image distances D = di, f − di, 0 (1) 111
+=
do di f
(Equation 26.6), where do is the object distance between the horse and the lens, and f is the
focal length of the lens. The horse canters at a constant speed of v = 7.0 m/s, and is initially
a distance do, 0 = 40.0 m from the lens. During each twosecond interval, the object distance
between the horse and the lens will decrease by (7.0 m/s)(2.0 s) = 14.0 m. The images distances are determined by the thinlens equation SOLUTION a. Solving 111
+=
(Equation 26.6) for 1/di and taking the reciprocal of both sides
do di f yields 1 1 1 do − f
=−
=
di f d o
do f di = or do f (2) do − f Applying Equation (2) to both the final and initial positions of the horse and its image
during this interval, we obtain
di, f = do, f f and do, f − f di, 0 = do, 0 f (3) do, 0 − f Substituting Equations (3) into Equatio...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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