Unformatted text preview: e and image 1
is d o – d i = 2 .00 m . Equation 25.4 gives the magnification m as m = – d i / d o = 4 .00
As a result c so that h d o – d i = d o – – 4 . 00 d o = 2 . 00 m The image distance is b d i = – 4 .00 d o
or d o = 0. 400 m g d i = – 4 .00 d o = – 4 .00 0.400 m = –1.60 m
We can now obtain the focal length from the mirror equation: 1
1
1
1
1
+
=
+
=
d o d i 0.400 m –1.60 m f or f = 0 .533 m Chapter 25 Problems 1303 32. REASONING The focal length f of the convex mirror determines the first image distance
1
1
1
di1 via =
+
(Equation 25.3), where do = 15.0 cm is the distance between the candle
f do di1
and the convex mirror. The second image, formed by the plane mirror, is located as far
behind the mirror as the object is in front of the mirror. Therefore, when the plane mirror
replaces the convex mirror, the second image distance di2 is given by
d i2 = − d o = −15.0 cm (1) The negative sign in Equation (1) arises because the image lies behind the plane mirror and,
therefore, is a virtual image.
When the plane mirror replaces the convex mirror, the image moves a distance x farther
away from the mirror, so the initial and final image distances are related by
d i2 = d i1 − x (2) The negative sign in Equation (2) occurs because the image moves farther from the mirror,
thus making the second image distance di2 a negative number of greater magnitude than the
first image distance di1. SOLUTION Substituting Equation (2) into Equation (1) and solving for di1 yields
d i1 − x = − d o
Substituting Equation (3) into or d i1 = x − d o (3) 11
1
=
+
(Equation 25.3), we obtain
f do di1
1
1
1
1
1
=
+
=
+
f do di1 d o x − do (4) Taking the reciprocal of both sides of Equation (4), we find that the focal length of the
convex mirror is
f= 1
1
=
= −17 cm
1
1
1
1
+
+
do x − do 15.0 cm 7.0 cm − 15.0 cm 1304 THE REFLECTION OF LIGHT: MIRRORS 33. REASONING AND SOLUTION The magnification is m = −di /do = +1/4, so that
di = −(1/4)do. Thus,
1
1
1
1
1
−3
=
+
=
+
=
1
f
do di
do
do
− 4 do ch Therefore, f = −(1/3)do, for the convex side. Since the radius of curvature is the same in
each case, the focal length of the concave side is f ′ = + 1 d ′ . So,
3 1
1
1
+
=
=
f′
′
′
do
di 1
1
3 so that d o′ o 1
d i′ = 2
d′ The magnification of the concave side is, therefore, m' = −di'/do' = o –1 / 2 . 34. REASONING In both cases, the image and object distances (di and do) are related to the
111
focal length f of the mirror by the mirror equation
+=
(Equation 25.3). Letting the
d o di f
subscript 1 denote the first situation (convex side of the mirror) and the subscript 2 denote
the second situation (concave side of the mirror), we have that 1
1
1
+
=
d o di2 f 2 and 1
1
1
+
=
d o di1 f1 (1) In Equations (1), we have made use of the fact that the man maintains the same distance of
do = 45 cm between his face and the mirror in both cases. Because the mirror is doublesided, the focal lengths of each side differ only by sign: that of the convex side is negative,
and that of the concave side is positive. Therefore, we have that f2 = −f1, where f1 is the
focal length of the convex side and f2 is the focal length of the concave side. Given this
relation, we can rewrite the first of Equations (1) as 1
1
1
+
=−
d o di2
f1 (2) We will determine the image distance di1 from the magnification m1 of the first image by
d
means of m = − i (Equation 25.4), the magnification equation:
do m1 = − di1
do (3) Chapter 25 Problems 1305 After using Equation 25.4 to determine di1, we will use mirror equation to determine the
image distance di2 when the man looks into the concave side of the mirror. If the image
distance is positive, it will appear in front of the mirror; if the image distance is negative, the
image will appear behind the mirror.
SOLUTION Solving Equation (2) for 1
, we obtain
d i2 1 1
1
11
=− −
= − +
f d
di2
f1 d o
o
1 (4) In this result, we can replace the term 1/f1 by using the second of Equations (1). In this way
we find that
1 1
1
2
1
1
1
1
= − + = − +
+ = − + f d d d di2
o
1 o di1 do o di1 (5) Taking the reciprocal of Equation (5), we obtain
di2 = − 1 (6) 2
1
+ d o di1 Solving Equation (3) for di1 yields di1 = − m1do = −0.20d o . Substituting this result into
Equation (6) yields
d i2 = − 1
2
1
+ d o d i1 =− 1
2
1
− d o 0.20 d o =− 1
1
( 2 − 5.0)
do In the preceding equation, we have used the fact that =− do
+45 cm
=
= +15 cm
3.0
− 3.0 1
= 5.0 .
0.20 1306 THE REFLECTION OF LIGHT: MIRRORS 35. REASONING
a. The image of the spacecraft appears a distance di beneath the surface of the moon, which
we assume to be a convex spherical mirror of radius R = 1.74×106 m. The image distance di
is related to the focal length f of the moon and the object distance do by the mirror equation 111
+=
d o di f (25.3) The spacecraft is the object, so the object distance do is equal to the altitude of the
spacecraft: do = 1.22×105 m. The moon is assumed to be a conv...
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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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