Physics Solution Manual for 1100 and 2101

# 5 and 2614 chapter 26 problems 1351 chapter 26 the

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Unformatted text preview: .600 f − do −27 .0 cm − 18.0 cm b b gb g gb g 27. SSM WWW REASONING When paraxial light rays that are parallel to the principal axis strike a convex mirror, the reflected rays diverge after being reflected, and appear to originate from the focal point F behind the mirror (see Figure 25.16). We can treat the sun as being infinitely far from the mirror, so it is reasonable to treat the incident rays as paraxial rays that are parallel to the principal axis. SOLUTION a. Since the sun is infinitely far from the mirror and its image is a virtual image that lies behind the mirror, we can conclude that the mirror is a convex mirror . b. With d i = –12.0 cm and d o = ∞ , the mirror equation (Equation 25.3) gives 1 1 1 11 1 = + = + = f do di ∞ di di Therefore, the focal length f lies 12.0 cm behind the mirror (this is consistent with the reasoning above that states that, after being reflected, the rays appear to originate from the focal point behind the mirror). In other words, f = –12.0 cm . Then, according to 1 Equation 25.2, f = – 2 R , and the radius of curvature is R = –2 f = –2(–12.0 cm) = 24.0 cm 1300 THE REFLECTION OF LIGHT: MIRRORS 28. REASONING The focal length f of the water drop is given by the mirror equation 111 = + (Equation 25.3), where do and di are, respectively, the object distance and the f d o di image distance. The object distance (do = 3.0 cm) is given, and we will determine the image hi d = − i (Equation 25.4), where ho is the ho do diameter of the flower and hi is the diameter of its image. The water drop acts as a convex spherical mirror, so the image is upright. Therefore, the image height hi is positive, and we expect the focal length f to be negative. distance di from the magnification equation m = SOLUTION Solving hi d = − i (Equation 25.4) for di and taking the reciprocal, we obtain ho do di = − Substituting Equation (1) into do hi h 1 =− o di do hi or ho (1) 111 = + (Equation 25.3) yields f d o di h 111 1 1 h = += − o = 1 − o f do di do do hi do hi (2) Taking the reciprocal of Equation (2), we find that the focal length of the water drop is 1 f = do h 1− o hi do 3.0 cm = = −0.16 cm = ho 2.0 cm 1− 1− 0.10 cm hi 29. SSM REASONING We need to know the focal length of the mirror and can obtain it from the mirror equation, Equation 25.3, as applied to the first object: 1 1 1 1 1 + = + = d o1 d i1 14.0 cm –7.00 cm f or f = –14.0 cm According to the magnification equation, Equation 25.4, the image height hi is related to the c h object height ho as follows: hi = mho = – d i / d o ho . Chapter 25 Problems 1301 SOLUTION Applying this result to each object, we find that hi2 = hi1 , or F d Ih = F d Ih – – GJ GJ d d HK HK F IF I d h d =d G J GJ d H Kh K H i2 i1 o2 o1 o2 Therefore, o1 i1 o1 o1 i2 o2 o2 Using the fact that ho 2 = 2 ho1 , we have d i2 = d o2 F IF I = d F cm IFh I = – 0.250 d d h –7.00 G J h J G cm J 2 h J GK H G d 14.0 KK HK H H i1 o1 o1 o2 o1 o2 o2 o1 Using this result in the mirror equation, as applied to the second object, we find that 1 1 1 + = d o2 d i2 f or 1 1 1 + = d o2 – 0.250 d o2 –14.0 cm Therefore, d o2 = + 42 . 0 cm 30. REASONING a. You need a concave mirror to make the measurement. Since you must measure the image distance and image height of the tree, the image must be a real image. Only a concave mirror can produce a real image. The image distance of the sun is equal to the focal length of the mirror. The sun is extremely far away, so the light rays from it are nearly parallel when they reach the mirror. According to the discussion in Section 25.5, light rays near and parallel to the principal axis are reflected from a concave mirror and converge at the focal point. With numerical values for the focal length f of the mirror and the image distance di, the mirror equation (Equation 25.3) can be used to find the object distance do: 1 11 =− do f di 1302 THE REFLECTION OF LIGHT: MIRRORS b. The height ho of the tree is related to the height hi of its image by the magnification m; ho = hi/m (Equation 25.4). However, the magnification is given by the magnification equation (Equation 25.4) as m = −di/do, where di is the image distance and do is the object distance, both of which we know. SOLUTION a. The distance to the tree is given by the mirror equation as 1 11 1 1 =−= − do f di 0.9000 m 0.9100 m so do = 82 m b. Since ho = hi/m and m = −di/do, we have that ho = hi hi = m di − d o d = hi − o di Now hi = −0.12 m, where the minus sign has been used since the image is inverted relative to the tree (see Figure 25.18b). Thus, the height of the tree is d ho = hi − o di 82 m = 11 m = ( −0.12 m ) − 0.9100 m 31. REASONING AND SOLUTION To obtain the focal length, we will use the mirror equation. To do so, it is necessary to obtain values for the object and image distances. Since the image distance for virtual image 1 in Figure 25.20b is a negative quantity, – d i will be a positive quantity. Therefore, the distance between the readout devic...
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