E40-35 If the two lenses “pass” the same amount of light then the solid angle subtended by each lens as seen from the respective focal points must be the same. If we assume the lenses have the same round shape then we can write this as do /f o = de /f e . Then de fo = = mθ , do fe or de = (72 mm)/36 = 2 mm. E40-36 (a) f = (0.25 m)/(200) ≈ 1.3 mm. Then 1/f = (n − 1)(2/r) can be used to find r; r = 2(n − 1)f = 2(1.5 − 1)(1.3 mm) = 1.3 mm. (b) The diameter would be twice the radius. In effect, these were tiny glass balls. E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b) i = f = 2.5 cm, even though f 0 6= f . Solving, 1 1 1 1 = + = . f (36 cm) (2.5 cm) 2.34 cm (b) The effective radii of curvature must have decreased. E40-38 (a) s = (25 cm) − (4.2 cm) − (7.7 cm) = 13.1 cm. (b) i = (25 cm) − (7.7 cm) = 17.3 cm. Then 1 1 1 1 = − = o (4.2 cm) (17.3 cm) 5.54 cm. The object should be placed 5.5 − 4.2 = 1.34 cm beyond F1 . (c) m = −(17.3)/(5.5) = −3.1. (d) mθ = (25 cm)/(7.7 cm) = 3.2. (e) M = mmθ = −10. E40-39 Microscope magnification is given by Eq. 40-33. We need to first find the focal length of the objective lens before we can use this formula. We are told in the text, however, that the microscope is constructed so the at the object is placed just beyond the focal point of the objective lens, then f ob ≈ 12.0 mm. Similarly, the intermediate image is formed at the focal point of the eyepiece, so f ey ≈ 48.0 mm. The magnification is then m=
−s(250 mm) (285 mm)(250 mm) =− = 124. f ob f ey (12.0 mm)(48.0 mm)
A more accurate answer can be found by calculating the real focal length of the objective lens, which is 11.4 mm, but since there is a huge uncertainty in the near point of the eye, I see no point in trying to be more accurate than this. P40-1 The old intensity is Io = P/4πd2 , where P is the power of the point source. With the mirror in place there is an additional amount of light which needs to travel a total distance of 3d in order to get to the screen, so it contributes an additional P/4π(3d)2 to the intensity. The new intensity is then In = P/4πd2 + P/4π(3d)2 = (10/9)P/4πd2 = (10/9)Io .
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