Chapter 25 Solutions of Selected Exercises. 15. (a) The patient has to take them
off to see distant objects. This is because the positive contact lens will alter the ...
Chapter 25 Solutions of Selected Exercises 15.
(a) The patient has to take them off to see distant objects. This is because the positive contact lens will alter the person’s far point. (b) The lens is to form an image, of an object at 25 cm, at the near point (100 cm). do = 25 cm = 0.25 m,
di = -100 cm = -1.0 m (virtual image on object side).
1 1 1 1 1 P = f = d + d = 0.25 cm + -1.0 m = +3.0 D . o i 19.
First find his new near point.
do = 33 cm = 0.33 m,
1 P = f = +2.0 D.
1 1 1 1 − − do = +2.0 D 0.33 m = -1.0 D, so di = -1.0 m. Therefore the near point is 1.0 m. di = f To bring this near point to 25 cm, the power of the new lenses must be 1 1 1 P' = f' = 0.25 m + -1.0 m = +3.0 D .
35.
1 1 (a) f = P = 3.0 D = 0.333 m = 33.3 cm. With a near point of 25 cm, m = 1 +
25 cm 25 cm = 1 + 33.3 cm = 1.8× . f
10 cm (b) With a near point of 10 cm, m = 1 + 33.3 cm = 1.3× .
41.
From the thin-lens equation, di =
do f . do − f
The lateral magnification of the objective is di f 1 1 1 Mo = - d = = = = = −4.0×. f − do 1 − do/f 1− doD 1 − (5.0 × 10−3 m)(250 D) o Mtotal -100× = 25× . The total magnification is Mtotal = Mome, so me = M = -4.0× o 55.
(a) For greater magnification, the focal length of the objective should be as long as possible and fo the focal length of the eyepiece should be as short as possible because m = - f . So the answers e are: maximum magnification: 60.0 cm and 0.80 cm ; minimum magnification: 40.0 cm and 0.90 cm . fo1 60.0 cm (b) m1 = - f = 0.80 cm = - 75× , e1
40.0 cm m2 = - 0.90 cm = - 44× .
67.
(a) The eye obtain the maximum resolution (smallest minimum angle of resolution) for objects of blue color. This is because the minimum angle of resolution is proportional to wavelength, θmin =
1.22λ D , and blue has the shorter wavelength.
(b) θmin =
1.22λ 1.22(550 × 10−9 m) = = 9.6 × 10−5 rad . D 7.0 × 10−3 m
