p35_035.pdf

inthethinlensequationtoobtain1 /i =1 /f ,where f is the focal length of the

relaxed eﬀective lens. Thus, i = f =2 . 50cm. When the eye focuses on closer objects, the image

distance i remains the same but the object distance and focal length change. If p is the new object

distance and f is the new focal length, then

∞

p + 1

i = 1

We substitute i = f and solve for f :

f =

f + p = (40 . 0cm)(2 . 50cm)

40 . 0cm+2 . 50cm =2 . 35 cm .

(b) Consider the lensmaker’s equation

f =( n

−

1) 1

r 1 −

r 2

where r 1 and r 2 are the radii of curvature of the two surfaces of the lens and n is the index of

refraction of the lens material. For the lens pictured in Fig. 35-34, r 1 and r 2 have about the same

magnitude, r 1 is positive, and r 2 is negative. Since the focal length decreases, the combination

(1 /r 1 )

−

(1 /r 2 ) must increase. This can be accomplished by decreasing the magnitudes of both

radii.

35. (a) When the eye is relaxed, its lens focuses far-away objects on the retina, a distance i behind the

lens. We set p =