A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 59.82 diopters, and the lens-to-retina distance is 1.686 cm. (a) How far is the blackboard from her eyes? (b) If the writing on the blackboard is 4.81 cm high, what is the height of the image on her retina?
Physics - drwls, Sunday, March 24, 2013 at 6:22am
59.82 diopters refractive power means that 1/f = 59.82 m-1 = 0.5982 cm^-1
Her eye's focal length is thus f = 1.67 cm
Let Do be the object (blackboard) distance.
(a) 1/1.686 + 1/Do = 0.5982
1/Do = 5.07*10^-3
Do = 196 cm
(b) image height on retina:
4.81 cm*(1.686/196) = 0.041 cm