Question

he far point of a nearsighted person is 6.3 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 19.3 m away and 1.7 m high. (a) When she looks through the contacts at the tree, what is its image distance? (b) How high is the image formed by the contacts?

Answer 1

(a)
far point = 6.3m
to see distant objects clearly he needs a diverging lens
the power of the lens = p =1/f= -1/6.3 = -0.1587D

the focal length of the lens he needs =-6.3m
object distance=19.3m
image distance=?

we have the lens equation
1/v-1/u=1/f

1/v=1/f+1/u= (-1/6.3) + (1/-19.3)= -4.75m, which is your answer for part a.


(b) How high is the image formed by the contacts?

magnification = v/u =4.75/19.3 = .246

size of the object * magnification = size of the image.

so size of the image = 0.246 * 1.7 = 0.4182m, the answer for part b.