Question

The far point of a nearsighted person is 8.8 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 16.3 m away and 2.3 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

As we know that;

a)

To see distant objects clearly he needs a diverging lens,

= The power of the lens = p =1/f

= -1/8.8

= -0.1136 D

= The focal length of the lens he needs =-8.8m

= object distance=16.3 m

= image distance=?

Now, we use the lens equation,

= 1/v-1/u=1/f

= 1/v=1/f+1/u

= (-1/8.8) + (1/-16.3)

= v = - 5.71 cm

Then,

(b)

= magnification = v/u =5.71/16.3

= 0.35

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

So,

= size of the image = 0.35 * 2.3

= 0.805 m

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