Human Eye Optics: Near-Sightedness, Tree Image Distance & Height

[runfast220]

Homework Statement

The far point of a nearsighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 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?

Homework Equations

m= hi/ho=-di/do
1/do + 1/di = 1/f

The Attempt at a Solution

I'm not sure how to start the problem. Can someone point me in the right direction?

 

[tiny-tim]

Hi runfast220!

The far point of a nearsighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 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?

The contact is a lens that makes an object at ∞ have an image at 6.0 m.

So calculate its f, and then apply that to an object at 18.0 m.

 

[runfast220]

ok so...

1/f = 1/infin + 1/6
so f=6.0

1/di = 1/6 - 1/18 = .11
di = 9.0 m

hi/ho = - di/do
hi/2 = -9/18

hi = -1.0 m

does that look right?

 

[tiny-tim]

Hi runfast220!

… does that look right?

Nooo … you've got the tree focussing at 9.0 m, which is too far away for the nearsighted person to see it!

Hint: is the contact a converging or a diverging lens?

 

[runfast220]

Its a converging lens so does that make the di a virtual image thus making it -9.0m?

 

[tiny-tim]

No, it's a diverging lens …

a converging lens wold make parallel lines (from ∞) come closer, but these contacts make them diverge, so that they appear to come from 6.0 m in front of the person.

And (from the PF Library on lens …)

f is positive for converging (eg. biconvex and plano-convex) lenses, and negative for diverging (eg. biconcave and plano-concave) lenses.