# Optics Question

1. Apr 9, 2007

### cheechnchong

1. The problem statement, all variables and given/known data

A student is reading a lecture written on a blackboard. Contact lenses in her eyes have a refractive power of 57.50 diopters; the lens-to-retina distance is 1.750 cm. (a) How far (in meters) is the blackboard from her eyes? (b) If the writing on the blackboard is 5.00 cm high, what is the size of the image on her retina?

2. Relevant equations

1/f = 1/di + 1/do

3. The attempt at a solution

(a) I know that once you get f (meters) using the given diopters measurement, you can use that to get an image. f = .0174 m...and then i get di = 0.328 m. (b) i substract 5 cm from 1.75 cm to get object distance...then i use f = .0174 m...finally, i plug in the equation and get di = 27 m.

I am not sure if I did this right because the negative and positive stuff really bother me. anyways, help me out if you see something wrong and let me know if i "might" right too! thanks.

2. Apr 9, 2007

### daniel_i_l

(a) Here you have to find do, the distance between the object and the lens. di is given so once you have f you just use you're equation to get do. Are you confusing do and di? Other than that it looks like you did it right (i didn't check the math though)
(b) I'm not sure what you did here. You have do, di and ho already, what equation can you use to get hi?

3. Apr 10, 2007

### cheechnchong

hey i appreciate the help...

i resolved (b) using the do, di, and ho

m = - (0.33)/ (.0175) = -18.9 m

and then i plug it into m = hi/ho

hi = (-18.9 m)(.05 m) = -0.945 m

do you think i approached this part right??

4. Apr 10, 2007

### daniel_i_l

Almost!
But it's a little confusing because you're using di for the distance of the object which is really do. So when you wrote - (0.33)/ (.0175) that really is
-do/di and it should be -di/do (di = 0.175 not do!)