farsightedness and nearsightedness?

Question

For this problem I know do (object distance) equals 23.3 cm and object height equals .23 cm. Now, I believe I'm supposed to find image height but I don't think I have enough information. What am I missing? Or do I have something wrong?

A person holds a book 25 cm in front of the effective lens of her eye; the print in the book is 2.3 mm high. If the effective lens of the eye is located 1.7 cm from the retina, what is the size (including the sign) of the print image on the retina?

I made it 23.3 because I was doing other problems that required subtracting a small amount from the object distance, but this problem was different so I confused myself. Thanks.

Answer 1

Why did the object distance (Do) become 23.3? It should be 25 cm. My solution is:

Do = 25 cm
Di = 1.7 cm
So = 0.23 cm
Si= ?

Di/Do = - Si/So
1.7/25 = - Si/0.23
Si = - 0.01564 cm

Answer 2

if L is the position of the lens,
R position of the retina ,
B the position of the book,
o the size of the object ,
i the size of the image,

you have the formula

i/o = LR/LB

i = o LR/Lb= 2.3*1.7/23.3 =0.168mm

as i remember the image is reversed

Answer 3

1