Question

The contacts worn by a farsighted person allow her to see objects clearly that are as close as 25.0 cm, even though her uncorrected near point is 91.0 cm from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of 208 cm from her eyes. (a) How far away is the poster actually located? (b) If the poster is 0.330 m tall, how tall is the image formed by the contacts?

Answer 1

Uncorrected Near point, = 62 cm

The object distance is =25cm

Finding Focal Length -
1/f=1/di + 1/do
1/f = -1/91 + 1/25
Solving for f
f = 34.47cm

Now,
di = - 208 cm
di is Neagtive because image is on the side of Object.
do = ?

1/f = 1/di + 1/do
1/34.47 = - 1/208 + 1/do
Solving for do
do = 29.57 cm
Far away is the poster actually located, do = 29.57 cm

(b)
From Lateral Magnification , M = hi/ho = - di/do
Where,
hi = ?
ho = 0.330 m
di = -208 cm
do = 29.57 cm
hi/0.330 = - (-208)/29.57
hi = 2.32 m
Height of image, hi = 2.32 m