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 75.0 cm from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of 162 cm from her eyes. (a) How far away is the poster actually located? (b) If the poster is 0.310 m tall, how tall is the image formed by the contacts?

Answer 1

If the near point is 75.0 cm, then d_i = – 75.0 cm, and d_o = 25.0 cm. Using the thin-lens equation, we find that the focal length of the correcting lens is

$\small f=\frac{d_od_i}{d_o+d_i}=\frac{(25.0cm)(-75.0cm)}{(25.0cm)+(-75.0cm)}=+37.5cm$

a) The distance $\small d_o^'$ to the poster can be obtained as follows:

$\small \frac{1}{d'_o}=\frac{1}{f}-\frac{1}{d_i'}=\frac{1}{37.5cm}-\frac{1}{-162cm}$ or   $\small d'_o=30.45cm$

b) The image size is

$\small h_i=h_o\left (- \frac{d'_i}{d'_o} \right )=(0.310m)\left (-\frac{-162cm}{30.45cm} \right )=1.649m$