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 81.0 cm from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of 237 cm from her eyes. (a) How far away is the poster actually located? (b) If the poster is 0.400 m tall, how tall is the image formed by the contacts?

Answer 1

Given

   the near point is 81.0 cm then d_i = -81 cm and d_o = 25 cm

from the relation 1/d_o +1/d_i = 1/f

       1/f = 1/25-1/81 ====> f = 36.16 cm

a. When she is looking at a poster, the contacts form an image of the poster at a distance of 237 cm from her eyes is

  

   1/d1_o = 1/f - 1/d1_i
       = 1/36.16 - 1/(-237)

   d1_0 = 31.373 cm
  

b) If the poster is 0.400 m tall, how tall is the image formed by the contacts

from lateral magnification m = y'/y = -d1_i/d1_o

               y' = -(-277/31.373)(0.4) m

               y ' = 3.532 m

  
the height of the image is y' = 3.532 m