Question

A farsighted man uses contact lenses with a refractive power of 2.20 diopters. Wearing the contacts, he is able to read books held no closer than 0.285 m from his eyes. He would like a prescription for eyeglasses to serve the same purpose. What is the correct prescription for the eyeglasses if the distance from the eyeglasses to his eyes is 0.030 m? in dipoters ??? not 1.67

A diverging lens (f1 = −13.5 cm) is located 23.0 cm to the left of a converging lens (f2 = 36.5 cm). A 3.0-cm-tall object stands to the left of the diverging lens, exactly at its focal point. What is the height of the final image (including proper algebraic sign)? in cm ??? not 51.97 cm

Answer 1

For diverging lens :

f1 = focal length = - 13.5 cm

do = object distance = 13.5 cm

di = image distance = ?

ho = height of the object = 3 cm

hi = height of the image

Using the lens equation

1/f1 = 1/do + 1/di

1/(- 13.5) = 1/(13.5) + 1/di

di = - 6.75 cm

Using the equation for magnification

hi/ho = - di/do

hi/3 = - (- 6.75)/13.5

hi = 1.5 cm

D = distance between the two lenses = 23 cm

For converging lens :

f2 = focal length = 36.5 cm

d'o = object distance = D + hi = 23 + 6.75 = 29.75 cm

d'i = image distance = ?

h'o = height of the object = 1.5 cm

h'i = height of the image

Using the lens equation

1/f2 = 1/d'o + 1/d'i

1/(36.5) = 1/(29.75) + 1/d'i

d'i = - 160.87 cm

Using the equation for magnification

h'i/h'o = - d'i/d'o

h'i/1.5 = - (- 160.87)/29.75

h'i = 8.11 cm