In: Physics
A converging lens has a focal length of 9.0 cm. Locate the images for the object distances of (a) 20.0 cm, (b) 10.0 cm, and (c) 5.00 cm, if they exist. For each case, state whether the image is real or virtual, upright or inverted, and find the magnification.
Given that :
focal length of converging lens, f = 9 cm
(a) At object distance, d0 = 20 cm
using a lens formula, 1 / d0 + 1 / di = 1 / f
or 1 / di = 1 / f - 1 / d0
inserting the values in above eq.
1 / di = 1 / (9 cm) - 1 / (20 cm)
1 / di = (11 / 180) cm
di = 16.3 cm
"positive image" which means real image formed to the right of the lens by a real object.
magnification, m1 = - di / d0
m1 = - (16.3 / 20) cm
m1 = - 0.815
(b) At object distance, d0 = 10 cm :
1 / di = 1 / f - 1 / d0
1 / di = 1 / (9 cm) - 1 / (10 cm)
1 / di = (1 / 90) cm
di = 90 cm
"positive image" which means real image formed to the right of the lens by a real object.
magnification, m2 = - di / d0
m2 = - (90 / 10) cm
m2 = - 9
(c) At object distance, d0 = 5 cm :
1 / di = 1 / f - 1 / d0
1 / di = 1 / (9 cm) - 1 / (5 cm)
1 / di = (- 4 / 45) cm
di = - 11.2 cm
"negative image" which means virtual image formed to the left of the lens by a real object.
magnification, m2 = - di / d0
m2 = - (-11.2 / 5) cm
m2 = 2.24