Question

Two converging lenses, each of focal length 14.9 cm, are placed 39.9 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed?

The image is located x cm  ---Location--- in front of the second lens.

What is the distance?

What is the magnification of the system?

Answer 1

here,

focal length of first lens, f1 = 14.9 cm
focal length of second lens, f2 = 39.9 cm
object distance, s = 30 cm

For First lens ,
from thin lens formula,
1/f1 = 1/s + 1/s'
(s' is image distance, s is object distance)
1/14.9 = 1/30 + 1/s'
s' = 29.60 cm

For Second Lens,
s' = s = 29.60 cm
again by using thin lens equation,
1/39.9 = 1/29.60 + 1/s'

s' = -114.664 cm
(The negative sign implies that the second and final image is formed in front of the second lens)

The magnification of the first lens
m1 = -s'/s = - 30.0 /30.0
m1 = - 1
(The negative sign says that the image of the first lens is inverted)

The magnification of the second lens
m2 = -s'/s = -(-114.664)/29.60
m2 = 3.87

Net magnification of configuration,
m = m1*m2 = 3.87 *- 1
m = -3.87
(The negative sign says that the final image is inverted)