Question

An object is placed 50 cm in front of a converging lens having a focal length of magnitude 30 cm. A double-concave lens that has equal radii of curvature of 13 cm and an index of refraction 1.64 is placed 40 cm past the first lens.

a) what are the nature and the lateral magnification of the image?

b) where is the image formed?

Answer 1

for the first lens,
f1 = 30 cm
object distance, u1 = 50 cm
let v1 is the image distance.
use, 1/u1 + 1/v1 = 1/f1
1/v1 = 1/f1 - 1/u1
1/v1 = 1/30 - 1/50
v1 = 75 cm

magnification, m1 = -v1/u1

= -75/50

= -1.50

for second lens,
using lens makers formula,
1/f2 = (n - 1)*(1/R1 - 1/R2)
1/f2 = (1.64 - 1)*(1/(-13) - 1/13)

f2 = -10.16 cm
object distance, u2 = -(75 - 40)

= -25 cm
let v2 is the image distance.
use, 1/u2 + 1/v2 = 1/f2
1/v2 = 1/f2 - 1/u2
1/v2 = 1/(-10.16) - 1/(-25)
v2 = -17.1 cm
magnification, m2 = -v2/u2

= -(-17.1)/(-25)

= -0.684

a) The image virtual and upright

magnificationb, m = m1*m2

= -1.5*(-0.684)

= 1.026

b) The image is formed at 17.1 cm to the left of double concave lens.