Final image of convex and concave lenses

Use the Cartesian sign convention with 1/v − 1/u = 1/f.

Given: convex lens L1: f₁ = +50 cm; concave lens L2: f₂ = −50 cm; separation d = 10 cm; object u₁ = −20 cm.

Step 1: first lens (convex)

1/v₁ = 1/f₁ + 1/u₁ = 1/50 − 1/20 = −0.03
v₁ = −33.33 cm
m₁ = v₁/u₁ = −33.33/−20 = +1.667

The object lies inside the focal length of L1, so I₁ is virtual, erect and magnified, 33.33 cm to the left of L1 (on the object side).

Step 2: second lens (concave). The distance of I₁ from L2 is 33.33 + 10 = 43.33 cm on its left side, so u₂ = −43.33 cm.

1/v₂ = 1/f₂ + 1/u₂ = −1/50 − 1/43.33 = −0.02 − 0.02308 = −0.04308
v₂ = −23.21 cm
m₂ = v₂/u₂ = −23.21/−43.33 = +0.536

Step 3: net magnification

m = m₁ m₂ = 1.667 × 0.536 = +0.893

 I1      O     I     L1     L2
 |       |     |     |      |
-33.3   -20  -13.2   0     +10     (cm, measured from L1)

The final image is at −23.21 cm from L2, i.e. at 10 − 23.21 = −13.21 cm from L1. It lies on the same side as the object, between the object and L1.

Answer: the final image is virtual and erect, formed 23.2 cm in front of the concave lens (13.2 cm in front of the convex lens, on the object side); its size is 0.89 times the object (m ≈ +0.89).