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Image and magnification, 3 cm and 4 cm lenses

Short answer

First lens produces v1 = 12 cm (magnification m1 = −3). The image serves as a virtual object for the second lens at u2 = +10 cm, giving v2 = 2.86 cm (m2 = 0.2857). Net lateral magnification m = −0.857; the final image is real, inverted, and 2.86 cm right of the second lens.

Two thin converging lenses of focal length 3cm and 4cm respectively are placed coaxially in air and separated by distance of 2cm. An object is placed 4cm in front of the first lens. Find the position and nature of the image and its lateral magnification.

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

Given: f₁ = +3 cm, f₂ = +4 cm, d = 2 cm, u₁ = −4 cm.

Step 1: first lens

1/v₁ = 1/3 − 1/4 = 1/12
v₁ = +12 cm            (real image I₁, 12 cm right of L1)
m₁ = v₁/u₁ = 12/(−4) = −3

Step 2: second lens. I₁ is 12 − 2 = 10 cm beyond L2, a virtual object: u₂ = +10 cm.

1/v₂ = 1/f₂ + 1/u₂ = 1/4 + 1/10 = 7/20
v₂ = +2.857 cm
m₂ = v₂/u₂ = 2.857/10 = 0.2857

Step 3: net lateral magnification

m = m₁ × m₂ = (−3)(0.2857) = −0.857

 O --4cm-- L1 ---2cm--- L2 -2.86cm- I
   (object)  I1 at 12 cm from L1 (virtual object for L2)

Answer: the final image is 2.86 cm to the right of the second lens; it is real and inverted; the lateral magnification is −0.857 (the image is 0.857 times the size of the object).

✓ Answer checked by Claude Opus 5.5 (high effort)

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