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Image by two lenses 30 cm and 40 cm

Short answer

First lens: v₁=120 cm, m₁=-3. For second lens, u₂=+100 cm (virtual), v₂=28.6 cm, m₂=0.286. Total magnification -0.86. Final image is 28.6 cm right of second lens, real, inverted, diminished.

Two thin converging lenses of focal lengths 30 cm and 40 cm respectively are placed co-axially in air separated by a distance of 20 cm. An object is placed 40 cm in front of the first lens. Find the position and nature of the image.

Use the lens formula 1/v − 1/u = 1/f with the Cartesian sign convention (distances measured from the lens, positive in the direction of light).

Given: f₁ = +30 cm, f₂ = +40 cm, separation d = 20 cm, object at u₁ = −40 cm.

 O                L1        L2
 |------40cm-----|---20cm---|

Step 1: image by the first lens

1/v₁ = 1/f₁ + 1/u₁ = 1/30 − 1/40 = 1/120
v₁ = +120 cm   (image I₁, 120 cm right of L1)
m₁ = v₁/u₁ = 120/(−40) = −3

Step 2: I₁ is the object for the second lens. L2 is 20 cm from L1, so I₁ lies 120 − 20 = 100 cm beyond L2. It is a virtual object, u₂ = +100 cm.

1/v₂ = 1/f₂ + 1/u₂ = 1/40 + 1/100 = 0.035
v₂ = +28.57 cm
m₂ = v₂/u₂ = 28.57/100 = 0.2857

Step 3: total magnification

m = m₁ × m₂ = (−3)(0.2857) = −0.857

Check by the equivalent-lens method: 1/F = 1/30 + 1/40 − 20/(30×40) = 1/24, so F = 24 cm. The principal points are 12 cm right of L1 and 16 cm left of L2. The object is 52 cm from H₁, so the image is at 44.57 cm from H₂, which is 28.57 cm from L2. This agrees.

Answer: the final image is 28.6 cm to the right of the second lens; it is real, inverted and diminished (m ≈ −0.86).

✓ Answer checked by Claude Opus 5.5 (high effort)

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