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Principal points and image, two lenses

Short answer

Equivalent focal length F = 0.15 m. H1 is 0.05 m behind the first lens; H2 is 0.075 m in front of the second lens (0.025 m behind the first lens). The final image is 0.12 m beyond the second lens, real and inverted (magnification -0.3).

Two thin converging lenses of focal lengths 0.2m and 0.3m are placed coaxially 0.10m apart in air. An object is located 0.6m in front of the lens of smaller focal length. Find the position of the two principal points and that of the image.

Given: f₁ = 0.2 m (smaller), f₂ = 0.3 m, d = 0.10 m, object u = −0.6 m in front of the 0.2 m lens (first lens).

Equivalent focal length

1/F = 1/0.2 + 1/0.3 − 0.1/(0.2×0.3)
    = 5 + 3.333 − 1.667 = 6.667 m⁻¹
F   = 0.15 m

Principal points

α = F d / f₂ = 0.15 × 0.10 / 0.3 = 0.05 m   (H₁ is 5 cm right of L1)
β = F d / f₁ = 0.15 × 0.10 / 0.2 = 0.075 m  (H₂ is 7.5 cm left of L2)

H₂ is 0.10 − 0.075 = 0.025 m (2.5 cm) to the right of L1.

Position of the image

Lens by lens:

Lens 1: 1/v₁ = 1/0.2 − 1/0.6 = 3.333  ->  v₁ = +0.30 m
Lens 2: object I₁ is 0.30 − 0.10 = 0.20 m beyond L2, u₂ = +0.20 m
        1/v₂ = 1/0.3 + 1/0.2 = 8.333  ->  v₂ = +0.12 m
m = (0.30/−0.6)(0.12/0.20) = (−0.5)(0.6) = −0.30

Check by principal points: U = −(0.6 + 0.05) = −0.65 m; 1/V = 1/0.15 − 1/0.65 gives V = 0.195 m from H₂, i.e. 0.195 − 0.075 = 0.12 m from L2. This agrees.

Answer: F = 0.15 m; H₁ is 0.05 m behind the first lens; H₂ is 0.075 m in front of the second lens (0.025 m behind the first lens); the final image is 0.12 m (12 cm) beyond the second lens, real and inverted (m = −0.3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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