Ramsden's eyepiece cardinal points

Take two thin lenses of equal focal length f separated by d = 2f/3, with L1 the field lens (facing the object) and L2 the eye lens.

Formulas for two thin lenses in air (distances positive to the right):

• Equivalent focal length: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂)
• First principal point H₁ from L1: α = F d / f₂ (to the right of L1)
• Second principal point H₂ from L2: β = F d / f₁ (to the left of L2)
• Nodal points coincide with principal points (same medium on both sides).

Step 1: equivalent focal length

1/F = 1/f + 1/f − (2f/3)/f²
    = 2/f − 2/(3f) = 4/(3f)
F   = 3f/4

Step 2: principal points

α = F d / f₂ = (3f/4)(2f/3)/f = f/2     H₁ is f/2 right of L1
β = F d / f₁ = f/2                      H₂ is f/2 left of L2

Since L2 is 2f/3 from L1, H₂ is at 2f/3 − f/2 = f/6 right of L1.

Step 3: focal points

F₂ (second focal point) from L2 = F − β = 3f/4 − f/2 = f/4  (right of L2)
F₁ (first focal point)  from L1 = α − F = f/2 − 3f/4 = −f/4 (f/4 left of L1)

 F1    L1    H2        H1     L2     F2
 |-----|-----|---------|------|------|
-f/4    0    f/6       f/2   2f/3   11f/12
        (distances from L1)

The focal points lie outside the lens system, f/4 from each lens, which is the reason a Ramsden eyepiece can be used with a cross-wire or scale placed at the first focal point.

Answer: F = 3f/4; H₁ is f/2 behind L1; H₂ is f/2 in front of L2 (f/6 behind L1); F₁ is f/4 in front of L1; F₂ is f/4 behind L2; nodal points coincide with H₁ and H₂.