The eyepiece of a telescope is a system of two biconvex lenses L1 and L2, of focal length 30 cm and 20 cm respectively, with a separation of 2 cm between them. Lens L2 faces the eye. This eyepiece is to be replaced by a single bi-convex lens L. Which one among the following is the correct combination of the focal length (F in cm) of the lens L and its distance (δ in cm) from the position of lens L1 ?
Equivalent focal length F:
1/F = 1/f₁ + 1/f₂ – d/(f₁f₂)
1/F = 1/30 + 1/20 – 2/(30 * 20)
1/F = (2 + 3)/60 – 2/600
1/F = 5/60 – 1/300
1/F = 1/12 – 1/300
1/F = (25 – 1)/300 = 24/300 = 2/25
F = 25/2 = 12.5 cm.
This confirms F = 12.5 cm, eliminating options C and D.
The question asks for the distance (δ) of the equivalent lens L from the position of lens L₁. For a system of lenses, the equivalent single lens producing the same effect as the combination is placed at one of the principal planes. Since L₂ faces the eye, the light exits towards the eye after passing through L₂. The equivalent lens is typically placed at the second principal plane (H₂) of the system, as measured from the lens where light exits (L₂). The position of H₂ is measured relative to L₂.
The distance of the second principal plane H₂ from L₂ is given by β = -d * (F / f₁).
β = -2 cm * (12.5 cm / 30 cm) = -2 * (12.5/30) = -25/30 = -5/6 cm.
The negative sign indicates that H₂ is to the left of L₂ (towards L₁).
The distance of H₂ from L₁ is the distance L₁-L₂ plus the distance L₂-H₂.
Distance L₁-H₂ = d + β = 2 cm + (-5/6 cm) = 2 – 5/6 = (12 – 5)/6 = 7/6 cm.
7/6 cm ≈ 1.1667 cm.
The distance δ from L₁ is approximately 1.17 cm.
This matches option A (F = 12.5 and δ = 1.17).