Practice question
Question
A convex lens (\( f = 20 \, \text{cm} \)) and a concave lens (\( f = 40 \, \text{cm} \)) are in
contact. What is the effective focal length?
Explanation
**Telescope resolving power** depends on objective aperture, magnification limited by objective diffraction, terrestrial telescope adds erecting lens, Galilean uses diverging eyepiece for erect image, reflecting telescope uses concave mirror objective avoiding chromatic aberration. f₁ = 20 cm , f₂ = -40 cm . (1/f) = (1/f₁) + (1/f₂) = (1/20) + (1/-40) = (2 - 1/40) = (1/40) . f = 40 cm (converging system). Substituting values gives 40 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.