Skip to content

Question

A convex lens (\( f = 50 \, \text{cm} \)) and a concave lens (\( f = 25 \, \text{cm} \)) are in
contact. What is the effective focal length?

Options

Choose one · Correct answer highlighted

Explanation

**Lens formula** 1/f = 1/v - 1/u, f focal length (m), u object distance, v image distance, sign: u negative if object left of lens in Cartesian convention, magnification m = v/u, real image inverted m negative, virtual erect m positive. Power P = 1/f (diopters D), f in meters, +5 D means f=0.2 m=20 cm converging. f₁ = 50 cm , f₂ = -25 cm . (1/f) = (1/f₁) + (1/f₂) = (1/50) + (1/-25) = (1/50) - (2/50) = (1 - 2/50) = (-1/50) . f = -50 cm (diverging system). Substituting values gives -50 cm, which matches expected image position and magnification

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.