Practice question
Question
A double convex lens of refractive index \( 1.5 \) has radii of curvature \( 12 \, \text{cm} \) and \(
-12 \, \text{cm} \). What is its focal length?
Explanation
**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.5 , R₁ = 12 cm , R₂ = -12 cm . (1/f) = (1.5 - 1) ( (1/12) - (1/-12) ) = 0.5 ( (1/12) + (1/12) ) = 0.5 × (2/12) = (1/12) . f = 12 cm . Substituting values gives 12 cm, which matches expected image position and magnification from
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.