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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?

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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

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