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Question

A double convex lens of refractive index \( 1.55 \) has both radii of curvature equal to \( 25 \,
\text{cm} \). 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 = 1.55 , R₁ = 25 cm , R₂ = -25 cm . (1/f) = (1.55 - 1) ( (1/25) - (1/-25) ) = 0.55 ( (1/25) + (1/25) ) = 0.55 × (2/25) = (1.1/25) . f = (25/1.1) ≈ 22.73 cm . Substituting values gives 22.7 cm, which matches expected image position and

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