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Question

A double convex lens has radii of curvature \( 30 \, \text{cm} \) and \( -30 \, \text{cm} \) with
refractive index \( 1.6 \). 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 = 1.6 , R₁ = 30 cm , R₂ = -30 cm . (1/f) = (1.6 - 1) ( (1/30) - (1/-30) ) = 0.6 ( (1/30) + (1/30) ) = 0.6 × (2/30) = (1.2/30) = (1/25) . f = 25 cm . Substituting values gives 25 cm, which matches expected image position and

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