Practice question
Question
A double convex lens of refractive index \( 1.55 \) has radii of curvature \( 22 \, \text{cm} \) and \(
-22 \, \text{cm} \). What is its focal length?
Explanation
**Lens maker's formula** 1/f = (n-1)(1/R₁ - 1/R₂), n refractive index, R₁,R₂ radii of curvature (m), sign convention R positive if surface convex towards incident light. For double convex R₁=12 cm, R₂=-12 cm, n=1.5, 1/f=(0.5)(1/12 -1/(-12))=(0.5)(2/12)=1/12, f=12 cm, converging. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.55 , R₁ = 22 cm , R₂ = -22 cm . (1/f) = (1.55 - 1) ( (1/22) - (1/-22) ) = 0.55 ( (1/22) + (1/22) ) = 0.55 × (2/22) = (1.1/22) = (1/20) . f = 20 cm . Substituting values gives 20 cm, which matches
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