Practice question
Question
A double convex lens of refractive index \( 1.5 \) has radii of curvature \( 18 \, \text{cm} \) and \(
-18 \, \text{cm} \). 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 = 1.5 , R₁ = 18 cm , R₂ = -18 cm . (1/f) = (1.5 - 1) ( (1/18) - (1/-18) ) = 0.5 ( (1/18) + (1/18) ) = 0.5 × (2/18) = (1/18) . f = 18 cm . Substituting values gives 18 cm, which matches expected image
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