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Question

A lens has a power of \( +3 \, \text{D} \). What is its focal length in centimeters?

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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. Power: P = (1/f) (in meters). P = +3 D ⇒ 3 = (1/f) ⇒ f = (1/3) ≈ 0.333 m ≈ 33.3 cm . Substituting values gives 33 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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