Lensmaker's Equation

Question

An equiconvex lens is ground from glass of refractive index 1.5, and each of its two spherical surfaces has a radius of curvature equal to 20 cm. Determine the focal length of this lens in air.

RankGuru · Accepted Answer

The lensmaker's equation predicts a thin lens focal length from the glass index and the surface curvatures: $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$. For an equiconvex lens the first surface bulges toward the object, so $R_1=+20$ cm, while the second surface curves the opposite way, giving $R_2=-20$ cm. With $n=1.5$, $\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}ight)=0.5	imes\left(\frac{1}{20}+\frac{1}{20}ight)=0.5	imes\frac{2}{20}=0.05$, so $f=20$ cm. The equation embodies two independent influences on focusing power: the material term $(n-1)$ measures how strongly the glass slows light relative to its surroundings, and the geometric term $\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$ measures the net curvature of the two surfaces. For an equiconvex lens both surfaces curve outward symmetrically, so their contributions add rather than cancel, doubling the effective curvature compared with a plano-convex lens of the same radius. The value 10 cm is wrong because it forgets that $R_2$ is negative, halving the bracket. The value 40 cm is wrong as it uses only one surface's curvature. The value 15 cm has no consistent derivation from the given data. This NCERT relation links manufacturing geometry to optical power and underlies all lens design. A sign check confirms a converging lens emerges with positive focal length when both bulging surfaces are accounted for.

Lensmaker's Equation

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