Lens Maker's Formula

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A biconvex lens is ground from glass of refractive index 1.5 with both surfaces having a radius of magnitude 20 cm, so what is its focal length in air?

RankGuru · Accepted Answer

The lens maker's formula in NCERT gives the focal length from the glass index and the two surface radii as $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$, where the Cartesian sign convention fixes the signs of the radii. For a biconvex lens the first surface bulges toward the incoming light with $R_1=+20$ cm, while the second surface curves the other way with $R_2=-20$ cm. Substituting, $\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}ight)=0.5	imes\left(\frac{1}{20}+\frac{1}{20}ight)=0.5	imes\frac{2}{20}=0.5	imes0.1=0.05$, so $f=20$ cm. The value 10 cm is wrong because it forgets the $(n-1)$ factor of 0.5 and doubles the result. The value 40 cm is wrong because it uses only one surface, dropping the second curvature. The value 15 cm has no consistent derivation and is a distractor. As stated in NCERT Class 12, Chapter 9, a symmetric biconvex lens with equal radii and index 1.5 has a focal length equal to that common radius. A magnitude check confirms $f=20$ cm equals the surface radius, exactly as expected for this classic symmetric case.

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