How this calculation works
Lens curvature and index contrast determine paraxial focusing power. A curved transparent object need not focus strongly if its index closely matches the surrounding material. This calculation derives focal length from two signed spherical surface radii. It answers a manufacturing-geometry question rather than an image-position question. Once a focal length is known, object distance is still needed to determine where an image forms.
1/f = (nlens/nmedium − 1)(1/R1 − 1/R2)A worked example
A lens index of 1.5 in air, with radii +100 and −100 mm, has reciprocal focal length 0.01 per millimetre. Its focal length is 100 mm, corresponding to 10 inverse metres. If the surrounding index becomes 1.25 while the geometry remains unchanged, relative index falls from 1.5 to 1.2. The focal length increases to 250 mm, showing why the medium cannot be ignored.
Good to know
This is the thin spherical lensmaker relation. Centre thickness, separation between refracting surfaces and principal-plane displacement are not included. No wavelength dispersion, surface tolerance or aberration model is supplied. The refractive indices are user inputs, so the page does not identify glass types or look up material properties.
Reference: OpenStax — lensmaker equation (opens in a new tab) ↗
What this calculation answers
Lens curvature and index contrast determine paraxial focusing power. A curved transparent object need not focus strongly if its index closely matches the surrounding material. This calculation derives focal length from two signed spherical surface radii.
It answers a manufacturing-geometry question rather than an image-position question. Once a focal length is known, object distance is still needed to determine where an image forms.
The Thin lens real and virtual image magnification calculator needs both focal length and object distance. Lens curvature alone does not specify the image location.
Use the right input conventions
Light travels left to right. A surface radius is positive when its centre of curvature lies to the right of that surface and negative when the centre lies to the left. The two surfaces of a biconvex lens therefore normally have opposite signs.
Use the same external medium at both faces. Neither radius may be zero; a very large radius can approximate a plane face within the entered range. Refractive indices must refer to the same wavelength and conditions.
Interpret the result
The result labelled reciprocal focal length is 1/f in inverse metres. In a surrounding medium other than air, this is not the same convention as the reduced optical power n/f used in some optics texts.
When the lens and medium indices match, or the signed curvatures cancel, the ideal element is afocal. The page reports that state explicitly instead of assigning an infinite numeric focal length.
Work through the example
A lens index of 1.5 in air, with radii +100 and −100 mm, has reciprocal focal length 0.01 per millimetre. Its focal length is 100 mm, corresponding to 10 inverse metres.
If the surrounding index becomes 1.25 while the geometry remains unchanged, relative index falls from 1.5 to 1.2. The focal length increases to 250 mm, showing why the medium cannot be ignored.
Where the model stops
This is the thin spherical lensmaker relation. Centre thickness, separation between refracting surfaces and principal-plane displacement are not included.
No wavelength dispersion, surface tolerance or aberration model is supplied. The refractive indices are user inputs, so the page does not identify glass types or look up material properties.
A few questions, answered
Why do equal convex curvatures use opposite signs?
Their centres of curvature lie on opposite sides under the left-to-right radius convention.
What changes in the comparison?
Keep curvature fixed and immerse the lens in a higher-index common medium.