Solve Focal Length - Optics Step by Step

A solve-focal-length rearrangement scan holds the object distance fixed and varies image distance across rows, inferring the focal length from the rearranged lens equation f = uv / (u + v). The arithmetic is exact only for the stated object and image distances, focal lengths, magnifications, signed heights, and rows in the ideal thin-lens model. Real measurements add alignment, lens aberrations, finite aperture, thickness, screen/readout calibration, sign-convention bookkeeping, finite precision, and measurement uncertainty.

Use object and image distances to infer which focal length produced the ray geometry. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Use object and image distances to infer focus

Now suppose the object distance is fixed at 30 metres and the image screen is moved. Each screen position implies a different focal length.

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Rearrange for focal length

Adding the two reciprocals gives the focal reciprocal. Inverting that sum gives the focal length.

f=\frac{uv}{u+v}

Three image positions imply three lenses

Hold object distance and object height fixed at 4 metres. A farther image position requires a longer focal length, and it also increases the magnitude of the signed magnification.

\begin{array}{c|c|c|c|c}u&v&f&m&h_{\text{image}}\\30\ \text{m}&20\ \text{m}&12\ \text{m}&\tfrac{-2}{3}&\tfrac{-8}{3}\ \text{m}\\30\ \text{m}&30\ \text{m}&15\ \text{m}&-1&-4\ \text{m}\\30\ \text{m}&60\ \text{m}&20\ \text{m}&-2&-8\ \text{m}\\\end{array}

The inferred focus controls the ray bend

For the middle row, object and image distances are equal. The inferred focal length is half that distance, and the image is same-size inverted.

f=15\ \text{m}\qquad m=-1\qquad h_{\text{image}}=-4\ \text{m}