Lens Equation Inference Scans

Solve Object Distance

infer where the object was

Rearrange the lens equation to infer object distance from a known image distance and focal length. 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 image distance to infer object distance

Sometimes the image screen position is known first. Keep the image distance at 30 metres and ask what object distance would make each focal length land there.

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

Rearrange for the object distance

Move the image-distance reciprocal across the equation. The object distance is set by the remaining reciprocal budget.

1u=1f−1v⇒u=fvv−f\frac{1}{u}=\frac{1}{f}-\frac{1}{v}\qquad\Rightarrow\qquad u=\frac{fv}{v-f}

Three focal lengths need three object positions

Hold the image distance and object height fixed at 4 metres. A longer focal length leaves less reciprocal room for the object, so the object must sit farther from the lens.

fuvmhimage10 m15 m30 m−2−8 m15 m30 m30 m−1−4 m20 m60 m30 m−12−2 m\begin{array}{c|c|c|c|c}f&u&v&m&h_{\text{image}}\\10\ \text{m}&15\ \text{m}&30\ \text{m}&-2&-8\ \text{m}\\15\ \text{m}&30\ \text{m}&30\ \text{m}&-1&-4\ \text{m}\\20\ \text{m}&60\ \text{m}&30\ \text{m}&\tfrac{-1}{2}&-2\ \text{m}\\\end{array}
Inferring object distanceThe middle row is the checked ray diagram.FFlensobjectimage

The inferred row must match the rays

For the middle row, the object and image distances both sit at 30 metres. The magnification is -1, so a positive object height becomes an equal-size inverted image.

u=30 mv=30 mm=−1u=30\ \text{m}\qquad v=30\ \text{m}\qquad m=-1
Inferring object distanceThe checked rays intersect at the table's image distance.FFlensobjectimage