Lens Equation and Magnification

Solve Image Distance

clean reciprocals

Use the lens equation with clean distances to solve for where the image forms.

optics The clean numbers are chosen so the reciprocal arithmetic lands exactly on a whole-number image distance.

Example

Use the lens equation with clean distances to solve for where the image forms. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start with focal length and object distance

Use focal length 10 metres and object distance 15 metres. The image distance is the unknown.

f=10 mu=15 mf = 10\ \text{m}\qquad u = 15\ \text{m}
Solving image distanceChecked rays meet at the solved image point.FFlensobjectimage

A nearby object makes a farther image

With the focal length fixed, compare three object placements. Outside the focus, moving closer pushes the real image farther away; inside the focus, the sign changes and the image becomes virtual.

fuvm10 m20 m20 m−110 m15 m30 m−210 m5 m−10 m2\begin{array}{c|c|c|c}f&u&v&m\\10\ \text{m}&20\ \text{m}&20\ \text{m}&-1\\10\ \text{m}&15\ \text{m}&30\ \text{m}&-2\\10\ \text{m}&5\ \text{m}&-10\ \text{m}&2\\\end{array}
Solving image distanceThe table compares distances while this ray case is checked.FFlensobjectimage

Subtract the reciprocals

The focal reciprocal minus the object reciprocal gives the image reciprocal.

1v=110 m−115 m=130 1/m\frac{1}{v} = \frac{1}{10\ \text{m}} - \frac{1}{15\ \text{m}} = \tfrac{1}{30}\ 1/\text{m}

Invert the reciprocal

The reciprocal is one over 30 metres, so the image distance is 30 metres on the far side of the lens.

v=30 mv = 30\ \text{m}
Solving image distanceChecked rays meet at the solved image point.FFlensobjectimage