Trace the upright virtual image of a diverging lens

Give the focal length its sign

The focal-magnitude control is positive, but this lesson uses a negative signed focal length. At magnitude 20 cm the model uses f = −20 cm. A ray incident parallel to the axis leaves spreading away from it; its backward extension passes through the incident-side focal point. The negative sign describes this optical behavior, not a negative physical length of glass.

At the default real-object distance dₒ = 30 cm, the image calculation is dᵢ = (−20 × 30)/(30 − (−20)) = −12 cm. Its location is between the object at x = −30 cm and the lens at zero. Magnification is −(−12)/30 = +0.4, so image height is +0.20 cm for the +0.50 cm object. It is upright, reduced and virtual.

Check the rays rather than discarding a negative result

The outgoing solid lines diverge. Their dashed backward extensions intersect at the predicted virtual image tip. The central ray continues through the lens center; the initially parallel ray bends away from the axis. The additional sampled ray gives the same intersection when extended backward. These lines test the signed equation geometrically.

A negative image distance is not an impossible answer. It specifies which side contains the virtual construction. It does not mean a screen behind the object will receive an image or that the lens reverses the transmitted light.

Prove the real-object pattern

Let F be the positive focal magnitude, so f = −F, and let dₒ be positive. Then dᵢ = −Fdₒ/(dₒ + F). The denominator and numerator magnitude are positive, so dᵢ is always negative. Its magnitude is smaller than both F and dₒ. The image lies between the lens and the incident-side focal point, and is closer to the lens than the object.

Magnification is m = F/(dₒ + F). It is positive and less than one, proving upright and reduced behavior for these real-object cases. These conclusions are scoped to a single diverging lens with a real object; an incoming converging bundle or a multi-lens system needs a different setup.

Capture dₒ = 10, 30 and 60 cm at F = 20 cm. The signed image distances are −6.666667, −12 and −15 cm; magnifications are 0.666667, 0.4 and 0.25. As the object moves farther away, the virtual image approaches the incident-side focal point while shrinking. At a finite object distance it has not reached that focal point exactly.

Compare two kinds of virtual image

A positive lens with the object inside its focal distance can give an upright enlarged virtual image. This negative lens with a real object gives an upright reduced virtual image. Thus “virtual” tells you about ray convergence and apparent origin, not whether the image is larger or smaller.

The plot’s vertical scale is exaggerated 16 times relative to horizontal. Use the reported height ratio for magnification. The model calculates linear magnification, not visual angular magnification, eyeglass prescriptions or image brightness.

Transfer with a different lens

For f = −15 cm, dₒ = 45 cm and hₒ = +0.80 cm, calculate dᵢ, m and hᵢ and place the virtual image relative to the lens and left focal point.

Check the diverging-lens transfer

dᵢ = −15 × 45/(45 + 15) = −11.25 cm. Magnification is +0.25 and image height is +0.20 cm. The virtual image is upright and reduced, between the lens at zero and the left focal point at −15 cm, closer to the lens than the object at −45 cm.

OpenStax: thin lenses explains the signed image convention. Continue to measuring focal length for a converging-lens screen experiment and a labeled synthetic alternative.