How this calculation works
A thin lens can form either a real image beyond the lens or a virtual image on the object side. The focal length and object distance determine which case applies, while object height sets the image scale. This page solves for signed image distance and transverse magnification together. Keeping the signs visible makes it possible to distinguish an inverted real image from an upright virtual image instead of reporting only an unsigned distance.
1/f = 1/u + 1/v; magnification = −v/uA worked example
A 50 mm converging lens viewing an object 150 mm away gives an image distance of 75 mm. Magnification is −0.5, so a 10 mm upright object has a −5 mm image height. Moving the same object inside the focal length changes the sign of the image distance. At 25 mm object distance, the virtual image is 50 mm before the lens and its upright magnification is two.
Good to know
The model is paraxial and treats the lens as one plane in a common surrounding medium. It does not locate the principal planes of a thick or internally focused photographic lens. Aberrations, finite aperture, diffraction and focus uncertainty are excluded. Use the sign conventions before comparing with a textbook that measures distances using a different coordinate convention.
What this calculation answers
A thin lens can form either a real image beyond the lens or a virtual image on the object side. The focal length and object distance determine which case applies, while object height sets the image scale.
This page solves for signed image distance and transverse magnification together. Keeping the signs visible makes it possible to distinguish an inverted real image from an upright virtual image instead of reporting only an unsigned distance.
The Lensmaker signed curvature and surrounding medium calculator derives a focal length from curvature and refractive-index contrast; this page instead starts with focal length to locate an image.
Use the right input conventions
Enter a positive real-object distance measured from the ideal lens plane. A converging lens has positive focal length; a diverging lens has negative focal length. The image-distance convention is positive on the outgoing side.
All three dimensional inputs use millimetres. Zero focal length has no meaning in this model and is rejected. An object exactly at the converging focal plane produces collimated output, displayed as an image at infinity rather than an enormous finite number.
Interpret the result
Negative magnification means the image is inverted relative to the upright object. Its absolute value gives the linear size ratio, which is different from an instrument’s angular magnification. A negative image height therefore carries orientation information.
Moving an object close to the focal plane makes image distance highly sensitive to small input changes. The displayed digits are arithmetic precision, not a claim that a real lens can be focused or measured to that accuracy.
Work through the example
A 50 mm converging lens viewing an object 150 mm away gives an image distance of 75 mm. Magnification is −0.5, so a 10 mm upright object has a −5 mm image height.
Moving the same object inside the focal length changes the sign of the image distance. At 25 mm object distance, the virtual image is 50 mm before the lens and its upright magnification is two.
Where the model stops
The model is paraxial and treats the lens as one plane in a common surrounding medium. It does not locate the principal planes of a thick or internally focused photographic lens.
Aberrations, finite aperture, diffraction and focus uncertainty are excluded. Use the sign conventions before comparing with a textbook that measures distances using a different coordinate convention.
A few questions, answered
Why is a virtual image distance negative?
Its rays diverge after the lens and only their backward extensions meet on the object side.
What changes in the comparison?
Move the object inside the focal length and compare image type and orientation.