Locate real images with a converging lens

Find a meeting point, not just a focus label

A converging lens can bring rays from one object point together. Start at the default object distance 30 cm and focal length +20 cm. The object tip is 0.50 cm above the axis. Follow three rays from that same tip: after the lens they meet 60 cm to its right, 1.00 cm below the axis. This is an inverted real image of that tip. A screen at the image plane can intercept the converging light.

The two marked focal points are 20 cm from the lens. They are not the image location in this case. The focal point on the right is where rays incident parallel to the optical axis converge. Rays from a nearby object point arrive with different slopes, so the image is generally elsewhere.

Declare the sign convention

The lens lies at coordinate x = 0. Light travels left to right. The real object lies at x = −dₒ, but its object distance dₒ is positive. Image distance dᵢ is positive for a real image on the right and negative for a virtual image on the object side. Focal length f is positive for this converging lens. These are signed optical distances; do not substitute the object’s negative coordinate for its positive object distance.

The thin-lens equation is 1/f = 1/dₒ + 1/dᵢ. Every distance must use the same unit, so all terms have units of inverse centimetres here. Rearranging gives dᵢ = fdₒ/(dₒ − f) when the denominator is nonzero.

For dₒ = 30 cm and f = 20 cm, 1/dᵢ = 1/20 − 1/30 = 1/60 per centimetre. Therefore dᵢ = +60 cm. It is not 1/60 cm: that number was the reciprocal distance.

Read magnification as size and orientation

Linear magnification m = hᵢ/hₒ = −dᵢ/dₒ is dimensionless. Heights hₒ and hᵢ are measured from the axis, positive upward. Here hₒ = +0.50 cm, m = −60/30 = −2, and hᵢ = −1.00 cm. The minus sign means inverted; magnitude two means twice the object’s height. It does not describe brightness or how large an image looks to an eye at a specified position.

At dₒ = 40 cm, dᵢ = 40 cm and m = −1. At dₒ = 60 cm, dᵢ = 30 cm and m = −0.5. Capture all three conditions. The same lens makes enlarged, equal-size and reduced real images depending on object placement. “Magnified” is not a synonym for “virtual.”

Connect the diagram to the calculation

The top incident ray is parallel to the axis; after the lens it passes through the right focal point. The ray through the lens center continues without changing its slope in this approximation. A third selected ray meets the same predicted image point. The third ray is a general ray through height −0.50 cm at the lens, not always a ray through a focal point.

Optional numerical check: a ray arriving at lens height y with incoming slope s leaves with slope s − y/f. Slope is vertical change divided by horizontal change, in cm/cm. The model uses the same rule for all three rays. At the default setting their different outgoing lines all have height −1.00 cm at x = 60 cm.

The plot uses fixed axes and exaggerates the vertical display scale by 16 relative to horizontal. Read coordinates and slopes from the ledger; do not measure physical angles from the picture. Rays leaving the plot are clipped, not absorbed. This is an ideal thin-lens paraxial model, meaning rays near the axis with small slopes; it does not predict diffraction, aperture effects or real-lens aberration.

Transfer to a different positive lens

For f = +15 cm, dₒ = 45 cm and hₒ = +0.60 cm, predict image distance, magnification and height before opening the result.

Check the real-image transfer

dᵢ = 15 × 45 / (45 − 15) = +22.5 cm. Magnification is −22.5/45 = −0.5, so hᵢ = −0.30 cm. The image is real, inverted and reduced. This transfer uses values outside the slider grid.

OpenStax: thin lenses develops the ray and image equations. Next, move through the focal plane to test where the finite-image calculation fails.