Use ray optics to predict where a lens forms an image, whether the image is inverted, and what must move to bring it into focus.
The engineering question
Given a lens and an object, where should a camera sensor or image plane be placed? Ray optics answers this question when the relevant features are much larger than the wavelength and diffraction is not the dominant limitation.
Model
Ideal thin lens
Represent the lens as a zero-thickness element at . It redirects paraxial rays but does not model aberrations, diffraction, finite aperture, or the physical thickness of the lens.
The signed magnification is
Coordinates and signs
Put the lens at and define positive in the nominal direction of light propagation. A real object placed to the left of the lens has . A real image formed to the right has . A converging lens has .
In the meridional – plane, define positive upward from the optical axis. The signed transverse coordinates and are the object and image heights at their respective planes.
Symbol
Meaning
Usual real-image case
object coordinate
image coordinate
object height (signed transverse coordinate)
image height (signed transverse coordinate)
focal length
lateral magnification
State a physical separation in ordinary language, then translate it before calculating. “The sample is 150 mm to the left of the lens” means .
Three ray rules
Figure 1.1. An ideal converging thin lens forms an inverted real image from a real object. The schematic uses signed axial and transverse coordinates and is not drawn to scale.
A ray parallel to the optical axis leaves a converging lens through the downstream focal point.
A ray directed through the upstream focal point leaves parallel to the optical axis.
A ray through the center of an ideal thin lens is undeviated.
When a thin lens makes a virtual image
For a converging lens, a real object placed between the lens and its upstream focal point produces and . The image is virtual, upright, and magnified.
A real image is where actual rays meet, so a sensor can be placed there. A virtual image is where backward extensions of diverging rays meet. No actual rays pass through that location.
Figure 1.2. When the object is inside the focal length of a converging lens, the outgoing rays diverge. Their backward extensions intersect at an upright virtual image on the object side of the lens.
The thin-lens equation still applies. Its negative result for tells us to draw a backward extension rather than to place a sensor at that coordinate.
Explore: the thin-lens model
Change the focal length and object position. Predict the signs of and , then use the explorer to test your prediction.
Worked examples
From guided calculation to model use.
Guided
Focus a close camera
A camera has a converging lens with . Its sample is 150 mm to the left of the lens. Find the sensor coordinate and magnification.
Hint / setup
Translate the physical location: .
Write the thin-lens equation before substituting numbers.
Use after finding the image coordinate.
Full solution
Substitute the signed object coordinate:
Therefore . The sensor goes 300 mm to the right of the lens. The magnification is : the image is inverted and twice the object height in magnitude.
Scaffolded
Focus at infinity
A camera with photographs a distant landscape. Where is the image plane, and what does the limiting magnification mean?
Hint / setup
For a distant object, use .
Full solution
The object term approaches zero, so . Also . A finite object still forms an image; the limit means a fixed physical object occupies a very small angle and therefore a very small sensor height.
Independent
Choose the object location
A camera sensor is fixed 50 mm to the right of a converging lens with . At what object coordinate will the camera focus, and what magnification results?
Hint / setup
Here . Solve the thin-lens equation for , then check whether the sign represents a real object.
Full solution
The object must be 200 mm to the left of the lens. The magnification is , so the image is inverted and one-quarter of the object height.
Boundary case
Virtual image inside the focal length
A converging lens has . An upright object with is located at . Find , , and . Can a camera sensor form a sharp image at the predicted image coordinate?
Hint / setup
Use the thin-lens equation, then determine whether the signs of and describe a real or virtual image.
Full solution
Rearrange the thin-lens equation before substituting:
Therefore . The magnification and image height are
The image is virtual, upright, and magnified. A sensor cannot record a sharp image at , because no actual rays meet there.
Before you trust the answer
Engineering checks
Have all distances and focal lengths been expressed in the same unit?
Did you translate left/right physical locations into signed coordinates before substitution?
Does a real image have a downstream coordinate, and does the sign of match the stated orientation?
Does the distant-object limit give ?
Are the thin-lens, paraxial, and negligible-diffraction assumptions plausible for this system?
