Thin Lenses and Image Formation

A converging lens brings rays parallel to its principal axis to the far principal focus. A ray through the optical centre of a thin lens continues approximately straight. A ray through the near focus emerges parallel to the axis. Use any two of these rays from the top of an object to locate its image.

A diverging lens makes an incident parallel ray diverge as though it originated at the near focus. Draw the actual outgoing rays as solid lines and their backward extensions as dashed lines. The focal length is the distance from the lens to a principal focus; the two thin-lens focal points are equally distant in the same surrounding medium.

Scroll the diagram to see the full view.

A real image forms where actual rays converge and can be projected onto a screen placed there. A virtual image is where backward extensions meet; a screen at that apparent position cannot collect converging rays into the image. Both image types can be seen by an eye receiving the outgoing light.

For a real object beyond a converging lens’s focal point, the image is real and inverted. Between lens and focus, the image is virtual, upright and magnified. A diverging lens with a real object produces a virtual, upright and diminished image. An object at the focus of an ideal converging lens gives parallel emerging rays, with no finite image distance.

Edexcel uses the real-is-positive convention: a real object distance u is positive, a real image distance v is positive, and a virtual image distance v is negative. A converging focal length f is positive and a diverging focal length is negative. Measure object and image distances from the thin lens, not from each other.

Substitute signs before solving. The magnitude of image size divided by object size is |v/u|. With this convention, v/u is the signed ratio in the formula listing; use the ray diagram to state orientation explicitly. Do not interpret a positive real-image distance as proof that the image is upright.

Lens power is reciprocal focal length in metres, measured in dioptres (m⁻¹). A stronger converging lens has a shorter positive focal length and a greater positive power. A diverging lens has negative power. Converting centimetres to metres before taking the reciprocal avoids a factor-of-100 error.

For thin lenses sufficiently close together, add their signed powers. A converging and diverging lens can partly cancel. The reciprocal of their total power gives the effective focal length; do not add focal lengths. For substantially separated lenses the simple in-contact rule is insufficient.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.