Optical Image Formation

An optical image forms when light rays from one point converge at another point. The converged rays create contrast against the background. The source point either emits light or reflects it.

The brightness of the image depends on 2 things:

  • how much light the object reflects
  • how large the lens opening is

Lens Equation

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
Here:
ff
focal distance of the lens
uu
distance to the object
vv
distance to the image

A lens is a curved section of glass. The tightness of the curve sets ff.

For a given focal length, only one object distance gives a perfectly focused image. In practice, images form between the optical center and the focal point.

A larger lens lets more light reach the image. But a shorter focal length needs a tighter curve, which leaves less surface area.

Principal Rays

The path of 3 rays locates the image formed by a thin convex lens.

  • A ray parallel to the optical axis refracts through the focal point on the image side.
  • A ray through the focal point on the object side refracts parallel to the optical axis.
  • A ray through the optical center continues approximately without deviation.

For an object beyond the focal length, a convex lens forms a real, inverted image. Image size depends on the object distance. It is not always smaller than the object.

Depth of Field

Depth of field is the range of object distances that appear in focus.

The lens equation makes only one plane perfectly sharp. Objects nearer or farther than that plane are out of focus. The blur stays invisible until it spans a full pixel.

This tolerance for small blur creates a sharp zone wider than the single focused plane.

Circle of Confusion

When light from a point does not converge onto the image plane, it spreads into a small blur spot. This spot is the circle of confusion.

A spot smaller than one pixel looks sharp. Once it covers several pixels, the fuzziness becomes visible.

Worked Example

Image Distance

A lens has f=50 mmf = 50\text{ mm}. An object is u=2 mu = 2\text{ m} from the lens. Find vv.

Solution Reveal Hide
1v=1f−1u=150−12000=392000\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{50} - \frac{1}{2000} = \frac{39}{2000}v=200039 mm≈51.3 mmv = \frac{2000}{39}\text{ mm} \approx 51.3\text{ mm}

The image plane is approximately 51.3 mm51.3\text{ mm} behind the lens.

Written by November 3, 2026 2 min read
Was this helpful?