AcademyGeometric Optics
Academy
Cameras
Level 1 - Physics topic page in Geometric Optics.
Principle
A camera forms a real inverted image on a sensor by placing the sensor at the lens image plane.
Notation
\(f\)
camera lens focal length
\(\mathrm{m}\)
\(s\)
object distance
\(\mathrm{m}\)
\(s'\)
sensor distance from lens
\(\mathrm{m}\)
\(m\)
image magnification
1
\(D\)
aperture diameter
\(\mathrm{m}\)
\(N_f\)
f-number
1
Method
Derivation 1: Focus condition
The sensor must be located where the lens forms the real image.
Lens equation
\[\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}\]
Distant object
\[s\rightarrow\infty\Rightarrow s'\approx f\]
Image size
\[m=-\frac{s'}{s}\]
Derivation 2: Aperture
The f-number compares focal length with aperture diameter.
F-number
\[N_f=\frac{f}{D}\]
Aperture area scaling
\[A\propto D^2\]
Rules
Focus condition
\[\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}\]
Camera magnification
\[m=-\frac{s'}{s}\]
F-number
\[N_f=f/D\]
Examples
Question
A
\[50\,\mathrm{mm}\]
camera lens images a distant object. Where is the sensor?Answer
For a distant object,
\[s'\approx f=50\,\mathrm{mm}\]
Checks
- Camera images on the sensor are real and inverted.
- Close focusing requires sensor distance greater than the focal length.
- Smaller f-number means larger aperture diameter.