AcademyGeometric Optics
Academy
Thin Lenses
Level 1 - Physics topic page in Geometric Optics.
Principle
Thin lenses form images by refracting rays twice with negligible separation between surfaces.
Notation
\(f\)
lens focal length
\(\mathrm{m}\)
\(s\)
object distance
\(\mathrm{m}\)
\(s'\)
image distance
\(\mathrm{m}\)
\(m\)
lateral magnification
1
\(P\)
lens power
D
\(R_1,R_2\)
surface radii
\(\mathrm{m}\)
Method
Derivation 1: Thin-lens equation
For paraxial rays, a thin lens maps object distance to image distance through its focal length.
Thin-lens equation
\[\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}\]
Magnification
\[m=-\frac{s'}{s}\]
Power
\[P=\frac{1}{f}\]
Derivation 2: Lensmaker form
The focal length depends on index and surface curvature.
Lensmaker equation
\[\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]
Rules
Thin-lens equation
\[\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}\]
Magnification
\[m=-\frac{s'}{s}\]
Lens power
\[P=1/f\]
Lensmaker equation
\[\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]
Examples
Question
An object is
\[30\,\mathrm{cm}\]
from a converging lens with \[f=10\,\mathrm{cm}\]
Find \[s'\]
Answer
\[\frac{1}{s'}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15}\]
\[s'=15\,\mathrm{cm}\]
Checks
- Converging lenses have positive focal length.
- Diverging lenses have negative focal length.
- Negative image distance means a virtual image on the object side.