Spherical Lens

calendar_today July 30, 2026 article 262 words access_time 2 min replay Modified August 16, 2026

Table of Contents

Ray crosses the first boundary at height $y$ with angle $\theta_1$
$\theta_2$ is obtained via
Theorem (Property 1).
$$ \theta_{2} \approx \frac{n_{1}}{n_{2}} \theta_{1} - \frac{n_{2}-n_{1}}{n_{2}} \frac{y}{R} $$
- Measures how much the ray changes direction at the $z$-axis

and the resulting ray is extended until the second boundary

$\theta_2$ then becomes $\theta_1$ and the property is applied again for the exiting ray

If the lens is thin, we may assume the emergent ray has approximately the same $y$ as the incident ray. In addition to simplifying things (we only have to track two angles, not four), this yields the following properties:

Theorem (Angle Relationships and Focal Length).

$$ \theta_{2} = \theta_{1} - \frac{y}{f} \quad \quad \quad \frac{1}{f} = (n-1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) $$

Proof of Angle Relationships and Focal Length. (EXERCISE 1.2-4 [Saleh & Teich, Fundamentals of Photonics, p. 181])

Theorem (Imaging Equation and Magnification).

For $P_1 \coloneqq (y_1, z_1)$ and $P_2 \coloneqq (y_2, z_2)$, we have:

$$ \frac{1}{z_{1}} + \frac{1}{z_{2}} = \frac{1}{f} \quad \quad \quad y_{2} = -\frac{z_{2}}{z_{1}} y_{1} $$

These results are identical to the Spherical Mirror:
- There is a one-to-one correspondence between the conjugate planes $(z=z_1, z=z_2) \mapsto (P_1, P_2)$ with magnification factor $m=-z_2/z_1$
- $m=1 \iff z_1=z_2=2f$

Proof of Imaging Equation and Magnification. (EXERCISE 1.2-4 [Saleh & Teich, Fundamentals of Photonics, p. 181])

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