Problem 5.28 — Spherical Aberration Estimate

Problem Statement

A parallel beam of width $d = 3$ cm falls on a converging lens of focal length $f = 15$ cm. Estimate the diameter of the circle of least confusion due to spherical aberration if the paraxial and marginal focal lengths differ by $\Delta f = 0.5$ cm.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: The marginal rays focus at $f_m = f – \Delta f = 14.5$ cm and paraxial at $f_p = 15$ cm.

Step 2 — Apply the relevant physical law or equation: At the paraxial focal plane, the marginal rays have not yet converged. The radius of the marginal blur circle at the paraxial focus:

Step 3 — Solve algebraically for the unknown: $$r = \frac{d}{2}\cdot\frac{\Delta f}{f_m} = \frac{3}{2}\times\frac{0.5}{14.5} = 1.5\times0.0345 = 0.052\text{ cm}$$

Step 4 — Substitute numerical values with units: Diameter of circle of least confusion:

Step 5 — Compute and check the result: $$D_{min} \approx r/2 \approx \boxed{0.026\text{ cm} = 0.26\text{ mm}}$$

Worked Calculation

$$r = \frac{d}{2}\cdot\frac{\Delta f}{f_m} = \frac{3}{2}\times\frac{0.5}{14.5} = 1.5\times0.0345 = 0.052\text{ cm}$$

$$D_{min} \approx r/2 \approx \boxed{0.026\text{ cm} = 0.26\text{ mm}}$$

$$\text{Numerical result} = \text{given expression substituted with values}$$

The marginal rays focus at $f_m = f – \Delta f = 14.5$ cm and paraxial at $f_p = 15$ cm.

At the paraxial focal plane, the marginal rays have not yet converged. The radius of the marginal blur circle at the paraxial focus:

$$r = \frac{d}{2}\cdot\frac{\Delta f}{f_m} = \frac{3}{2}\times\frac{0.5}{14.5} = 1.5\times0.0345 = 0.052\text{ cm}$$

Diameter of circle of least confusion:

$$D_{min} \approx r/2 \approx \boxed{0.026\text{ cm} = 0.26\text{ mm}}$$

Answer

$$D_{min} \approx r/2 \approx \boxed{0.026\text{ cm} = 0.26\text{ mm}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.