Problem 5.140 — Circular Aperture: Intensity on Axis

Problem Statement

A circular aperture of radius $a = 1.0$ mm is illuminated by a plane wave ($\lambda = 500$ nm). Find the on-axis intensity at $z = 1.0$ m as a multiple of the unobstructed intensity $I_0$.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.

• $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
• $F_c = mv^2/R$ — net centripetal force needed
• Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
• Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: Number of Fresnel zones: $N_F = a^2/(\lambda z) = (10^{-3})^2/(5\times10^{-7}\times1) = 2$.

Step 2 — Apply the relevant physical law or equation: With 2 open Fresnel zones ($N = 2$, even), the amplitudes nearly cancel:

Step 3 — Solve algebraically for the unknown: For $N = 2$: amplitude $A \approx |a_1 – a_2| \approx 0$ (nearly dark).

Step 4 — Substitute numerical values with units: More precisely: $A = a_1 – a_2 + a_3 – \ldots$. For $N = 2$: $A = a_1-a_2 \approx a_2(a_1/a_2 – 1) \approx 0$.

Step 5 — Compute and check the result: $$\boxed{I \approx 0 \text{ (minimum — even number of zones cancel)}}$$

Worked Calculation

$$\boxed{I \approx 0 \text{ (minimum — even number of zones cancel)}}$$

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

$$\boxed{\boxed{I \approx 0 \text{ (minimum — even number of zones cancel)}}}$$

Number of Fresnel zones: $N_F = a^2/(\lambda z) = (10^{-3})^2/(5\times10^{-7}\times1) = 2$.

With 2 open Fresnel zones ($N = 2$, even), the amplitudes nearly cancel:

For $N = 2$: amplitude $A \approx |a_1 – a_2| \approx 0$ (nearly dark).

More precisely: $A = a_1 – a_2 + a_3 – \ldots$. For $N = 2$: $A = a_1-a_2 \approx a_2(a_1/a_2 – 1) \approx 0$.

$$\boxed{I \approx 0 \text{ (minimum — even number of zones cancel)}}$$

Answer

$$\boxed{I \approx 0 \text{ (minimum — even number of zones cancel)}}$$

Physical Interpretation

The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.