Diffraction by a Circular Aperture — Question 4

Section 36-5 · Physics pages 5938–5939

Question 4 Two Rayleigh-separated Airy patterns have a calculable central dip For one point source, P(u) = [2J1(u)/u]2 with P(0) = 1. The first zero is u1 = 3.831705970. Two equal incoherent sources are separated by one first-zero radius. Use J1(u1/2) = 0.580724582. Each isolated peak is I0. midpoint Schematic; labels, not drawing lengths, specify the data. (a) Calculate the combined intensity at either source image center. (b) Calculate the midpoint intensity and its fractional drop relative to the intensity at a source image center.
Show solutionHide solution
Question 4 - Solution Two Rayleigh-separated Airy patterns have a calculable central dip midpoint (a) Method and calculation Icenter = I0[P(0) + P(u1)] = I0 . P(u1) = 0 because J1(u1) = 0. (b) Interpretation and further calculation Imid = 2I0 [2J1(u1/2) u1/2 ]2 = 0.735032I0 . 1 −Imid/Icenter = 0.264968 ≃26.5% . Physical check The intensities add because the sources are incoherent; adding their field amplitudes would give a phase-dependent result.

Question 4

Two Rayleigh-separated Airy patterns have a calculable central dip

For one point source, P(u)=[2J1(u)/u]2P(u)=[2J_1(u)/u]^2 with P(0)=1P(0)=1. The first zero is u1=3.831705970u_1=3.831705970. Two equal incoherent sources are separated by one first-zero radius. Use J1(u1/2)=0.580724582J_1(u_1/2)=0.580724582. Each isolated peak is I0I_0.

Diagram shown in the original SVG worksheet.

  1. Calculate the combined intensity at either source image center.

  2. Calculate the midpoint intensity and its fractional drop relative to the intensity at a source image center.

Question 4 - Solution

Two Rayleigh-separated Airy patterns have a calculable central dip

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Icenter=I0[P(0)+P(u1)]=I0.I_{\mathrm{center}}=I_0[P(0)+P(u_1)]=\boxed{I_0}.P(u1)=0becauseJ1(u1)=0.P(u_1)=0\quad\text{because}\quad J_1(u_1)=0.

(b) Interpretation and further calculation

Imid=2I0[2J1(u1/2)u1/2]2=0.735032I0.\boxed{I_{\mathrm{mid}}=2I_0\left[\frac{2J_1(u_1/2)}{u_1/2}\right]^2=0.735032I_0}.1−Imid/Icenter=0.264968≃26.5%.\boxed{1-I_{\mathrm{mid}}/I_{\mathrm{center}}=0.264968\simeq26.5\%}.

Physical check

The intensities add because the sources are incoherent; adding their field amplitudes would give a phase-dependent result.