Intensity in Single-Slit Diffraction, Quantitatively — Question 7

Section 36-4 · Physics pages 5924–5925

Question 7 A rectangular aperture factorizes into two independent sinc amplitudes A uniform rectangular aperture has widths ax, ay. Its far-field amplitude is the double integral of A0e−i(qxu+qyv) over the rectangle. Define βx = qxax/2, βy = qyay/2. dark lines Schematic; labels, not drawing lengths, specify the data. (a) Factor the integral and derive the normalized intensity. (b) Evaluate the intensity at βx = π/2, βy = π/3, and describe what happens along βx = π for arbitrary βy.
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Question 7 - Solution A rectangular aperture factorizes into two independent sinc amplitudes dark lines (a) Method and calculation F = A0 (∫ax/2 −ax/2 e−iqxudu ) (∫ay/2 −ay/2 e−iqyvdv ) . I/Ic = (sin βx βx )2 ( sin βy βy )2 . (b) Interpretation and further calculation I/Ic = 4 π2 27 4π2 = 27 π4 = 0.277182 . βx = π =⇒I = 0 for every βy . Physical check Independent aperture dimensions generate dark lines, not merely isolated dark points, in the ideal separable model.

Question 7

A rectangular aperture factorizes into two independent sinc amplitudes

A uniform rectangular aperture has widths ax,aya_x,a_y. Its far-field amplitude is the double integral of A0e−i(qxu+qyv)A_0e^{-i(q_xu+q_yv)} over the rectangle. Define βx=qxax/2\beta_x=q_xa_x/2, βy=qyay/2\beta_y=q_ya_y/2.

Diagram shown in the original SVG worksheet.

  1. Factor the integral and derive the normalized intensity.

  2. Evaluate the intensity at βx=π/2\beta_x=\pi/2, βy=π/3\beta_y=\pi/3, and describe what happens along βx=π\beta_x=\pi for arbitrary βy\beta_y.

Question 7 - Solution

A rectangular aperture factorizes into two independent sinc amplitudes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

F=A0(∫−ax/2ax/2e−iqxudu)(∫−ay/2ay/2e−iqyvdv).F=A_0\left(\int_{-a_x/2}^{a_x/2}e^{-iq_xu}du\right)\left(\int_{-a_y/2}^{a_y/2}e^{-iq_yv}dv\right).I/Ic=(sin⁡βxβx)2(sin⁡βyβy)2.\boxed{I/I_c=\left(\frac{\sin\beta_x}{\beta_x}\right)^2\left(\frac{\sin\beta_y}{\beta_y}\right)^2}.

(b) Interpretation and further calculation

I/Ic=4π2274π2=27π4=0.277182.\boxed{I/I_c=\frac4{\pi^2}\frac{27}{4\pi^2}=\frac{27}{\pi^4}=0.277182}.βx=π⇒I=0for every βy.\beta_x=\pi\Longrightarrow\boxed{I=0\ \text{for every }\beta_y}.

Physical check

Independent aperture dimensions generate dark lines, not merely isolated dark points, in the ideal separable model.