Intensity in Single-Slit Diffraction, Quantitatively — Question 9

Section 36-4 · Physics pages 5928–5929

Question 9 A triangular amplitude profile broadens the central lobe and reduces tails A triangular aperture amplitude is A(u) = A0(1 −2|u|/a) for |u| ≤a/2. It is a scaled convolution of two equal rectangles of width a/2. Use the convolution theorem and define β = qa/2. u A Schematic; labels, not drawing lengths, specify the data. (a) Find the normalized far-field amplitude and intensity. (b) Compare its first-zero location with a uniformly illuminated slit of the same full width and evaluate its normalized intensity at β = 3π.
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Question 9 - Solution A triangular amplitude profile broadens the central lobe and reduces tails u A (a) Method and calculation F(0) = A0a/2, F(q)/F(0) = [sin(β/2) β/2 ]2 . I/Ic = [sin(β/2) β/2 ]4 . (b) Interpretation and further calculation β1,tri = 2π = 2β1,rect . β = 3π : I/Ic = 16 81π4 = 0.00202785 . Physical check Smoother illumination suppresses distant light at the cost of a broader central lobe; the comparison uses separately normalized axial intensities.

Question 9

A triangular amplitude profile broadens the central lobe and reduces tails

A triangular aperture amplitude is A(u)=A0(1−2|u|/a)A(u)=A_0(1-2|u|/a) for |u|≤a/2|u|\leq a/2. It is a scaled convolution of two equal rectangles of width a/2a/2. Use the convolution theorem and define β=qa/2\beta=qa/2.

Diagram shown in the original SVG worksheet.

  1. Find the normalized far-field amplitude and intensity.

  2. Compare its first-zero location with a uniformly illuminated slit of the same full width and evaluate its normalized intensity at β=3π\beta=3\pi.

Question 9 - Solution

A triangular amplitude profile broadens the central lobe and reduces tails

Diagram shown in the original SVG worksheet.

(a) Method and calculation

F(0)=A0a/2,F(q)/F(0)=[sin⁡(β/2)β/2]2.\boxed{F(0)=A_0a/2,\quad F(q)/F(0)=\left[\frac{\sin(\beta/2)}{\beta/2}\right]^2}.I/Ic=[sin⁡(β/2)β/2]4.\boxed{I/I_c=\left[\frac{\sin(\beta/2)}{\beta/2}\right]^4}.

(b) Interpretation and further calculation

β1,tri=2π=2β1,rect.\boxed{\beta_{1,\mathrm{tri}}=2\pi=2\beta_{1,\mathrm{rect}}}.β=3π:I/Ic=1681π4=0.00202785.\beta=3\pi:\quad\boxed{I/I_c=\frac{16}{81\pi^4}=0.00202785}.

Physical check

Smoother illumination suppresses distant light at the cost of a broader central lobe; the comparison uses separately normalized axial intensities.