Intensity in Single-Slit Diffraction, Quantitatively — Question 8

Section 36-4 · Physics pages 5926–5927

Question 8 A finite pixel records a nonzero average at a perfect diffraction zero A pixel of width w is centered on the first minimum at y1 = λf/a in a paraxial focal plane. Let ϵ = π(y −y1)/y1, so β = π + ϵ. Assume w ≪y1 and a uniform pixel response. β/π I/Ic pixel Schematic; labels, not drawing lengths, specify the data. (a) Expand the intensity near the minimum and average it over the pixel. (b) Find the residual recorded intensity for w = 0.10y1, as a fraction of Ic.
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Question 8 - Solution A finite pixel records a nonzero average at a perfect diffraction zero β/π I/Ic pixel (a) Method and calculation I Ic = sin2(π + ϵ) (π + ϵ)2 ≃ϵ2 π2 = (y −y1)2 y2 1 . I Ic ≃1 w ∫w/2 −w/2 z2 y2 1 dz = w2 12y2 1 . (b) Interpretation and further calculation w = 0.10y1 =⇒I/Ic ≃1/1200 = 8.3333 × 10−4 . I →0 as w →0, I ∝w2 to leading order . Physical check A filled recorded minimum can result from spatial averaging alone, even when the optical intensity is exactly zero at the pixel center.

Question 8

A finite pixel records a nonzero average at a perfect diffraction zero

A pixel of width w is centered on the first minimum at y1=λf/ay_1=\lambda f/a in a paraxial focal plane. Let ϵ=π(y−y1)/y1\epsilon=\pi(y-y_1)/y_1, so β=π+ϵ\beta=\pi+\epsilon. Assume w≪y1w\ll y_1 and a uniform pixel response.

Diagram shown in the original SVG worksheet.

  1. Expand the intensity near the minimum and average it over the pixel.

  2. Find the residual recorded intensity for w=0.10y1w=0.10y_1, as a fraction of IcI_c.

Question 8 - Solution

A finite pixel records a nonzero average at a perfect diffraction zero

Diagram shown in the original SVG worksheet.

(a) Method and calculation

IIc=sin⁡2(π+ϵ)(π+ϵ)2≃ϵ2π2=(y−y1)2y12.\frac{I}{I_c}=\frac{\sin^2(\pi+\epsilon)}{(\pi+\epsilon)^2}\simeq\frac{\epsilon^2}{\pi^2}=\frac{(y-y_1)^2}{y_1^2}.I¯Ic≃1w∫−w/2w/2z2y12dz=w212y12.\frac{\overline I}{I_c}\simeq\frac1w\int_{-w/2}^{w/2}\frac{z^2}{y_1^2}dz=\boxed{\frac{w^2}{12y_1^2}}.

(b) Interpretation and further calculation

w=0.10y1⇒I¯/Ic≃1/1200=8.3333×10−4.w=0.10y_1\Longrightarrow\boxed{\overline I/I_c\simeq1/1200=8.3333\times10^{-4}}.I¯→0asw→0,I¯∝w2to leading order.\boxed{\overline I\to0\ \text{as}\ w\to0,\quad\overline I\propto w^2\ \text{to leading order}}.

Physical check

A filled recorded minimum can result from spatial averaging alone, even when the optical intensity is exactly zero at the pixel center.