Intensity in Single-Slit Diffraction, Quantitatively — Question 4

Question 4

The central-lobe curvature sets a small-loss angular tolerance

Near the axis, use the Taylor series of $\sin\beta/\beta$ through fourth order. A slit has $a=0.200\,\mathrm{mm}$ and wavelength $\lambda=600\,\mathrm{nm}$. Define an approximate alignment tolerance by allowing at most one percent intensity loss using the quadratic term.

1. Obtain the intensity series through $\beta^4$.

2. Find the quadratic angular tolerance and estimate the size of the neglected quartic correction there.

Question 4 - Solution

The central-lobe curvature sets a small-loss angular tolerance

(a) Method and calculation

$$\sin\beta/\beta=1-\beta^2/6+\beta^4/120+O(\beta^6).$$$$\boxed{I/I_c=1-\beta^2/3+2\beta^4/45+O(\beta^6)}.$$

(b) Interpretation and further calculation

$$\beta^2/3\leq0.01\Longrightarrow\boxed{|\theta|\lesssim\frac{\sqrt{0.03}\lambda}{\pi a}=1.65399\times10^{-4}\,\mathrm{rad}}.$$$$\boxed{2\beta^4/45=4.00\times10^{-5}\quad\text{at }\beta^2=0.03}.$$

Physical check

The quartic term raises the intensity slightly above the quadratic estimate, so the stated tolerance is mildly conservative.