Diffraction by a Single Slit: Locating the Minima — Question 10

Question 10

Minimum-position sensitivity grows sharply near grazing angles

For fixed slit width a, screen distance L, and nonzero order m, the wavelength is inferred from $\lambda=(a/m)y/\sqrt{L^2+y^2}$. Assume a and L are exact and the measured y has small standard uncertainty $\sigma_y$.

1. Derive the relative wavelength uncertainty caused by $\sigma_y$.

2. Compare the sensitivity to relative y error at $y\ll L$ and at $y=L$, then derive the inverse sensitivity $d\ln y/d\ln\lambda$.

Question 10 - Solution

Minimum-position sensitivity grows sharply near grazing angles

(a) Method and calculation

$$\ln\lambda=\ln(a/m)+\ln y-\tfrac12\ln(L^2+y^2).$$$$\boxed{\frac{\sigma_\lambda}{\lambda}=\frac{L^2}{L^2+y^2}\frac{\sigma_y}{|y|}=\cos^2\theta\frac{\sigma_y}{|y|}}.$$

(b) Interpretation and further calculation

$$y\ll L:\ \boxed{\text{factor }1};\qquad y=L:\ \boxed{\text{factor }1/2}.$$$$\boxed{\frac{d\ln y}{d\ln\lambda}=\sec^2\theta\to\infty\quad\text{as}\ |\theta|\to90^\circ}.$$

Physical check

Near grazing, a small wavelength change moves a minimum a large distance, but the required screen size also grows without bound.