Diffraction by a Single Slit: Locating the Minima — Question 7

Question 7

A wire diameter can be inferred from its complementary slit minima

Use scalar complementarity: away from the direct beam, a thin opaque wire and a slit of the same width produce identical diffraction intensities. At wavelength $650\,\mathrm{nm}$, a wire gives third-order minima at $y=\pm39.0\,\mathrm{mm}$ on a screen $L=2.00\,\mathrm m$ away.

1. Infer the wire diameter using exact screen geometry.

2. Find the percentage error made by replacing sine of the diffraction angle with y/L.

Question 7 - Solution

A wire diameter can be inferred from its complementary slit minima

(a) Method and calculation

$$\sin\theta_3=\frac{y}{\sqrt{L^2+y^2}},\quad\boxed{a=\frac{3\lambda\sqrt{L^2+y^2}}y}.$$$$\boxed{a=100.019\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$a_{\mathrm{parax}}=3\lambda L/y=100.000\,\mu\mathrm m.$$$$\boxed{\frac{a-a_{\mathrm{parax}}}a=1-\frac L{\sqrt{L^2+y^2}}=0.0190071\%}.$$

Physical check

The complementary-pattern argument applies off axis; the unobstructed forward field must still be treated separately.