Diffraction by a Single Slit: Locating the Minima — Question 3

Question 3

Two minimum positions can determine both slit width and screen distance

The first two positive single-slit minima occur at $y_1=0.100\,\mathrm m$ and $y_2=\sqrt5\,y_1$. The wavelength is $500\,\mathrm{nm}$; both slit width a and screen distance L are unknown. Use far-field path differences and exact screen geometry.

1. Let $s=\lambda/a$ and derive s from $R=y_2/y_1$.

2. Find a and L numerically.

Question 3 - Solution

Two minimum positions can determine both slit width and screen distance

(a) Method and calculation

$$y_m=\frac{Lms}{\sqrt{1-m^2s^2}}\Longrightarrow R^2=\frac{4(1-s^2)}{1-4s^2}.$$$$\boxed{s^2=\frac{R^2-4}{4(R^2-1)}=\frac1{16},\quad s=1/4}.$$

(b) Interpretation and further calculation

$$\boxed{a=\lambda/s=2.00\,\mu\mathrm m}.$$$$\boxed{L=y_1\frac{\sqrt{1-s^2}}s=0.100\sqrt{15}\,\mathrm m=0.387298\,\mathrm m}.$$

Physical check

In the small-angle limit the ratio would be exactly two and could not separately determine a and L; the nonlinear geometry supplies the extra information.