Numerical 1 — YDSE Complete Analysis

Problem: $d = 0.5\,\text{mm} = 5\times10^{-4}\,\text{m}$; $D = 1\,\text{m}$; $\lambda = 600\,\text{nm} = 6\times10^{-7}\,\text{m}$. Find: (a) fringe width, (b) 3rd bright fringe position, (c) 2nd dark fringe position.

Solution: (a) Fringe width: $\beta = \frac{\lambda D}{d} = \frac{6\times10^{-7}\,\text{m}\times1\,\text{m}}{5\times10^{-4}\,\text{m}} = \frac{6\times10^{-7}}{5\times10^{-4}}\,\text{m} = 1.2\times10^{-3}\,\text{m} = \mathbf{1.2\,mm}$

(b) 3rd bright fringe ($n = 3$): $y_3 = n\beta = 3\times1.2\,\text{mm} = \mathbf{3.6\,mm} \text{ from center}$

(c) 2nd dark fringe ($n = 2$): $y_2 = \left(n - \frac{1}{2}\right)\beta = \left(2 - \frac{1}{2}\right)\times1.2\,\text{mm} = \frac{3}{2}\times1.2 = \mathbf{1.8\,mm} \text{ from center}$

Verification: 2nd dark fringe lies between 1st bright (1.2 mm) and 2nd bright (2.4 mm). ✓