A diffraction pattern is formed by a single slit of width $a=0.2\text{ mm}$ using light of wavelength $\lambda =500\text{ nm}$. The distance to the screen is $D=1\text{ m}$. Calculate the distance between the first and third minima on one side of the central maximum.

Solution

1. Identify the given values: slit width $$\begin{array}{rl}a& =0.2\text{ mm}\\ & =2\times 10^{-4}\text{ m}\end{array}$$, wavelength $$\begin{array}{rl}\lambda & =500\text{ nm}\\ & =5\times 10^{-7}\text{ m}\end{array}$$, and distance $D=1\text{ m}$.
2. The position of the $m$-th minimum for a single slit is given by $y_m=\frac{m\lambda D}{a}$.
3. The distance between the first ($m=1$) and third ($m=3$) minima is $$\begin{array}{rl}\Delta y& =y_3-y_1\\ & =(3-1)\frac{\lambda D}{a}\\ & =\frac{2\lambda D}{a}\end{array}$$.
4. Substitute the values: $$\begin{array}{rl}\Delta y& =\frac{2\times (5\times 10^{-7}\text{ m})\times (1\text{ m})}{2\times 10^{-4}\text{ m}}\\ & =5\times 10^{-3}\text{ m}\\ & =5.0\text{ mm}\end{array}$$.
5. Hence the answer is (B).