In a single slit diffraction experiment, the width of the slit is $a=0.2\text{ mm}$ and the screen is at a distance of $D=2\text{ m}$. If the wavelength of light is $\lambda =600\text{ nm}$, what is the width $w$ of the central maximum?

Solution

1. The angular position of the first minimum is given by $\sin (\theta )=\lambda /a$.
2. For small angles, $\theta \approx \lambda /a$, so the linear distance is $y=D\lambda /a$.
3. The total width of the central maximum is $$\begin{array}{rl}w& =2y\\ & =2D\lambda /a\end{array}$$.
4. $$\begin{array}{rl}w& =2\times 2\times 600\times 10^{-9}/(0.2\times 10^{-3})\\ & =12\times 10^{-3}\text{ m}\end{array}$$.
5. Hence the answer is (D).