If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^\circ$ in a single slit diffraction pattern recorded using $628\ \text{nm}$ light, then the width of the slit is ______ $\mu\text{m}$.
Numerical value type. Enter your answer.
Answer: 6
The $n$th minimum is at $a\sin\theta_n = n\lambda$. The second minimum on one side and the third on the other are separated by $\theta_2 + \theta_3 = 30^\circ$.
Using the small-angle approximation $\theta_n \approx n\lambda/a$:
$$\frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{\pi}{6} \Rightarrow a = \frac{30\lambda}{\pi} = \frac{30\times628\times10^{-9}}{3.14} = 6\times10^{-6}\ \text{m}$$
So $a = 6\ \mu\text{m}$. (Solving with the exact $\sin^{-1}$ terms gives $a \approx 6.07\ \mu\text{m}$, which also rounds to 6.)
Solution by Sreeraj P, M.Sc Physics