Question

How wide does a single slit have to be so that 530-nm light passing through the slit has its first dark fringe 30.0? from the center of the interference pattern?

Answer 1

The slit width is given by, \(d \sin \theta=n \lambda\)

$$ \begin{aligned} d &=\frac{n \lambda}{\sin \theta} \\ &=\frac{(1)\left(530 \times 10^{-9} \mathrm{~m}\right)}{\sin 30^{\circ}} \\ &=1.06 \times 10^{-6} \mathrm{~m} \end{aligned} $$

Therefore, the slit width is \(1.06 \times 10^{-6} \mathrm{~m}\).