Answer
$691\ nm$
Work Step by Step
It is given that:
width of slit $a=22\ \mu m$
incident angle $\theta = 1.80^{\circ}$
Also, m=1
Condition for the single slit diffraction is;
$a\ \sin\theta = m\lambda$
We rearrange the formula to find the wavelength:
$\lambda = \frac{a\ \sin\theta}{m}$
$\lambda = \frac{220\mu m\ \times \sin 1.8^{\circ}}{1}$
$\lambda=6.91\times10^{-7}$
$\lambda = 691\ nm$
