Solution
Given,
Fringe width (β) = 1 mm = 10^-3 m,
Slit width (d) = 1.0 mm = 10^-3 m,
Distance (D) = 2m
Wavelength (λ) = ?
We have,
\(\beta=\frac{\lambda D}d\)
Or, \(\lambda=\frac{\beta d}D\)
= \(\frac{10^{-3}\times10^{-3}}2\)
= 5 x 10-7 m
Hence the required wavelength is 5 x 10^-7 m.