Solution

Given,

Distance between slit (d) = 0.03 cm = 0.03 x 10^-3 m,
Distance between slit and screen (D) = 1.5 m,

Distance between central bright fringe and fourth bright fringe is,

\(4\beta=10^{-2}\) m
Or, \(\beta=0.25\times10^{-2}\)

Wavelength (λ) = ?,

Now, we have

\(\beta=\frac{\lambda D}d\)
Or, \(\lambda=\frac{\beta d}D\)
= \(\frac{0.25\times10^{-2}\times0.03\times10^{-2}}{1.5}\)
= 5 × 10^-7 m

Hence, the required wavelength is 5x10^-7 m.