(i) \(x =\frac{n\,\lambda \,D}{d}\)
\(x = \frac{2\times600\times1}{0.6\times10^{-3}}\)
x = 2000 x 103
x = 2 x 106 nm
(ii) we consider that nth bright fringe of \(\lambda\) and (n -1)th bright fringe of wavelength \(\lambda_1\) consider with each other
\(n\lambda_2 = (n -1)\lambda_1\)
480n = (n - 1) 600
480n = 600n - 600
600 = 120n
n = \(\frac{600}{120}\)
n = 5
The least distance from the central maxima
\(x' = n\lambda_2\frac Dd\)
\(x' = 5\times\frac{480\times1}{0.6\times10^{-3}}\)
x' = 400 x 103
x' = 4 x 106 nm