Solution

Given,

The width of slit (d) = 0.01 mm = 10^-5 m,
The distance between the slit and the screen (D) = 3.5 m,
The wavelength of light used (λ) = 400 nm = 500×10^-9 m, 

Now, the width of the central diffraction peak is given by

\(x=\frac{2\lambda D}d\),
\(=\frac{2\times500\times10^{-9}\times3.5}{d10^{-5}}\)
= 0.35 m.

Hence, the required width of the central diffraction peak is 0.35 m