Solution
Given,
Width of slit (d) = 0.01 mm = 10-5 m,
Distance between slit and the screen (D) = 5 m,
Wavelength of the light used (λ) = 500 nm = 500×10-9 m,
Width of the central diffraction peak (x) = ?
We know that,
\(x=\frac{2\lambda D}d\)
\(=\frac{2\times500\times510^{-9}}{10^{-5}}\)
Hence, the peak of central diffraction is 5 m wide on the screen.
