# Explain two features to distinguish between the interference pattern in Young double slit experiment

A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.
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Hassan Watkins
The wavelength of incident light is given as
The width of a single slit is given as
The angular width of central maxima is,
$w=\frac{2\lambda }{a}\phantom{\rule{0ex}{0ex}}=\frac{2×500×{10}^{-9}}{0.2×{10}^{-3}}\phantom{\rule{0ex}{0ex}}=5×{10}^{-3}m$
Thus, the angular width of the central maxima obtained on the screen is 5 mm.
Given the width of fringe in the young double-slit experiment is $d=0.5×{10}^{-3}m$
Let there are n fringes made by a double-slit in the central maxima of the diffraction,
Then,
$n=\frac{{B}^{\prime }}{B}$
where ${B}^{\prime }=\frac{2\lambda D}{a}$ is the width of central maxima in the diffraction and $B=\frac{\lambda D}{d}$ is the fringe width by the double slit.
Substituting the known values,
$n=\frac{\left(\frac{2\lambda D}{a}\right)}{\left(\frac{\lambda D}{d}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2\lambda D}{a}×\frac{d}{\lambda D}\phantom{\rule{0ex}{0ex}}=2\frac{d}{a}\phantom{\rule{0ex}{0ex}}=\frac{2×0.5×{10}^{-3}}{0.2×{10}^{-3}}\phantom{\rule{0ex}{0ex}}=5$
Thus, the number of fringes obtained is 5.