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  4. Light of wavelength 2 × 10-3 m falls on a slit of width 4 × 10-3 m. The angular dispersion of the central maximum will be

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Q. Light of wavelength $2 × 10^{-3}\, m$ falls on a slit of width $4 × 10^{-3}\, m$. The angular dispersion of the central maximum will be

Wave Optics

Solution:

Angular dispersion of central maximum = angular dispersion of 1st minimum $( = 2\theta)$
From, $sin\,\theta=\frac{\lambda}{d}=\frac{2\times10^{-3}}{4\times10^{-3}}=\frac{1}{2}$
or $\theta=30^{°}$
$\therefore 2\theta=2\times30^{°}=60^{°}$