Q. In a sngle slit diffraction pattern the angular width of a central maxima is $30^{\circ}$ when the slit is illuminated by light of wavelength $6000\,\mathring{A}$. Then width of the slit will be approximately given as:

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Solution:

Angular width of central maxima in diffraction pattern is given as
$2 \theta=30^{\circ}$
Where we have
$\sin 30^{\circ}=\frac{\lambda}{b}$
$\Rightarrow b=\frac{\lambda}{\sin 30^{\circ}}=\frac{6000 \times 10^{-10}}{1 / 2}$
$=12000 \times 10^{-10} \,m$
$\Rightarrow b=12 \times 10^{-7} \,m$