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(C) In single-slit diffraction,the condition for the minima is given by $b \sin 	heta = n\lambda$,where $n = \pm 1, \pm 2, \dots$.For small angles,$\

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$\frac{2 \lambda}{b}$

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(C) In single-slit diffraction,the condition for the minima is given by $b \sin 	heta = n\lambda$,where $n = \pm 1, \pm 2, \dots$.For small angles,$\sin 	heta \approx 	heta$,so $	heta = \frac{n\lambda}{b}$.The central maximum lies between the first minima at $	heta = -\frac{\lambda}{b}$ and $	heta = \frac{\lambda}{b}$.The secondary maxima are located approximately midway between the minima,i.e.,at $	heta = \pm \frac{3\lambda}{2b}, \pm \frac{5\lambda}{2b}, \dots$.The angular width of any secondary maximum is the distance between the two minima surrounding it.Angular width $= 	heta_{n+1} - 	heta_{n-1} = \frac{(n+1)\lambda}{b} - \frac{(n-1)\lambda}{b} = \frac{2\lambda}{b}$.

$A$ plane wave front of wavelength $\lambda$ is incident on a single slit of width $b$. What is the angular width of the secondary maximum?

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