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(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $	heta = \frac{\lambda}{d}$.For the first case: $	heta =

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$q: 1$

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(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $	heta = \frac{\lambda}{d}$.For the first case: $	heta = \frac{\lambda}{d} \implies d = \frac{\lambda}{	heta}$.For the second case: $q	heta = \frac{p\lambda}{d'} \implies d' = \frac{p\lambda}{q	heta}$.The half-angular width of the $n^{th}$ secondary maxima is given by $	heta_n = \frac{(2n+1)\lambda}{2d}$.For the first secondary maxima $(n=1)$,$	heta_{s1} = \frac{3\lambda}{2d}$ and $	heta_{s2} = \frac{3(p\lambda)}{2d'}$.The ratio is $\frac{	heta_{s1}}{	heta_{s2}} = \frac{3\lambda / 2d}{3p\lambda / 2d'} = \frac{1}{d} \cdot \frac{d'}{p} = \frac{d'}{pd}$.Substituting $d' = \frac{p\lambda}{q	heta}$ and $d = \frac{\lambda}{	heta}$,we get $\frac{	heta_{s1}}{	heta_{s2}} = \frac{p\lambda / q	heta}{p(\lambda / 	heta)} = \frac{1}{q}$.Thus,the ratio is $1:q$,which corresponds to the inverse of the given options if interpreted as $q:1$ or $1:q$. Given the standard form,the ratio of widths is $1:q$.

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