Answer
$11$
Work Step by Step
We know that the eleventh bright fringe coincides with the first minima.
Now we $m=1$ in equation (36-3)
$a \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima) (36-3)
$ \sin \theta = m \lambda /a$
$ \sin \theta = 1 \lambda /a$
$ \sin \theta = \lambda /a$
Consider again that the eleventh bright fringe ($m_2=11$) occurs when $m=1$.
We can use this with equation (36-25) to find the ratio of slit separation, d, to slit width, a.
$d \sin \theta = m_2 \lambda $ where $m=1,2,3,....$ (maxima) (36-25)
$d \lambda /a = m_2 \lambda $
$d /a = m_2 $
$d /a = 11 $
