Answer
six total minima
Work Step by Step
To find the total number of minima, we can use the given values and the slit width from part 12a in equation (36-3).
$\alpha \sin \theta_{max} = m_{max} \lambda $ where $m=1,2,3,....$ (minima)
Therefore;
$\alpha \sin \theta_{max} / \lambda= m_{max} $
$2330nm *sin45/ 610nm= m_{max} $
$2.7= m_{max} $
$m_{max} \approx3 $
This suggests we can have a total of three minima on each side of the central maxima, for a total of six minima in the pattern
