Answer
$1.0mm$
Work Step by Step
We can use the given values in equation (36-3) to solve for $\theta$;
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
Therefore;
$ \theta =\arcsin( m \lambda /\alpha )$
$ \theta =\arcsin( 1 *590*10^{-9}m /(.40*10^{-3}m ))$
$ \theta =1.475 * 10^{–3}rad$
The distance to the screen, D, and the distance to the minima,y, are related by : $y=D \tan \theta$
$y=D \tan \theta$
$y=0.70 \tan 1.475 * 10^{–3}$
$y=1.0*10^{-3}$
$y=1.0mm$
The distance on the screen from the center of the diffraction pattern to the first minimum is 1.0mm
