For maximum intensity on the screen
dsin θ=nλ⇒sin θ=nλd=n(2000)7000⇒sin θ=n3.5
Since maximum value of sin θ is 1.
n=3.5
So n=0,1,2,3 are only possible values of n and they will on both sides which gives n=−1,−2 and −3
Thus only 7 maximas can be obtained on both sides of the screen.