Answer
Intensity is zero when $~~n = 2.2$
Work Step by Step
Note that the phase difference is $\frac{\lambda}{2}$ when $n = 1.4$ because the first minimum occurs when $n = 1.4$
We can find an expression for $L$:
$phase~difference = 1.4~L-L = \frac{\lambda}{2}$
$L = \frac{\lambda}{0.8}$
$L = 1.25~\lambda$
We can find $n$ when the phase difference is $(m+0.5)~\lambda$, where $m$ is an integer:
$phase~difference = n~L-L = (m+0.5) \lambda$
$n-1 = \frac{(m+0.5) \lambda}{L}$
$n = 1+\frac{(m+0.5) \lambda}{1.25~\lambda}$
$n = 1+0.8~(m+0.5)$
$n = 1.4, 2.2, 3.0,...$
Note that the range of $n$ is $1.0$ to $2.5$
Intensity is zero when $~~n = 2.2$
