Answer
Two
Work Step by Step
We know that
$m_{max}=\frac{1}{N\lambda_{min}}$
We plug in the known values to obtain:
$m_{max}=\frac{1}{(89\times 10^4m^{-1})(400\times 10^{-9}m)}$
$m_{max}=2.8$
Since the value of $m$ is less than 3, therefore, the highest order visible maximum is 2.
