Answer
$\Delta t_{min}=2.4\mu s$
Work Step by Step
We know that;
$\Delta t=\frac{\Delta x}{v}$
As the maximum speed is the speed of light so the minimum $\Delta t$ can be determined as;
$\Delta t_{min}=\frac{x_2-x_1}{v}$
We plug in the known values to obtain:
$\Delta t_{min}=\frac{1200-480}{3\times 10^8}=2.4\times 10^{-6}=2.4\mu s$
