Answer
$\Delta i=72\mu m$
Work Step by Step
We know that
$i_{m\lambda_1}=m\frac{\lambda_1D}{a}$......eq(1)
and $i_{m\lambda_2}=m\frac{\lambda_2D}{a}$.....eq(2)
Now,
$\Delta i= i_{m\lambda_2}-i_{m\lambda_1}$
Substituting values from eq(1) and eq(2), we obtain:
When $m=3$
$\implies \Delta i=3(\frac{(\lambda_2-\lambda_1)D}{a})$
We plug in the known values to obtain:
$\Delta i=3(\frac{(600\times 10^{-9}-480\times 10^{-9})(1.0)}{5.0\times 10^{-3}})=72\times 10^{-6}=72\mu m$
