Answer
The fringe spacing will be $~~1.2~mm$
Work Step by Step
We can write an expression for the fringe spacing:
$\Delta y = \frac{\lambda~L}{d}$
Note that: $~~\Delta y \propto \lambda$
We can find $\Delta y_2$:
$\frac{\Delta y_2}{\Delta y_1} = \frac{\lambda_2}{\lambda_1}$
$\Delta y_2 = (\frac{\lambda_2}{\lambda_1})(\Delta y_1)$
$\Delta y_2 = (\frac{420~nm}{630~nm})(1.8~mm)$
$\Delta y_2 = 1.2~mm$
The fringe spacing will be $~~1.2~mm$
