Answer
(a) Decreasing the wavelength $\lambda$ would cause the fringe spacing to decrease.
(b) Decreasing the slit spacing $d$ would cause the fringe spacing to increase.
(c) Decreasing the distance to the viewing screen $L$ would cause the fringe spacing to decrease.
(d) The path length difference is $1~\mu m$
Work Step by Step
We can write a general equation for the fringe spacing $y$:
$y = \frac{\lambda~L}{d}$
(a) Decreasing the wavelength $\lambda$ would cause the fringe spacing to decrease.
(b) Decreasing the slit spacing $d$ would cause the fringe spacing to increase.
(c) Decreasing the distance to the viewing screen $L$ would cause the fringe spacing to decrease.
(d) The path length difference between each fringe is one wavelength. Since fringe E is two fringes away from the central fringe, the path length difference is two wavelengths. Thus the path length difference is $2\lambda$ which is $1~\mu m$
