Uh.. what's the question..?
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by

(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern

is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,


while the distance between the first and the fifth minima is

(2)
If we use the formula to rewrite

, eq.(2) becomes

Which we can solve to find a, the width of the slit:
D is the answer trust me have faith it’s for the glory of humanity
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 4 s
Find: Δx and v
Δx = v₀ t + ½ at²
Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²
Δx = 24 m
v = at + v₀
v = (3 m/s²) (4 s) + 0 m/s
v = 12 m/s