(a) 184 mm
The formula for the diffraction from single slit is:
[tex]y=\frac{m\lambda D}{a}[/tex]
where
y is the distance of the mth-minimum from the central maximum
[tex]\lambda[/tex] is the wavelength
D is the distance of the screen from the slit
a is the width of the slit
In this problem, we have:
[tex]\lambda=633 nm = 633\cdot 10^{-9} m[/tex]
D = 8.0 m
[tex]y_1 = 23 mm = 0.023 m[/tex] for m = 1
Solving for a, we find the width of the slit:
[tex]a=\frac{m\lambda D}{y}=\frac{(1)(633\cdot 10^{-9})(8.0)}{0.023}=2.20\cdot 10^{-4} m = 0.22 mm[/tex]
Now we can find the distance of the 4th minimum from the central maximum:
[tex]y_4 = \frac{m \lambda D}{a}=\frac{(4)(633\cdot 10^{-9})(8.0)}{2.2\cdot 10^{-4}}=0.092 m = 92 mm[/tex]
So, the distance between the forth minima on both sides of the central maximum is
[tex]d=2y_4 = 2(92 mm)=184 mm[/tex]
(b) 5.90 mm
Again, we can use the formula
[tex]y=\frac{m\lambda D}{a}[/tex]
where in this situation:
[tex]\lambda = 633\cdot 10^{-9} m[/tex]
[tex]a=0.750 mm = 7.5\cdot 10^{-4} m[/tex]
D = 3.50 m
Solving for m = 1, we find the distance of the first minimum from the central bright fringe:
[tex]y_1=\frac{m\lambda D}{a}=0.00295 m = 2.95 mm[/tex]
And so, the distance between the two minima is
[tex]d=2 y_1 = 2(2.95 mm)=5.90 mm[/tex]
