A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is
[tex]sin \theta=\frac{n \lambda}{d}[/tex]
where
n is the order of the maximum
[tex]\lambda[/tex] is the wavelength
[tex]a[/tex] is the distance between the slits
In this problem,
n = 5
[tex]\lambda=525 nm =5.25\cdot 10^{-7} m[/tex]
[tex]a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m[/tex]
So we find
[tex]\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}[/tex]
And given the distance of the screen from the slits,
[tex]D=75.0 cm = 0.75 m[/tex]
The distance of the 5th bright fringe from the central bright fringe will be given by
[tex]y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm[/tex]
B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:
[tex]sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}[/tex]
To find the angle corresponding to the 8th dark fringe, we substitute n=8:
[tex]\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}[/tex]
And the distance of the 8th dark fringe from the central bright fringe will be given by
[tex]y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm[/tex]
