(a) [tex]4.2\cdot 10^{-4} m[/tex]
The formula for the diffraction from a single slit is:
[tex]y=\frac{n\lambda D}{d}[/tex]
where
y is distance of the nth-minimum from the centre
n is the order of the minimum
[tex]\lambda[/tex] is the wavelength
D is the distance of the screen from the slit
d is the slit width
In this problem:
[tex]D=2.50 m[/tex]
The width of the central maximum is 6.00 m; for n=1, y represents half the width of the central maximum, so
[tex]y=3.00 mm=0.003 m[/tex]
[tex]\lambda= 500 nm = 5\cdot 10^{-7} m[/tex]
Solving the equation for d, we find the slit width:
[tex]d=\frac{n\lambda D}{y}=\frac{(1)(5.0\cdot 10^{-7} m)(2.50 m)}{0.003 m}=4.2\cdot 10^{-4} m[/tex]
(b) 0.042 m
This time, the wavelength of the radiation is
[tex]\lambda=50.0 \mu m= 50.0\cdot 10^{-6} m[/tex]
and again the value of y is
[tex]y=3.00 mm=0.003 m[/tex]
so the slit width this time is
[tex]d=\frac{n\lambda D}{y}=\frac{(1)(50\cdot 10^{-6} m)(2.50 m)}{0.003 m}=0.042 m[/tex]
(c) [tex]4.2\cdot 10^{-7} m[/tex]
This time, the wavelength of the x-rays is
[tex]\lambda=0.500 nm= 5\cdot 10^{-10} m[/tex]
and again the value of y is
[tex]y=3.00 mm=0.003 m[/tex]
so the slit width this time is
[tex]d=\frac{n\lambda D}{y}=\frac{(1)(5\cdot 10^{-10} m)(2.50 m)}{0.003 m}=4.2\cdot 10^{-7} m[/tex]
