Answer:
0.004 m
Explanation:
For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by
[tex]y=\frac{n\lambda D}{d}[/tex]
where
[tex]\lambda[/tex] is the wavelength
D is the distance of the screen from the slit
d is the width of the slit
Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):
[tex]w=2\frac{\lambda D}{d}[/tex]
where we have
[tex]\lambda=500 nm = 5\cdot 10^{-7}m[/tex] is the wavelength
D = 2.0 m is the distance of the screen
[tex]d=0.50 mm=5\cdot 10^{-4}m[/tex] is the width of the slit
Substituting, we find
[tex]w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m[/tex]
