The formula for the single slit diffraction (in the small angle approximation) is:
[tex]y=\frac{n\lambda D}{d}[/tex]
where
y is the distance of the n-minimum from the centre of the screen (so, for n=1 it corresponds to half the width of the central maximum)
n is the order of the minimum
D is the distance between the slit and the screen
d is the width of the slit
[tex]\lambda[/tex] is the wavelength
So, the width of the central bright spot is equal to 2y when n=1:
[tex]w=2y = \frac{2\lambda D}{d}[/tex]
using this formula, let's analyze the different situations:
(a) The width doubles
The initial width of the central bright spot is
[tex]w= \frac{2\lambda D}{d}[/tex]
Here the wavelength is doubled, so
[tex]\lambda' = 2\lambda[/tex]
Therefore, the new width of the central spot will be
[tex]w'=\frac{2 \lambda' D}{d}=\frac{2(2\lambda) D}{d}=2(\frac{2\lambda D}{d})=2w[/tex]
so, the width has doubled.
(b) The width halves
The initial width of the central bright spot is
[tex]w= \frac{2\lambda D}{d}[/tex]
Here the slit width is doubled, so
[tex]d'=2d[/tex]
Therefore, the new width of the central spot will be
[tex]w'=\frac{2 \lambda D}{d'}=\frac{2\lambda D}{2d}=\frac{1}{2}(\frac{2\lambda D}{d})=\frac{w}{2}[/tex]
so, the width has halved.
(c) The width doubles
The initial width of the central bright spot is
[tex]w= \frac{2\lambda D}{d}[/tex]
Here the distance from the slit to the screen is doubled, so
[tex]D' = 2 D[/tex]
Therefore, the new width of the central spot will be
[tex]w'=\frac{2 \lambda D'}{d}=\frac{2\lambda (2D)}{d}=2(\frac{2\lambda D}{d})=2w[/tex]
so, the width has doubled.
