In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

Hint The size of the band and the intensity of the central band can be determined by using the size of the slit formula and the intensity of the slit formula. And by using the given information, the relation of the solution can be found easily.
Useful formula
The size of the slit can be given as,
$\beta = \dfrac{{2D\lambda }}{d}$
Where, $\beta $ is the size of the slit, $D$ is the distance between the slit and the source, $\lambda $ is the wavelength of the source and $d$ is the width of the slit.
The interference can be given as,
$I = k{a^2}$
Where, $I$ is the interference of the light, $k$ is the constant and $a$ is the width of the slit.

Complete step by step solution
Given that,
The width of the slit is doubled.
Now,
The size of the slit can be given as,
$\beta = \dfrac{{2D\lambda }}{d}\,....................(1)$
In the above equation if the width is doubled, then the equation (1) is written as,
$\beta = \dfrac{{2D\lambda }}{{2d}}$
Now, the above equation is written as,
$ \Rightarrow \dfrac{\beta }{2}$
So, when the width of the slit is doubled, then the size of the slit is halved.
Now,
The interference can be given as,
$I = k{a^2}\,...............\left( 2 \right)$
If the width of the slit is doubled, then the above equation is written as,
$I = k{\left( {2a} \right)^2}$
By squaring the terms in the above equation, then the above equation is written as,
$I = 4 \times k{a^2}$

So, when the width of the slit is doubled, then the interference of the slit is four times.

Note The size of the slit is directly proportional to the distance between the slit and the screen, wavelength of the light. The size of the slit is inversely proportional to the width of the slit. As the width of the slit is increased, then the size of the slit is decreased.