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 fringe width is inversely proportional to the slit width. The intensity is directly proportional to the square of the slit width.

Formula used: In this solution we will be using the following formula;
 $ y = \dfrac{{2\lambda D}}{d} $ where $ y $ is the width of the central maximum in a single slit diffraction, $ D $ is the distance of the slit to the screen, $ \lambda $ is the wavelength of the light, and $ d $ is the slit width

Complete step by step answer
In a single slit experiment, the width of the central maximum is given by
 $ y = \dfrac{{2\lambda D}}{d} $ where $ y $ is the width of the central maximum in a single slit diffraction, $ D $ is the distance of the slit to the screen, $ \lambda $ is the wavelength of the light, and $ d $ is the slit width
Hence, when the when the width is doubled, while every other variable is kept constant we have
 $ {y_2} = \dfrac{{2\lambda D}}{{2d}} = \dfrac{{\lambda D}}{d} $
Hence, by comparison to the first equation, we can see that
 $ {y_2} = \dfrac{y}{2} $
Hence, we can conclude that the width of the maximum central decreases.
For the intensity, it is intensity is directly proportional to the square of the fringe with as in
 $ I \propto {d^2} $
Then $ I = k{d^2} $ where $ k $ is a constant of proportionality,
Hence, when the width of the slit is doubled (and allowing all the variables making up the constant), we have
 $ {I_2} = k{\left( {2d} \right)^2} $ then,
 $ {I_2} = k4{d^2} $ , hence, by comparison with the original intensity, we see that
 $ {I_2} = 4I $
Hence, it is four times as intense as the original intensity.

Note
For clarity, the intensity is directly proportional to the square of the slit width because the intensity is directly proportional to the square of the amplitude which in turn is directly proportional to the slit width. This is because; increasing the fringe width increases the amount of light which passes onto the screen.