Step 1: Set up the geometry:
A parallel beam of wavelength $\lambda$ passes through a slit of width $a$. The screen is at distance $D$.
Waves from the two edges of the slit have a path difference $a\sin\theta$ in direction $\theta$.
Dark fringes need $a\sin\theta = n\lambda$.
For small angles $\sin\theta \approx \theta$, so the minima are at $\theta_n = \dfrac{n\lambda}{a}$.
Step 2: Part (i), describe the intensity curve:
Draw the axes with $\theta$ along the horizontal line and $I$ upward.
At $\theta = 0$ all wavelets are in phase, so the curve peaks at the maximum value $I_0$.
The intensity is given by $I = I_0\left(\dfrac{\sin\beta}{\beta}\right)^2$ with $\beta = \dfrac{\pi a\sin\theta}{\lambda}$.
It becomes zero at $\beta = \pm\pi, \pm 2\pi, \ldots$, that is at $\theta = \pm\dfrac{\lambda}{a}, \pm\dfrac{2\lambda}{a}, \ldots$
Between two successive zeros there is a small secondary maximum. Its height falls quickly: about $4.5\%$ of $I_0$, then $1.6\%$, and so on.
So the sketch is a bell-shaped central hump, flanked on both sides by much lower and narrower humps.
Step 3: Part (ii), find the linear width:
The first dark fringe is at $x_1 = D\tan\theta_1 \approx \dfrac{D\lambda}{a}$ from the centre.
The central bright fringe extends from $-x_1$ to $+x_1$.
Hence $W = 2x_1 = \dfrac{2D\lambda}{a}$.
Step 4: Use proportionality:
Take the ratio of the new width to the old width: $\dfrac{W_2}{W_1} = \dfrac{D_2}{D_1}$, where the subscript 2 marks the new values.
If $D_2 < D_1$, then the width decreases in the same ratio.
So the central maximum becomes narrower on the screen. The angular spread stays at $\dfrac{2\lambda}{a}$.
Final Answer:
The intensity pattern has a broad central maximum with weak symmetric side maxima. The linear width $W = \dfrac{2D\lambda}{a}$ decreases when the slit-to-screen separation $D$ is decreased.