I'm getting conflicting answers from various sources on the internet, and stackexchange isn't helping. Unfortunately, I'm thinking R C Mishra is right and the accepted answer there is wrong (as of when this is posted). If I'm wrong, I'd like someone to point out why.
Initially, I made the following comment:
My naive approach is to think in terms of conservation of energy. When you halve the slit, only half the energy gets through, so the intensity is halved. Amplitude, on the other hand, does not obey any conservation law.
Emphasis on naive here. Most of what I said was true, but the conclusion about intensity is false. My comment was incomplete because it left out the effects of single-slit diffraction.
Short Intuitive Answer.
Amplitude is additive, so doubling the slit width doubles the amplitude at the middle of the projected pattern.
Intensity does not follow any straightforward additivity law! Instead, you need to use conservation of energy (which is not the same thing). In the case of a single-slit, you have one factor of 2x because the slit is twice as wide, and in addition there is another 2x factor because the diffraction pattern is twice as narrow. Therefore, conservation of energy + diffraction considerations tell you there is 4x more intensity at the middle of the pattern.
I find it really interesting how both reasonings, despite being so distinct, end up giving you the exact same conclusion.
Longer Intuitive Answer.
For amplitude, remember that amplitude is additive. This means that if region 1 contributes amplitude $a_{1}$ to point $P$, and region 2 contributes amplitude $a_{2}$ to point $P$, the result will be amplitude $a_{1} + a_{2}$ at point $P$. (Note $a_{1}, a_{2}$ can be negative or positive, so it's not as straightforward as you might expect since cancellation may occur.)
Since the light from the single-slit is projected to a wall that is very far away from the slits, we may assume all distances from each point at the slit to the middle point of the pattern are approximately the same. As a result, if a planewave is sent through the slit, all points contribute the same amplitude.
Since doubling the slit width means you're doubling the number of points at the slit, additivity of amplitude means you will double the amplitude at the middle point of the pattern.
Since amplitude (at the middle point of the pattern) is directly proportional to slit width, intensity (at the middle point of the pattern) is directly proportional to the square of the slit width. However, we can actually reason about intensity directly as well!
For intensity, we think in terms of conservation of energy (as I mentioned in my original comment), but also remember to factor in diffraction and interference.
When you double the slit, twice the energy gets through, so you'd expect the intensity at the middle of the pattern to be doubled. However, making the slit wider causes the diffraction pattern to become narrower, so the energy is twice as focused at the center. Therefore, there is another factor of 2x for intensity, and therefore the intensity (at the middle point of the pattern) is quadrupled.
Math Analysis.
When planar light passes through a slit, it creates a single-slit pattern.
(For reference, the double-slit pattern occurs when two single-slit patterns interfere and create dark fringes within the middle spot, as you can see below.)
We'll only look at the single-slit scenario.
The way to derive the single-slit pattern is to use the Fraunhofer diffraction equation (if you are interested, I can explain where this equation comes from).
Based on the derivation here, the amplitude function for a single-slit of width $w$ is
$$ A(\theta) = A_{0}w \operatorname{sinc} \left( \tfrac{\pi w\sin\theta}{\lambda} \right). $$
I changed a few symbols from the wikipedia page.
Here, $\theta$ is the angle from the slit to the point on the screen you are looking at, $A(\theta)$ is the amplitude function, $A_{0}$ is a constant, and $\lambda$ is the wavelength.
The middle of the screen is at $\theta = 0$. When we plug this into our function, we get $A(0) = A_{0}w$.
As we can see, the amplitude at the middle of the screen is directly proportional to the slit width.
The intensity is the square of the amplitude, so we have
$$ I(\theta) = A_{0}^{2}w^{2} \operatorname{sinc}^{2} \left( \tfrac{\pi w\sin\theta}{\lambda} \right). $$
At the middle of the screen we have $I(0) = A_{0}^{2}w^{2}$.
We can see the intensity is proportional to the square of the slit width.
