# Explanation of Width of Slit Being Proportional to Light Intensity

In this question Immortal Player states that:

The intensity of light due to a slit (source of light) is directly proportional to width of the slit.

The second answer posted by R C Mishra states that:

The amplitude should be proportional to the width.

While I am inclined to believe that the first statement is correct due to other sources I found online, I do not understand why the width of the slit would be proportional to the intensity. Intuitively, I would have thought that the width of the slit would be proportional to the amplitude, since a wider opening should allow a taller wave (greater amplitude) to pass through, and the height of the wave should be directly proportional to the slit width.

Could someone please explain to me what is wrong with my intuition, and why the width of the slit is actually proportional to the intensity of the light.

• You do not have to follow your intuition. You can do the math and see what comes out. – my2cts Jan 26 '20 at 0:12
• @MaximalIdeal thank you, this is a good way to think about it and makes it much clearer. – openset Jan 26 '20 at 0:36
• @openset Unfortunately, I think my initial comment was not correct (there's a reason I said "naive")! It's really frustrating. Let me know if you have any questions or if you want any clarifications for the answer I posted. – Maximal Ideal Jan 27 '20 at 10:47
• @openset If nothing else, this really goes to show you just how hard it is to get truth right. – Maximal Ideal Jan 27 '20 at 10:47
• openest: Are you asking about the amplitude/intensity of light in the diffraction pattern or the amplitude/intensity of light passing through the slit? – Not_Einstein Jan 27 '20 at 19:34

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.

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.

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.

• While this seems very convincing, I'm still slightly inclined to believe that R C Mishra was correct. If there are no other explanations by tomorrow I will accept your answer. – openset Jan 29 '20 at 14:54
• Just a quick thought, would the answer change in any way if it were electrons passing through the slit instead of light? If electrons were passing through the slit then intensity has units of current per unit area per unit time, and doubling the slit width would double the number of electrons passing through the slit - doubling the current/intensity. In this case it would seem that the slit width is directly proportional to the intensity. – openset Jan 29 '20 at 14:57
• @openset (I agree with R C Mishra, so I assume you meant to write Immortal Player.) If it were electrons, doubling the slit width would double the number of electrons passing through, but then the diffraction pattern would become twice as narrow, so the area density of electrons landing at the middle is quadrupled in total. Hence intensity is still 4x. – Maximal Ideal Jan 29 '20 at 23:11
• Thank you, and yes, I did mean Immortal Player – openset Jan 30 '20 at 0:57