How relative amplitudes relate to light intensities in visual optics

Two light waves walk into a lab, their brightness measured in numbers rather than watts. One shines with a relative intensity of 4, the other with 9. If you’re staring at those two little numbers and wondering, “What does that say about how strong each wave’s voice is?” you’re not alone. This little puzzle sits at the heart of how we compare waves in the first place.

Here’s the thing you’ll see in many introductory captions for light and sight: intensity and amplitude aren’t the same thing, but they’re tightly linked. The bridge between them is simple, elegant, and incredibly useful.

The key rule: intensity is proportional to the square of the amplitude

• I ∝ A^2

• I is the intensity, A is the amplitude.

• The proportionality constant doesn’t matter when you’re comparing two waves—it cancels out when you take a ratio.

Let me break down the steps with your numbers in mind.

Step-by-step: from intensities 4 and 9 to amplitudes

1. Write down the two expressions for the waves:

• I1 = k · A1^2

• I2 = k · A2^2

Here, k is the same proportionality constant for both waves (since we’re talking about the same medium and the same kind of wave).

2. Take the ratio to strip out k:

• I1/I2 = (A1^2)/(A2^2)

3. Solve for the amplitude ratio by taking the square root:

• A1/A2 = sqrt(I1/I2)

4. Plug in the given intensities (I1 = 4, I2 = 9):

• A1/A2 = sqrt(4/9) = 2/3

That tells you the relative amplitudes are in the ratio 2:3. If you want concrete numbers, you can pick a common scale factor. For example, A1 = 2 and A2 = 3 fits the 2:3 ratio, or any pair proportional to 2 and 3 would work. The important takeaway is that amplitudes scale with the square root of the intensities.

A quick mental shortcut you can keep in your pocket

• When you know two intensities and you want the relative amplitudes, just take the square root of their ratio.

• A1:A2 = sqrt(I1:I2). In your case, sqrt(4:9) = 2:3.

• This works no matter what the absolute units are, because we’re looking at a relative comparison.

Why this matters beyond a multiple-choice question

• In real experiments, you’ll often measure how bright a source is and want to know how “loud” its wave is in terms of amplitude. The amplitude matters because it governs how waves add or subtract when they meet (through interference). If you have two speakers or two light waves with different amplitudes, the resulting intensity pattern depends on those amplitudes, not just the raw brightness of each source.

• The square-law relationship also makes intuitive sense: doubling the “ups and downs” of the wave (the amplitude) makes the energy carried by the wave jump fourfold, which is what intensity captures in a simplified, proportional picture.

A little digression that clarifies the intuition

• Imagine two water waves in a calm basin. If one wave’s height is twice as big as the other, the energy it carries isn’t just twice—it's four times larger (because energy scales with the square of the height, in a rough analogy). Light follows the same kind of logic: doubling amplitude doesn’t just double intensity; it quadruples it. Of course, water and light aren’t the same, but the math nudges the same way: amplitude is the square-root of intensity.

Common pitfalls to avoid

• Don’t confuse relative amplitude with the actual strength of the source. Here we’re talking about a ratio, not an absolute measure.

• The sign of amplitude can flip depending on phase, but intensity uses A^2, so the sign doesn’t affect intensity. If you’re only comparing two waves by their intensities, you’re looking at magnitudes, not directions.

• Remember: the constant k cancels in a ratio. If you try to squeeze numbers into I1 = kA1^2 and I2 = kA2^2 without using a ratio, you’ll trip over that constant and lose the sense of the relative picture.

Connecting this to the broader world of light and measurement

• In labs and classrooms, you’ll see this relationship pop up again and again when calibrating detectors, comparing laser outputs, or analyzing interference fringes. If you know how bright a source is, you can infer its energy-carrying capability or how strongly it can drive a signal in an optical system.

• When two sources are identical in amplitude, the intensities add in straightforward ways; when they differ, the resulting pattern shifts as the amplitudes weigh in. The sqrt rule helps you predict those shifts quickly without getting bogged down in heavy algebra.

A few tips to keep your intuition sharp

• Work with amplitudes in the same units as your square root of intensity. If you’re using arbitrary units, denote them clearly and stick to ratios.

• Practice with a quick triad of numbers: if I1 = 1 and I2 = 4, then A1:A2 = sqrt(1:4) = 1:2. If I1 = 9 and I2 = 16, then A1:A2 = 3:4. Seeing these little patterns helps you see the rule in action.

• When you’re faced with a problem that gives you “relative intensities,” you’re usually meant to find a ratio of amplitudes. The square-root shortcut is your friend.

Back to the original question

The options you were given included A. 2 and 3, B. 2 and 4.5, C. 8 and 18, D. 16 and 81. With the relationship I ∝ A^2 and the ratio A1/A2 = sqrt(I1/I2), the correct choice is A: 2 and 3. The math lines up perfectly: sqrt(4/9) simplifies to 2/3, which corresponds to amplitudes in the 2:3 ratio.

A closing thought

If you’re a student who loves the clean elegance of physics, this tiny calculation is a perfect example of how a simple principle—intensity grows with the square of amplitude—can unlock quick, reliable insights about the world of light. It’s the kind of nugget that shows up again and again, in labs, in simulations, and in the way you think about signals passing through optical systems. The next time you see two light sources of different brightness, you’ll have a ready-made intuition: the quieter one has an amplitude that's the smaller square root of the intensity ratio, and the louder one takes the larger square root. Simple, satisfying, and quietly powerful.