Dispersion Through a Prism: Newton's Rainbow

In 1666, Isaac Newton darkened his room at Trinity College, Cambridge, bored a small hole in the window shutter, and let a thin beam of sunlight fall on a glass prism. What came out the other side wasn't white — it was a band of colors, spreading from red at one end to violet at the other. Newton had demonstrated that white light is not a single entity but a mixture of colors, and that a prism separates them because glass bends each color by a slightly different amount.

That separation is called dispersion, and it happens because the refractive index $n$ of a material isn't a single fixed number — it depends on the wavelength (or equivalently, the color) of the light passing through it. Shorter wavelengths (violet, blue) experience a slightly higher $n$ than longer wavelengths (red, orange). Since a higher $n$ means more bending via Snell's law, violet bends the most and red bends the least. The result: a prism fans white light out into its component colors.

Why $n$ depends on wavelength

At the atomic level, when light passes through glass, it interacts with the electrons in the material. The electrons are driven into oscillation by the light's electric field. The amount the electrons respond depends on how close the light's frequency is to the electrons' natural resonance frequency.

For most transparent materials, the natural resonance is in the ultraviolet — higher frequency than visible light. Violet light, being closest to that resonance, interacts most strongly with the electrons, producing the largest refractive index. Red light, farthest from resonance, interacts less and has a lower $n$.

The empirical relationship is captured by the Cauchy equation:

$n(\lambda) \approx A + \dfrac{B}{\lambda^{2}} + \dfrac{C}{\lambda^{4}}$

where $A$, $B$, $C$ are material-specific constants and $\lambda$ is wavelength. The $1/\lambda^{2}$ term dominates, confirming: shorter $\lambda$ → larger $n$.

For crown glass in the visible spectrum:

• Red ($\lambda \approx 700$ nm): $n \approx 1.514$
• Orange ($\lambda \approx 600$ nm): $n \approx 1.517$
• Yellow ($\lambda \approx 580$ nm): $n \approx 1.519$
• Green ($\lambda \approx 530$ nm): $n \approx 1.522$
• Blue ($\lambda \approx 470$ nm): $n \approx 1.527$
• Violet ($\lambda \approx 400$ nm): $n \approx 1.532$

The difference between red and violet is only 0.018 — less than 2%. But when you apply this twice (entering and exiting the prism), and the prism has a steep angle (60°), even this tiny difference fans the colors out into a visible spectrum.

Geometry of a prism

A triangular prism has an apex angle $A$ (the angle at the top of the triangle, where the two refracting faces meet). Light enters one face, refracts toward the base, crosses the interior, and exits the other face, refracting again.

The total angle of deviation $\delta$ is the angle between the original incoming ray and the final outgoing ray. For a given $n$ and $A$, the deviation depends on the angle of incidence. There's a special incidence angle — the minimum deviation angle — at which $\delta$ is smallest. At minimum deviation, the ray passes through the prism symmetrically (the refraction at each face is equal).

At minimum deviation:

$n = \dfrac{\sin\!\left(\dfrac{A + \delta_{\min}}{2}\right)}{\sin(A/2)}$

This formula is frequently used in laboratory measurements to determine the refractive index of a glass prism.

Solving the problem — deviation for red and violet

For the general case (not necessarily minimum deviation), we trace the ray through both refractions using Snell's law at each face:

At the first face (entering): $\sin\theta_{1} = n\,\sin r_{1}$, where $\theta_{1}$ is the incidence angle and $r_{1}$ is the refraction angle inside the prism.

Geometry constraint: $r_{1} + r_{2} = A$ (the internal angles must sum to the prism apex angle).

At the second face (exiting): $n\,\sin r_{2} = \sin\theta_{2}$, where $\theta_{2}$ is the exit angle.

Total deviation: $\delta = \theta_{1} + \theta_{2} - A$.

Let's compute for minimum deviation where $\theta_{1} = \theta_{2}$ and $r_{1} = r_{2} = A/2 = 30°$:

Red ($n = 1.514$): $\sin\theta = n \sin(A/2) = 1.514 \times \sin 30° = 1.514 \times 0.5 = 0.757$ $\theta = 49.19°$ $\delta_{\min} = 2\theta - A = 2(49.19°) - 60° = 38.38°$

Violet ($n = 1.532$): $\sin\theta = 1.532 \times 0.5 = 0.766$ $\theta = 49.99°$ $\delta_{\min} = 2(49.99°) - 60° = 39.98°$

Angular spread (dispersion): $\Delta\delta = 39.98° - 38.38° = 1.60°$

That 1.6° spread is what creates the visible rainbow on the wall. It may sound small, but projected over a distance of a few meters, it becomes several centimeters of spectral spread — easily visible.

Rainbows in nature

The rainbow in the sky is Nature's prism. Sunlight enters a spherical raindrop, refracts (Snell's law), reflects off the back wall (TIR — Problem 152), and exits with a second refraction. The two refraction events disperse the colors, and the geometry of a sphere means the rainbow always appears at approximately 42° from the antisolar point (the point directly opposite the sun). Red is on the outside (42.4°) and violet on the inside (40.7°) — a 1.7° spread, very similar to our prism calculation.

Real-world applications

• Spectroscopy: Prisms (and their modern replacement, diffraction gratings) are used to analyze the spectrum of light from stars, flames, LEDs, and other sources. Each element produces unique spectral lines.
• Rainbow projectors & Pink Floyd album covers: The iconic cover of The Dark Side of the Moon is literally a white beam entering a prism and exiting as a rainbow.
• Chromatic aberration in lenses: Simple lenses are prisms in disguise — the outer edges of a lens refract like a prism, causing different colors to focus at slightly different distances. Expensive camera lenses use multiple glass types (achromatic doublets) to cancel this.
• Fiber optic communications: Dispersion in glass fibers causes different wavelengths to travel at different speeds, blurring the signal over long distances. Special "dispersion-shifted" fibers are engineered to minimize this.

Common mistakes

• Thinking the prism "adds" color to white light. It doesn't — the colors are already in the white light. The prism only separates them.
• Confusing dispersion with diffraction. Dispersion is $n(\lambda)$ — refraction-based color separation. Diffraction is wave-interference-based (Problems 154–155). Both can produce spectra, but the mechanism is different.
• Using a single $n$ for all colors. In many problems, a single $n$ is appropriate (it simplifies the calculation). But in dispersion problems, using a single $n$ would give zero dispersion — the whole point is that $n$ varies with color.

Try it in the visualization

Adjust the prism apex angle and the angle of incidence. Watch the white beam split into a rainbow spectrum on the far side. Toggle "monochromatic" to see just one color at a time — notice how red (lowest $n$) deviates least and violet (highest $n$) deviates most. Turn on "min deviation" mode to snap to the symmetric passage condition.