Dispersion spectra and color perception

Dispersion of light is the dependence of a medium's refractive index on wavelength, $n=n(\lambda)$. At a boundary between two media, Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$ gives different refraction angles for different $\lambda$, because $n$ differs; this is why a prism spreads white light into a spectrum. In a diffraction grating, studied in interference and diffraction, maxima occur at $d\sin\theta=m\lambda$, so a spectrum can also be observed without a prism, and orders $m$ can overlap. Color perception begins after the spectrum: the retina reduces incoming radiation to three cone signals S, M, and L rather than measuring wavelength directly. Therefore, different spectra can produce the same S, M, L responses — this is metamerism; RGB displays and color printing rely on it. The practical picture is: a spectrometer separates wavelengths, and the visual system recombines them into color, so one must distinguish the physical spectrum from perceived color.

Light from air strikes a glass surface at angle $\theta_1=30^\circ$ to the normal. Refractive indices are $n_{\text{red}}=1.50$, $n_{\text{violet}}=1.55$; air $n_1=1.00$. Find the refraction angles for red and violet light and the angular separation between them inside the glass.

A diffraction grating has $500$ lines per millimeter. White light falls normally. Find the first-order angles for red ($\lambda=700$ nm) and violet ($\lambda=400$ nm) light, the angular distance between them, and check whether the first order overlaps with the second order of violet light.

A display mixes three primaries with wavelengths $450$ nm (B), $550$ nm (G), and $650$ nm (R). A target spectral yellow at $580$ nm has relative cone responses $(S,M,L)=(0.10,0.90,0.80)$. For the primaries the responses are: B $(0.95,0.10,0.02)$, G $(0.05,0.90,0.30)$, R $(0.00,0.25,0.95)$. Find relative intensities $I_B,I_G,I_R$ that make the summed responses equal the target responses, and explain why this is metamerism.

Safe self-study plan. Open a diffraction-grating or spectrometer simulation (for example, PhET or another online model) and a color-vision simulation. In the first, set $300$–$600$ lines/mm, normal incidence, and wavelengths $400$, $550$, and $700$ nm; record the first two order angles and check for overlap. In the second, adjust RGB sliders to obtain yellow and compare its graph spectrum with a monochromatic $580$ nm line. Do not point a laser or bright light into your eyes; for home observation use only diffuse light from a lamp or screen through a CD spectroscope, never look at the Sun, and do not use a laser.

Show the link between physical wavelength separation and three-component visual response: how a prism or grating creates a spectrum, and how the eye recombines it into color.

A two-panel schematic. Left: a prism or diffraction grating, an incoming white ray, outgoing rays for $400$, $550$, and $700$ nm, and angles $\theta$; label $d\sin\theta=m\lambda$. Right: a graph of S, M, L cone responses versus wavelength and three RGB bars whose sum gives yellow; a dashed vertical line at $580$ nm. An arrow between the panels says: “spectrum → three signals S, M, L → perceived color”.

The Sun is low, about $10^\circ$ above the horizon, behind you, and raindrops roughly $1\ \text{mm}$ across are still in the air. The refractive index $n$ of water depends on wavelength: $n(650\ \text{nm}) = 1.331$ for red and $n(400\ \text{nm}) = 1.343$ for violet. A ray enters a drop, is refracted, reflects once off the back surface, and exits, deviating from the Sun direction by $42.4^\circ$ for red light and $40.6^\circ$ for violet light. The roughly $2^\circ$-wide band is therefore spread into a colored spectrum: red on the outside, violet on the inside.

This exercises the core of the concept: the refractive index $n(\lambda)$ differs across wavelengths, so the same drop deflects red and violet rays through different angles. The deviation formula (two refractions plus one internal reflection) yields $42.4^\circ$ and $40.6^\circ$, so the eye sees not a white point but a continuous spectrum.

Tilt a CD or DVD under a desk lamp and colored bands sweep across its surface. The disc tracks form a reflective diffraction grating with period $d = 1.6\ \text{µm}$. White light contains wavelengths from $400$ to $700\ \text{nm}$, and for the first order $m = 1$ the maximum appears at an angle given by $d\sin\theta = m\lambda$. For red light at $650\ \text{nm}$, $\sin\theta = 650/1600 = 0.406$, so $\theta \approx 24^\circ$; for violet at $400\ \text{nm}$, $\theta \approx 14.5^\circ$. The rings are the lamp spectrum spread out in space, each wavelength arriving at its own angle.

The scenario shows how the wave picture (a grating with $d\sin\theta = m\lambda$) turns into a spectrum: different $\lambda$ peak at different angles. The grating resolution decides whether you see separate lines or smeared bands — the bridge from diffraction to spectrum measurement.

A phone screen with $400\ \text{ppi}$ is built from three subpixels per pixel: red near $630\ \text{nm}$, green near $530\ \text{nm}$, blue near $460\ \text{nm}$. In the retina, three cone types peak near $420$, $534$ and $564\ \text{nm}$. By varying the brightness of the red, green and blue subpixels, the screen makes the eye see yellow, white or pink even though the emitted light may contain no $580\ \text{nm}$ or $500\ \text{nm}$ waves at all. This is metamerism: different spectra produce the same sensation.

This uses the second half of the concept — color perception. The eye does not sample the spectrum point by point; it forms three integrals of the spectrum against the cone sensitivities, so any two spectra with the same three numbers are perceived as one color. This is why three subpixels can stand in for a continuous spectrum.