light dispersion colour and spectra

Light dispersion is the dependence of a medium's refractive index on wavelength: $n = n(\lambda)$. Because the speed of light in a medium is $v = c/n$, different colours travel at different speeds in glass and are refracted through different angles. In transparent media such as glass, $n$ decreases as $\lambda$ grows, so violet light ($\lambda \approx 400$ nm) is deviated more than red light ($\lambda \approx 700$ nm), and a white beam is spread into a coloured band — a spectrum.

A thin prism with apex angle $A$ separates two wavelengths by the angle

$$\Delta D = (n_2-n_1)A.$$

A hot, dense body emits a continuous spectrum, while a low-pressure gas gives separate bright lines (an emission spectrum) or dark lines on a continuous background (an absorption spectrum); the pattern of lines is a fingerprint of the chemical element. The colour of an object is set by the wavelengths it reflects: a red cloth absorbs almost everything except $\lambda \approx 600$–$700$ nm, so under blue light it looks black.

In glass the refractive index of red light is $n_{\text{red}} = 1.51$ and of violet light $n_{\text{violet}} = 1.53$. Calculate the speed of red and of violet light in the glass (${c = 3.00\times10^{8}}$ m/s) and state which colour a prism deviates more.

White light passes through a thin prism with apex angle $A = 30^\circ$ and falls on a screen $L = 2.0$ m away. For the glass $n(400\ \text{nm}) = 1.532$ and $n(700\ \text{nm}) = 1.513$. Find the angular separation of these two colours and the width of the spectrum on the screen.

A telescope records the spectrum of a star: a continuous band crossed by dark lines. In the laboratory, hydrogen gives bright lines at 486.1 nm and 656.3 nm, while in the star's spectrum the corresponding dark lines appear at 486.4 nm and 656.7 nm. (a) Identify the type of the star's spectrum. (b) Which element is present in the star's outer layers? (c) Estimate the star's speed along the line of sight and state whether it approaches or recedes.

Goal: compare a continuous spectrum and a line spectrum with your own eyes and link each to its light source.

You need: a compact disc (or a glass prism), a white LED flashlight, a fluorescent lamp, a sheet of white paper, a table for notes.

Procedure

1. Darken the room. Hold the disc shiny side towards the source and tilt it slowly until a rainbow band appears on the white paper.
2. Shine the white LED flashlight onto the disc and sketch the order of colours: which colour is closest to the direction of the incident light, which is farthest?
3. Repeat with the fluorescent lamp: instead of a continuous rainbow you will see a few bright lines on a dark background.
4. Fill in a table: source — type of spectrum (continuous or line) — how many lines are visible.
5. Conclusion: a continuous band comes from a hot dense body, separate lines come from a rarefied gas.

Safety: never look directly at the Sun and never direct its light into an eye with a mirror or disc; if you need sunlight, use only diffuse light from a window. Do not use a laser pointer.

Show why white light splits into colours: the deviation depends on wavelength, because $n$ depends on $\lambda$.

Diagram: a triangular prism, a white incident ray on the left, inside the prism a fan of rays from red (least deviated) to violet (most deviated), and on the right a coloured band on a screen with the wavelength $\lambda$ marked. Below the diagram, a graph of $n(\lambda)$: a monotonically decreasing curve with two marked points at $400$ nm and $700$ nm.

After a shower, with the Sun low, water droplets of radius about $0{,}5\ \text{mm}$ hang in the air. A ray of white light enters a spherical droplet, reflects once from its back surface and comes back out. Red light ($\lambda \approx 700\ \text{nm}$, $n_{\text{water}} = 1{,}331$) leaves at about $42{,}4^\circ$ to the direction of the incident sunlight, while violet light ($\lambda \approx 400\ \text{nm}$, $n_{\text{water}} = 1{,}344$) leaves at about $40{,}6^\circ$. The observer sees the red arc outside and the violet arc inside; the whole $400$–$700\ \text{nm}$ range is spread across the sky into a band about $2^\circ$ wide.

This is a direct test of the statement that the refractive index of a medium depends on wavelength. Each colour follows its own path, obeying the refraction law $n_1\sin\theta_1 = n_2\sin\theta_2$ on entering and leaving the droplet, so white light splits itself into a spectrum — the droplet acts as a dispersing system.

In a school laboratory a narrow white beam from a lamp passes through a glass prism with refracting angle $A = 60^\circ$. For red light $700\ \text{nm}$ the glass has $n = 1{,}513$, for violet $400\ \text{nm}$ it has $n = 1{,}532$; the difference of the deviation angles is about $1{,}1^\circ$. In diamond $n$ changes from $2{,}417$ ($700\ \text{nm}$) to $2{,}451$ ($400\ \text{nm}$), so a cut stone with $57$ facets scatters white light into bright coloured flashes.

Angular dispersion is at work here: the deviation of a ray by a prism, $\delta = \theta_1 + \theta_2 - A$, is computed from Snell's law separately for each wavelength, and it is precisely the difference $n(\lambda)$ that turns one beam into a continuous spectrum from $400$ to $700\ \text{nm}$.

A white pixel of a phone OLED screen mixes three narrow bands: red $630\ \text{nm}$, green $532\ \text{nm}$ and blue $465\ \text{nm}$. Looking at the screen through a cheap spectroscope with a diffraction grating of $600\ \text{lines/mm}$ one sees three bright lines instead of a continuous rainbow. A sodium streetlamp instead gives two yellow lines with wavelengths $589{,}0\ \text{nm}$ and $589{,}6\ \text{nm}$.

The example shows the difference between a continuous (thermal) spectrum and a line spectrum: the position of each line on the spectroscope screen is fixed by the grating condition $d\sin\theta = m\lambda$, so the angle gives the wavelength and links it to a specific light source.