1.5 Dispersion  (Page 2/4)

Dispersion of white light by flint glass

A beam of white light goes from air into flint glass at an incidence angle of $43.2\text{°}$ . What is the angle between the red (660 nm) and violet (410 nm) parts of the refracted light?

Strategy

Values for the indices of refraction for flint glass at various wavelengths are listed in [link] . Use these values for calculate the angle of refraction for each color and then take the difference to find the dispersion angle.

Solution

Applying the law of refraction for the red part of the beam

$n_{\text{air}}\text{sin}{\theta }_{\text{air}}=n_{\text{red}}\text{sin}{\theta }_{\text{red}},$

we can solve for the angle of refraction as

${\theta }_{\text{red}}={\text{sin}}^{\mathrm{-1}}\left(\frac{n_{\text{air}}\text{sin}{\theta }_{\text{air}}}{n_{\text{red}}}\right)={\text{sin}}^{\mathrm{-1}}\left[\frac{(1.000)\text{sin}43.2\text{°}}{(1.662)}\right]=27.0\text{°}.$

Similarly, the angle of incidence for the violet part of the beam is

${\theta }_{\text{violet}}={\text{sin}}^{\mathrm{-1}}\left(\frac{n_{\text{air}}\text{sin}{\theta }_{\text{air}}}{n_{\text{violet}}}\right)={\text{sin}}^{\mathrm{-1}}\left[\frac{(1.000)\text{sin}43.2\text{°}}{(1.698)}\right]=26.4\text{°}.$

The difference between these two angles is

${\theta }_{\text{red}}-{\theta }_{\text{violet}}=27.0\text{°}-26.4\text{°}=0.6\text{°}.$

Significance

Although $0.6\text{°}$ may seem like a negligibly small angle, if this beam is allowed to propagate a long enough distance, the dispersion of colors becomes quite noticeable.

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Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you look away from the Sun. Light enters a drop of water and is reflected from the back of the drop ( [link] ). The light is refracted both as it enters and as it leaves the drop. Since the index of refraction of water varies with wavelength, the light is dispersed, and a rainbow is observed ( [link] (a)). (No dispersion occurs at the back surface, because the law of reflection does not depend on wavelength.) The actual rainbow of colors seen by an observer depends on the myriad rays being refracted and reflected toward the observer’s eyes from numerous drops of water. The effect is most spectacular when the background is dark, as in stormy weather, but can also be observed in waterfalls and lawn sprinklers. The arc of a rainbow comes from the need to be looking at a specific angle relative to the direction of the Sun, as illustrated in part (b). If two reflections of light occur within the water drop, another “secondary” rainbow is produced. This rare event produces an arc that lies above the primary rainbow arc, as in part (c), and produces colors in the reverse order of the primary rainbow, with red at the lowest angle and violet at the largest angle.