# 25.3 The law of refraction  (Page 4/7)

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## The law of refraction

${n}_{1}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}={n}_{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$

## Take-home experiment: a broken pencil

A classic observation of refraction occurs when a pencil is placed in a glass half filled with water. Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation.

## Determine the index of refraction from refraction data

Find the index of refraction for medium 2 in [link] (a), assuming medium 1 is air and given the incident angle is $\text{30}\text{.}0º$ and the angle of refraction is $\text{22}\text{.}0º$ .

Strategy

The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus ${n}_{1}=1\text{.}\text{00}$ here. From the given information, ${\theta }_{1}=\text{30}\text{.}0º$ and ${\theta }_{2}=\text{22}\text{.}0º$ . With this information, the only unknown in Snell’s law is ${n}_{2}$ , so that it can be used to find this unknown.

Solution

Snell’s law is

${n}_{1}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}={n}_{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}.$

Rearranging to isolate ${n}_{2}$ gives

${n}_{2}={n}_{1}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}}{\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}}.$

Entering known values,

$\begin{array}{ll}{n}_{2}& =& 1\text{.}\text{00}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30}\text{.}0º}{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{22}\text{.}0º}=\frac{0\text{.}\text{500}}{0\text{.}\text{375}}\\ & =& 1.33.\end{array}$

Discussion

This is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.

## A larger change in direction

Suppose that in a situation like that in [link] , light goes from air to diamond and that the incident angle is $\text{30}\text{.}0º$ . Calculate the angle of refraction ${\theta }_{2}$ in the diamond.

Strategy

Again the index of refraction for air is taken to be ${n}_{1}=1\text{.}\text{00}$ , and we are given ${\theta }_{1}=\text{30}\text{.}0º$ . We can look up the index of refraction for diamond in [link] , finding ${n}_{2}=2\text{.}\text{419}$ . The only unknown in Snell’s law is ${\theta }_{2}$ , which we wish to determine.

Solution

Solving Snell’s law for sin ${\theta }_{2}$ yields

$\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}=\frac{{n}_{1}}{{n}_{2}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}.$

Entering known values,

$\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}=\frac{1\text{.}\text{00}}{2\text{.}\text{419}}\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30}\text{.}0º\text{=}\left(0\text{.}\text{413}\right)\left(0\text{.}\text{500}\right)=0\text{.}\text{207}.$

The angle is thus

${\theta }_{2}={\text{sin}}^{-1}0\text{.}\text{207}=\text{11}\text{.}9º.$

Discussion

For the same $30º$ angle of incidence, the angle of refraction in diamond is significantly smaller than in water ( $11.9º$ rather than $22º$ —see the preceding example). This means there is a larger change in direction in diamond. The cause of a large change in direction is a large change in the index of refraction (or speed). In general, the larger the change in speed, the greater the effect on the direction of the ray.

## Section summary

• The changing of a light ray’s direction when it passes through variations in matter is called refraction.
• The speed of light in vacuum $c=2\text{.}\text{9972458}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\approx 3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}.$
• Index of refraction $n=\frac{c}{v}$ , where $v$ is the speed of light in the material, $c$ is the speed of light in vacuum, and $n$ is the index of refraction.
• Snell’s law, the law of refraction, is stated in equation form as ${n}_{1}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}={n}_{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ .

## Conceptual questions

Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.

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