# 1.6 Huygens’s principle  (Page 2/4)

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## Refraction

The law of refraction can be explained by applying Huygens’s principle to a wave front passing from one medium to another ( [link] ). Each wavelet in the figure was emitted when the wave front crossed the interface between the media. Since the speed of light is smaller in the second medium, the waves do not travel as far in a given time, and the new wave front changes direction as shown. This explains why a ray changes direction to become closer to the perpendicular when light slows down. Snell’s law can be derived from the geometry in [link] ( [link] ).

## Deriving the law of refraction

By examining the geometry of the wave fronts, derive the law of refraction.

## Strategy

Consider [link] , which expands upon [link] . It shows the incident wave front just reaching the surface at point A , while point B is still well within medium 1. In the time $\text{Δ}t$ it takes for a wavelet from B to reach $B\text{′}$ on the surface at speed ${v}_{1}=c\text{/}{n}_{1},$ a wavelet from A travels into medium 2 a distance of $AA\text{′}={v}_{2}\text{Δ}t,$ where ${v}_{2}=c\text{/}{n}_{2}.$ Note that in this example, ${v}_{2}$ is slower than ${v}_{1}$ because ${n}_{1}<{n}_{2}.$

## Solution

The segment on the surface $AB\text{′}$ is shared by both the triangle $ABB\text{′}$ inside medium 1 and the triangle $AA\text{′}B\text{′}$ inside medium 2. Note that from the geometry, the angle $\text{∠}BAB\text{′}$ is equal to the angle of incidence, ${\theta }_{1}$ . Similarly, $\text{∠}AB\text{′}A\text{′}$ is ${\theta }_{2}$ .

The length of $AB\text{′}$ is given in two ways as

$AB\text{′}=\frac{BB\text{′}}{\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{1}}=\frac{AA\text{′}}{\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{2}}.$

Inverting the equation and substituting $AA\text{′}=c\text{Δ}t\text{/}{n}_{2}$ from above and similarly $BB\text{′}=c\text{Δ}t\text{/}{n}_{1}$ , we obtain

$\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{1}}{c\text{Δ}t\text{/}{n}_{1}}=\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{2}}{c\text{Δ}t\text{/}{n}_{2}}.$

Cancellation of $c\text{Δ}t$ allows us to simplify this equation into the familiar form

${n}_{1}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{1}={n}_{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{2}.$

## Significance

Although the law of refraction was established experimentally by Snell and stated in Refraction , its derivation here requires Huygens’s principle and the understanding that the speed of light is different in different media.

Check Your Understanding In [link] , we had ${n}_{1}<{n}_{2}$ . If ${n}_{2}$ were decreased such that ${n}_{1}>{n}_{2}$ and the speed of light in medium 2 is faster than in medium 1, what would happen to the length of $AA\text{′}$ ? What would happen to the wave front $A\text{′}B\text{′}$ and the direction of the refracted ray?

$AA\text{′}$ becomes longer, $A\text{′}B\text{′}$ tilts further away from the surface, and the refracted ray tilts away from the normal.

This applet by Walter Fendt shows an animation of reflection and refraction using Huygens’s wavelets while you control the parameters. Be sure to click on “Next step” to display the wavelets. You can see the reflected and refracted wave fronts forming.

## Diffraction

What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, we observe a sharp shadow of the doorway on the floor of the room, and no visible light bends around corners into other parts of the room. When sound passes through a door, we hear it everywhere in the room and thus observe that sound spreads out when passing through such an opening ( [link] ). What is the difference between the behavior of sound waves and light waves in this case? The answer is that light has very short wavelengths and acts like a ray. Sound has wavelengths on the order of the size of the door and bends around corners (for frequency of 1000 Hz,

$\lambda =\frac{c}{f}=\frac{330\phantom{\rule{0.2em}{0ex}}\text{m/s}}{1000\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}}=0.33\phantom{\rule{0.2em}{0ex}}\text{m,}$

about three times smaller than the width of the doorway).

If we pass light through smaller openings such as slits, we can use Huygens’s principle to see that light bends as sound does ( [link] ). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a wave characteristic and occurs for all types of waves. If diffraction is observed for some phenomenon, it is evidence that the phenomenon is a wave. Thus, the horizontal diffraction of the laser beam after it passes through the slits in [link] is evidence that light is a wave. You will learn about diffraction in much more detail in the chapter on Diffraction .

## Summary

• According to Huygens’s principle, every point on a wave front is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wave front is tangent to all of the wavelets.
• A mirror reflects an incoming wave at an angle equal to the incident angle, verifying the law of reflection.
• The law of refraction can be explained by applying Huygens’s principle to a wave front passing from one medium to another.
• The bending of a wave around the edges of an opening or an obstacle is called diffraction.

## Conceptual questions

How do wave effects depend on the size of the object with which the wave interacts? For example, why does sound bend around the corner of a building while light does not?

Does Huygens’s principle apply to all types of waves?

yes

If diffraction is observed for some phenomenon, it is evidence that the phenomenon is a wave. Does the reverse hold true? That is, if diffraction is not observed, does that mean the phenomenon is not a wave?

as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
explain l-s coupling
how can we say dirac equation is also called a relativistic equation in one word
what is the electronic configration of Al
what's the signeficance of dirac equetion.?
what is the effect of heat on refractive index
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Swati
yes
Chemist
1 millimeter is How many metres
1millimeter =0.001metre
Gitanjali
The photoelectric effect is the emission of electrons when light shines on a material.