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# 12.5: Refraction of Mechanical Waves

Difficulty Level: At Grade Created by: CK-12
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Practice Refraction of Mechanical Waves

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A straw in a glass of water seen from the side often appears broken, even though it is not. The apparent break is due to the bending of light rays leaving the straw; as the light passes from the water to the glass and from the glass to the air, the light rays are bent. Nonetheless, your eye traces the light ray backward as if the light has followed a straight path from its origin at the straw. Since the light appears to have come from a different place, your eye sees the straw as being broken.

### Refraction of Mechanical Waves

When any wave strikes a boundary between media, some of the energy is reflected and some is transmitted. When the wave strikes the media interface at an angle, the transmitted wave will move in a slightly different direction than the incident wave. This phenomenon is known as refraction.

Consider the image sketched above. Suppose that the waves represented here are water waves. The wave crests are represented by the black lines in the image. As such, the distance between two consecutive black lines is the wavelength. Let the red line represent a transition from deep to shallow water. This transition is called the media interface. As the waves hit the boundary, the waves slow down. The right side of the wave reaches the boundary before the left side of the wave, causing the left side to catch up and the angle of propagation to change slightly. This change in direction can be seen in the yellow line, which is slightly angled at the boundary.

The refraction of waves across boundaries operates similarly to the method by which tanks are steered. Tanks do not have a steering wheel. Instead, they have an accelerator to produce forward motion and separate brakes on each tread. The operator uses brakes on both treads at the same time in order to stop, but brakes on only one tread to turn the tank. By braking one side, the operator causes that side to slow down or stop while the other side continues at the previous speed, causing the tank to turn towards the slower tread.

This sketch shows a wave ray striking an interface between old medium and new medium.  A normal line has been drawn as a dotted line perpendicular to the interface. The angle between the incident ray and the normal line is called the angle of incidence, shown as θi, and the angle between the refracted ray and the normal line is called the angle of refraction, θr

We already understand that the change in the wave direction at the border depends on the difference between the two velocities. This relationship is conveniently expressed in a mathematical relationship:

sinθrsinθi=vrvi=λrλi\begin{align*}\frac{\sin\theta_r}{\sin\theta_i}=\frac{v_r}{v_i}=\frac{\lambda _r}{\lambda _i}\end{align*}

The ratio of the sine of the angle of refraction to the sine of the angle of incidence is the same as the ratio of the velocity of the wave in the new medium to the velocity of the wave in the old medium and equal to the ratio of wavelength (λ) in the old medium to the wavelength in the new medium.

Example Problem: A water wave with a wavelength of 3.00 m is traveling in deep water at 16.0 m/s.  The wave strikes a sharp interface with shallow water with an angle of incidence of 53.0\begin{align*}53.0^\circ \end{align*}.  The wave refracts into the shallow water with an angle of refraction of 30.0\begin{align*}30.0^\circ \end{align*}.  What is the velocity of the wave in shallow water and what is its wavelength in the new medium?

Solution: sinθrsinθi=vrvi\begin{align*}\frac{\sin\theta_r}{\sin\theta_i}=\frac{v_r}{v_i}\end{align*} so sin30sin53=vr16.0 m/s\begin{align*}\frac{\sin30^\circ }{\sin53^\circ}=\frac{v_r}{16.0 \ m/s} \end{align*} and vr=10.0 m/s\begin{align*}v_r=10.0 \ m/s\end{align*}.

vrvi=λrλi\begin{align*}\frac{v_r}{v_i}=\frac{\lambda _r}{\lambda _i}\end{align*} so 10.0 m/s16.0 m/s=λr3.00 m\begin{align*}\frac{10.0 \ m/s}{16.0 \ m/s}=\frac{\lambda _r}{3.00 \ m}\end{align*} and λr=1.88 m\begin{align*}\lambda _r=1.88 \ m\end{align*}.

Example Problem: The ratio of the sinθr\begin{align*}\sin\theta _r\end{align*} to sinθi\begin{align*}\sin\theta _i\end{align*} is 0.769 .  If the wavelength of a wave in a new medium is 5.00×109 m\begin{align*}5.00 \times 10^{-9} \ m\end{align*}, what is its wavelength in the original medium?

Solution:

0.769=λrλi so λi=5.00×109 m0.769=6.50×109 m\begin{align*}0.769=\frac{\lambda _r}{\lambda _i} \ so \ \lambda _i=\frac{5.00\times10^{-9} \ m}{0.769}=6.50\times10^{-9} \ m\end{align*}

#### Summary

• When any wave strikes a boundary between media, some of the energy is reflected and some is transmitted.
• When a wave strikes the media interface at an angle, the transmitted wave will move in a different direction than the incident wave. This phenomenon is known as refraction.
• At any media interface, sinθrsinθi=vrvi=λrλi\begin{align*}\frac{\sin\theta_r}{\sin\theta_i}=\frac{v_r}{v_i}=\frac{\lambda _r}{\lambda _i}\end{align*}

#### Practice

1. What causes refraction?
2. What doesn’t change during refraction?

#### Review

1. A laser beam passes through water and enters a glass block at an angle.  The ratio of the speed of the wave in glass to the speed in water is 0.866 .  If the angle of incidence to the interface is 60\begin{align*}60^\circ \end{align*}, what is the angle of refraction?
2. A ray of light is traveling from air into glass at an angle of 30.0\begin{align*}30.0^\circ \end{align*} to the normal line.  The speed of the light in air is 3.00×108 m/s\begin{align*}3.00 \times 10^8 \ m/s\end{align*} and in glass the speed drops to 2.00×108 m/s\begin{align*}2.00 \times 10^8 \ m/s\end{align*}.  What is the angle of refraction?
3. Which of the following change when a water wave moves across a boundary at an angle between deep water and shallow water?
1. frequency
2. wavelength
3. speed
4. wave direction
5. period
4. Which of the following change when a water wave moves across a boundary exactly along the media interface between deep water and shallow water?
1. frequency
2. wavelength
3. speed
4. wave direction
5. period
5. The speed of sound is 340 m/s.  A particular sound wave has a frequency of 320. Hz.
1. What is the wavelength of this sound in air?
2. If this sound refracts into water where the speed of sound is 4 times faster, what will be the new wavelength?
3. What will be the new frequency?
6. When a light ray passes from air into diamond, the angle of incidence is 45.0\begin{align*}45.0^\circ \end{align*} and the angle of refraction is 16.7\begin{align*}16.7^\circ \end{align*}.  If the speed of light in air is 3.00×108 m/s\begin{align*}3.00 \times 10^8 \ m/s\end{align*}, what is the speed of light in diamond?

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