Waves in 2D and 3D

Chapter 8.4

1D Waves

• We have really only talked about 1D waves like springs and ropes

Some Common 2D/3D Waves

• Water – A stone dropped in water creates circular

waves– An ocean wave creates waves with straight

wavefronts

• Sound

• Light

Reflection of Waves

• Waves reflecting from a rigid support are in 1D

• Waves reflecting on the surface of water are in 2D

• Electromagnetic waves (light, TV, radio) are in 3D.

• Sound waves are in 3D• We show the movement of these waves

with ray diagrams.

Key Terms• Incident ray - the ray that falls or strikes a barrier (initial wave/ray).

• Reflected ray - the ray that reflects or bounces off the barrier.

• Normal Line – a perpendicular line (90°) to the barrier surface

• Angle of incidence (θi) - the angle between the incident ray and the normal line.

• Angle of reflection (θr) - the angle between the normal line and the reflected ray.

Reflection

• If a straight wave meets a straight barrier, the wave reflects directly backward.

• If a straight wave meets a barrier at an angle, it reflects at an angle.

• http://www.physicsclassroom.com/mmedia/optics/lr.cfm

Law of Reflection

• The angle of incidence (θi) is equal to the angle of reflection (θr).

Law of Reflection

What does this mean???

• If the angle of incidence is large, so is the angle of reflection.

• If the angle of incidence is small, so is the angle of reflection.

• http://www.youtube.com/watch?v=2P3nKJHO2j0

Law of Reflection

• If the light falls on a very smooth surface, the rays undergo regular reflection.

Law of Reflection

• If the surfaces of an object are rough, the rays are reflected in many different directions, producing a diffuse reflection.

Why can’t you see your reflection on paper but you can in a mirror???• Because of the smoothness of the surface!• The smoother the surface, the better the

reflection!• Regular reflection happens in a mirror.• Diffuse reflection happens on paper and other

rough surfaces.• The Law of Reflection holds true for both though!

The light is still reflected where the angle of incidence = angle of reflection.

Quick Quiz

• http://www.physicsclassroom.com/curriculum/refln/refln2.pdf

• http://www.physicsclassroom.com/curriculum/refln/refln2.pdf

Diffraction

• The spreading of waves around the edge of a barrier or hole.

Diffraction

• If waves encounter a barrier with a hole in it, they do not pass straight through. They bend around the edges of the barrier, forming circular waves that spread out.

Single-slit diffraction of water waves

Diffraction

• If waves spread around 2 closely-spaced holes, diffraction occurs

Double-slit diffraction of

water waves

Diffraction

Diffraction of Sound Waves

• We notice sound diffracting around corners or through door openings, allowing us to hear others who are speaking to us from adjacent rooms.

• Owls are able to communicate across long distances due to the fact that their long-wavelength hoots are able to diffract around forest trees and carry farther than the short-wavelength tweets of songbirds.

• Low-pitched (long wavelength) sounds always carry further than high-pitched (short wavelength) sounds.

Echolocation

• Bats use high frequency (low wavelength) ultrasonic waves in order to enhance their ability to hunt.

• The typical prey of a bat is the moth - an object not much larger than a couple of centimeters. Bats use ultrasonic echolocation methods to detect the presence of bats in the air.

• As the wavelength of a wave becomes smaller than the obstacle that it encounters, the wave is no longer able to diffract around the obstacle, instead the wave reflects off the obstacle.

• Bats use ultrasonic waves with wavelengths smaller than the dimensions of their prey. These sound waves will encounter the prey, and instead of diffracting around the prey, will reflect off the prey and allow the bat to hunt by means of echolocation.

Questions for You

• Page 370• 1, 3• 1) Sketch the wave produced by dipping a finger into

water in a ripple tank. Add rays to your diagram to illustrate the directions of the wave movement.

• 3) Sketch the appearance of a straight wave after it has passed through a small opening in a straight barrier. Add rays to your diagram to illustrate the directions of movement.

• http://phet.colorado.edu/en/contributions/view/3043

Refraction

• Light bends when it passes through a new transparent medium (like water, glass, etc)

• Why?

• Light travels at different speeds in different materials. Therefore when the light (or wave) hits the new medium, it changes speed.

Refraction

• Example. If a toy wagon is rolling on the sidewalk and hits the grass, the wheels meet the grass at an angle.

The direction ofthe rollingwheels changeswhen one slowsdown before theother does.

Parts of the Wave

• Incident Wave

• Reflected Wave (find angle using Law of Reflection)

• Refracted Wave (find angle of refraction using Snell’s Law symbol = θR)

Refraction Note the r and R in wrong place!

r

R

Refraction of Light

• Just like with boundaries, waves behave differently when going from a less dense material to a more dense material (or vice versa).

• Light twists inward from less to more dense

• As light rays enter a more optically dense medium, they travel more slowly and the refracted rays will bend toward the normal line.

Refraction

• More dense to less dense

Light twists outward from more to less dense

• If light rays enter a less dense medium, they travel faster and the refracted rays bend away from the normal line.

How do I remember this?

• So the trick is, for the bottom medium which is denser or have a higher refractive index, just write the word 'denser' from the normal line.

• Do the same for the less dense medium only this time write the word 'less dense' as shown in the figure above.

• Now compare those pictures. • The medium which is denser is close to the normal line

and the less dense medium is further from the normal line.

• Why? Because when you write the words 'denser', it takes up less space compared to the word 'less dense'.

I don’t think you are dumb but…

• http://www.youtube.com/watch?v=8ZxG_vyZWCw

Snell

• A Dutch Mathematician• Discovered the relationship between refraction and

the incident angle with different materials

• When the incident medium is a vacuum or air, the constant in Snell’s relationship is actually the same value as the index of refraction, n, of the refracting medium (see page 397 in text).

n constant sin

sin

R

i

Example

• Light travels from air into an unknown liquid at an angle of incidence of 65.0’. The angle of refraction is 42.0’.

• A) Will it bend inward or outward?

• B) Make a sketch!

• C) Determine the index of refraction (n) of the unknown liquid.

Answer

• A) Inward (less to more dense)

• B)

• C) n= sin65’/sin42’ = 1.35

Try these…

• Page 400, q7, 8, 9 (note – round to 3 sig figs to determine what substance)

Answers

• 7) n= sin59’/sin41’ = 1.31• ICE• 8) rearrange

• sinθR = sinθi/n = sin72.0’/1.92 =0.4953

• Now solve for θR. θR = sin-1(0.4953) = 29.7’

• 9) rearrange • sinθi = (sinθR)n = (sin35’)(1.362) =0.7812

• Now solve for θi. θi = sin-1(0.7812) = 51’

Snell’s Law

• A ray of light bends in such a way that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

• Formula: nisinθi = nRsinθR

• where n = refracted index (no unit).

Indices of Refraction

Why does refraction occur?

• Because the speed of light depends upon the medium it travels through.

• The speed of light also depends upon the temperature of the substance (we will look at this later…).

Example

• Light travels from air into a ruby crystal at an angle of incidence of 45’.

• A) Sketch a diagram.

• B) Will the angle of refraction be larger or smaller than 45’?

• C) Determine the angle of refraction of the light of the ruby.

Answer

• Air to ruby is less dense to more dense.• A) B) Smaller angle

• C) Rearrange equation: sinθR = nisin θi/nR = (1.00029)(sin45’)/(1.54) = 0.4593

• θR = 27’

Practice Problems

• A) Page 405, questions 10 and 11

• B) ALSO: Do these make sense? Why?

• C) Draw a diagram for a beam of light at an angle of 60’ going through from air to glass and then back to air. Determine the refraction angle for the glass and then for the air.

• D) Draw a diagram for a beam of light at an angle of 60’ going through from air to water and then back to air. Determine the refraction angle for the water and then for the air.

Answers

• 10. 39.6’

• 11. 72.6’

• In 10, the angle will be smaller because it goes from less dense to more dense.

• In 11, the angle will be larger because it goes from more dense to less dense.

Index of Refraction

Index of Refraction

• The index of refraction is a measure of the amount the light bends.

• Units for index = none (they cancel)• Units for speed = m/s• Speed of light in a vacuum = 3.00 x 108 m/s

But wait… I thought we needed a medium for waves to travel in???

• Mechanical Waves – waves that travel through matter.

• Their speed depends on the medium only (not the size of the wave).

• Examples: Ocean Waves and Sound Waves

Electromagnetic Waves

• Electromagnetic Waves – they do not require a medium to travel through. (Light – some physicists still believe it doesn’t even behave as a wave because of this!).

• Example: Light, microwaves, heat waves, radiation, radio waves

• How do these waves work?• The charges create an electric field in space and

the field exerts a force on other charges (magnetic field).

Extra info for you

• http://interactagram.com/physics/optics/refraction/

• Notice how the light bends toward the normal when the light enters a medium of greater refractive index, and away from the normal when entering a medium of lesser refractive index.

• Then notice what happens when you move the flashlight to an angle close to 90 or -90 degrees in the medium with a higher refractive index.

• As you approach the critical angle the refracted light approaches 90 or -90 degrees and, at the critical angle, the angle of refractions becomes 90 or -90 and the light is no longer transmitted across the medium/medium interface. For angles greater in absolute value than the critical angle, all the light is reflected. This is called total internal reflection.

Example

• Calculate the speed of light in zircon (n=1.92).

• Answer: n = c/v

• Rearrange v = c/n = 3.00 x108 /1.92 = 1.56 x 108 m/s

Example 2

• What is the index of refraction of a liquid in which light travels at 2.50 x 108 m/s?

• Answer: n = c/v = 3.00 x 108/2.50 x 108 = 1.20

Questions

• 1. Light in air is incident upon a piece of glass at an angle of 30.0’. What is the angle of refraction?

• 2. The index of refraction of diamond is 2.42. What is the speed of light in diamond?

• 3. What is the speed of light in sodium chloride if the index of refraction is 1.53?

• 1. 19.2’

• 2. 1.24 x 108 m/s

• 3. 1.96 x 108 m/s

Complete worksheet on refraction!

Factors Related to the Speed of Waves?

• Sound waves are affected by the temperature.

• Formula: v = 331 + 0.59Tc

• Units: v is in m/s

• Tc is in ‘C (which is actually not an SI unit!)

Table 9.1 on page 387!!!

• Example Question:

• Suppose the room temperature was 22’C. Calculate the speed of sound in the classroom.

• Answer: v = 331 + 0.59 x 22

• v = 343.39 m/s = 340 m/s

Example 2

• The temperature was 4.0’C outside. If Mrs. Evans yells at her dog Pi to come inside and it takes Pi 0.015s to hear her, how far away is Pi?

• Answer: v = 331 + 0.59 x 4.0 = 333.36m/s• Assume the speed is constant!• v = d/t d = vt = 333.36 x 0.015 = 5.0 m• Pi is 5.0 m away!

Try These!

• Page 390, questions 1 to 6

• You may need to use the table!

• Spencer 4.1.3