Higher Physics – Unit 3

3.1 - Waves

a

a

λ

λ

crest

trough

Wave Theory

All waves transmit energy.

The energy of a wave depends on its amplitude.

a = amplitude

λ = wavelength

Frequency of a wave: the number of waves per second.

Period of a wave: time taken for one complete wave to pass a point/be transmitted.

T1

f period

(s)

frequency

(Hz)

frequency of wave = frequency of source

Wave speed: is the distance travelled in one second.

Wavelength: is the minimum distance in which a wave repeats.

λ fv

frequency

(Hz)

speed

(ms-1)

wavelength

(m)

In Phase

Two waves are in phase when the same part of both waves arrive at the same point in space (peak/trough) at the same time.

Out of Phase

Two waves are exactly out of phase when opposite parts of both waves arrive at the same point in space (peak and trough) at the same time.

Coherence

Two waves are coherent if they are:

• the same frequency ( same wavelength)

• the same speed

• in phase.

Wave Behaviour

All waves display characteristic behaviour.

They all show:

• reflection

• refraction

• diffraction

• interference.

Reflection

The law of reflection states:

angle of incidence = angle of reflection

mirror

normal

i r

incident ray

reflected ray

Refraction

Waves change direction when passing from one medium to another.

This is because waves move at a different velocity in each media.

ir

ir

Diffraction

Waves bend round corners.

long wavelength

short wavelength

Interference

Interference occurs when two or more waves are superimposed.

The total effect is the sum of the waves.

+ =

constructive interference (in phase)

+ =

destructive interference (exactly out of phase)

Interference is the test for a wave.

In order to prove that any form of energy travels as a wave, an interference pattern must be shown.

Interference Patterns

To show that a form of energy travels as a wave, an interference pattern must be shown.

Thomas Young did this in 1801 for light, proving that light is a wave motion.

s1

s2

two troughs meet

crest lines from s1

crest lines from s2

two crests meet

crest and trough meetTroughs are not drawn, but

they would be between two crests:

constructive interference

constructive interference

destructive interference

+ =

+ =

+ =

d

d

d

d

c

c

c

c

c

All the points of constructive interference join up to form a series of lines.

Similarly, all the points of destructive interference join up to form a series of lines.

The two sources of wave must have the same frequency and be in phase.

Young’s Slits Experiments

laser

central maxima

1st order maxima

2nd order maxima

1st order maxima

2nd order maxima

3rd order maxima

3rd order maximadouble slit

(splits light source into two)

The fringes are more widely spread when the slits are closer together.

maxima

minima

Bright spots are called maxima (waves must be arriving in phase).

Dark regions are called minima (waves arriving exactly out of phase).Path Difference for Maxima

FringePath

Difference

central maxima 0

1st maxima λ

2nd maxima 2λ

3rd maxima 3λ

nth maxima nλ

λ ndifference path

number fringe n where

s1

s2

Waves from s1 and s2 are travelling the same distance, so there is no path difference for the central maxima.

s1

s2

At the first order maxima, waves must arrive in phase.

The path difference must be one complete wavelength.

At the second order maxima, waves must arrive in phase.

The path difference must be two complete wavelengths.

s1

s2

path difference

path difference

Path Difference for Minima

FringePath

Difference

1st minima ½ λ

2nd minima 1½ λ

3rd minima 2½ λ

nth minima (n - ½) λ number fringe n where

λ 21

-ndifference path

At the first order minima, waves must arrive exactly out of phase.

The path difference must be ½ a wavelength.

At the second order minima, waves must arrive exactly out of phase.

The path difference must be 1½ wavelengths.

s1

s2

path difference

s1

s2

path difference

Example 1

Find the wavelength of the signals used in the following experiment, where x represents the 3rd minima.

51 cm

61 cm

x

λ 21

ndifference path

Minima

λ 21

35161

λ 21

210

cm 4λ

Example 2

Find the wavelength of microwaves used in the experiment shown.

75 cm

84 cm

maxmaxmaxmax

Maxima

λ n difference path

λ 3 75-84

n = 3 (since central maxima is n = 0)

λ 3 9

cm 3λ

Diffraction Grating

A diffraction grating is a slide with multiple slits equally spaced. (typically 400-800 per mm).

θ

θ

path difference

s1

s2

s3

d

d

to maxima λ n difference path

ddifference path

θ sin

dλ n

θ sin

sinθ dλ n

For Maxima

Example 1

Calculate the wavelength of light passed through a diffraction grating, when the first order maxima is produced at an angle of 21°.

The grating has 600 lines per mm.

?λ

600101

d3

m101.67 6

1n

21θ θ sin dλ n

21 sin 101.67λ1 -6

21 sin 101.67λ -6

m 105.97λ 7

nm597 λ

Example 2

Light of wavelength 700 nm is shone through a diffraction grating that has 1200 slits per millimetre.

Calculate the angle between the central and first order maxima.

θ sin dλ n nm 700λ

m 10 700 -9

1200101

d3

m108.33 7

1n

?θ

θ sin 108.3310700 1 7-9

7

-9

108.3310700

θ sin

0.84θ sin

0.84 sinθ -1

57.1θ

Measuring Wavelength

Aim

To calculate the wavelength of laser light.

Diagram

laser

screen

diffraction grating

θ

Method

• Measure distance from laser to screen.

• Measure distance between central and first order maxima.

• Using Pythagoras, calculate angle at which constructive interference occurs

• Slit separation is calculated by:

• For first order maxima, n = 1.

slits of no.slide of width

d

Results

θ

1m

0.05m

adjopp

θ tan

10.05

θ tan

0.05tanθ 1

2.8θ

?λ

1n

2.8θ

dθ sin dλ n

Effect of Colour on Gratings

Monochromatic light beams consist of one single frequency of light.

Here are some wavelengths of monochromatic light:

Colour Wavelength (nm)

red 700

green 550

blue 500

violet 400

need to know these values but

** NOT GIVEN IN EXAM **

TASK

Calculate the angle between the central and first order maxima for the four colours of light passing through a diffraction grating with 700 lines per mm.

Red

θ sin dλ n

θ sin 101.4310700 1 6-9

6

-9

101.4310700

θ sin

0.49θ sin

0.49 sinθ -129.3θ

Green

θ sin dλ n

θ sin 101.4310550 1 6-9

6

-9

101.4310550

θ sin

0.38θ sin

0.38 sinθ -122.6θ

Blue

θ sin dλ n

θ sin 101.4310500 1 6-9

6

-9

101.4310500

θ sin

0.35θ sin

0.35 sinθ -120.5θ

Violet

θ sin dλ n

θ sin 101.4310400 1 6-9

6

-9

101.4310400

θ sin

0.28θ sin

0.28 sinθ -116.2θ

Conclusion

The longer the wavelength the the maxima separation.

The shorter the wavelength the the maxima separation.

greater

smaller

White Light Spectra

Diffraction Grating

When a white light source is used to produce an interference pattern with a diffraction grating, the following is seen:

central maxima (white)

1st order maxima (visible spectrum)

1st order maxima (visible spectrum)white light

source

diffraction grating

Dispersion is caused by interference.

All other maxima are seen in the form of the visible spectrum.

Red light makes the greatest angle with the original ray of white light because it has the longest wavelength.

The longer the wavelength the greater the maxima separation.

More than one spectra is produced.

Prism

When light is dispersed using a prism, the following is seen:

prismwhite light

source

Dispersion is caused by refraction.

Violet light makes the greatest angle with the original ray of white light because it has the highest frequency.

The higher the frequency, the greater refraction that takes place.

Only one spectrum is produced.