A-LEVEL FOUNDATION PHYSICS

CAMBRIDGE MODULAR SCIENCES AND OCR

Wave motion and progressive waves

• A wave motion is a means of transferring energy from one point to another without there being any transfer of matter between the points.

• Waves that move through a material or a vacuum and distribute energy from the source to the surroundings are called progressive waves.

Classification of waves

•Mechanical waves

• Water waves • Sound waves• Shock waves • Waves in a stretched

string and spring.

• Electromagnetic waves

• Light• Radio• X-rays• Microwaves• Gamma rays• Infra-red• Ultraviolet

Describing waves (definitions)Describing waves (definitions)• Frequency (f) in Hz.• Wavelength () in m or

nm.• Displacement(x) in m.• Amplitude (a) in m.• Speed (v) in m/s.• Period (T) in s.• Phase difference in

degrees.

• Wave equation• V = f x • This equation applies

to any wave.• The wave speed is

measured in m, if frequency is measured in Hz and wavelength in m.

Frequency calculations• The calculation of the frequency of sound waves can

be done by using an oscilloscope and a signal generator with or without a speaker and microphone.

• On CRO, you set time to 0.5 ms and count the number of divisions on screen of one complete wave (distance between two crests)

• Then find the time period T = divisions x time• You must change time from milliseconds to seconds

before calculating T.• Then calculate f = 1/T in Hz.• You can also calculate the wavelength of sound

waves using the same method.

More on polarisation• Electromagnetic waves are transverse waves and

can therefore be polarised.• If a radio wave is transmitted from a vertical aerial

then the wave will have vertical polarisation and any receiving aerial must also be positioned vertically.

• The main transmitters in the U.K. send out signals which are horizontally polarised.

• Many areas of the country are now served by mixed polarisation transmitters.

• Examples: Sunglasses reduce the glare of light, LCD displays on calculators are usually polarised.

The laws of reflection• The angle of incidence is equal to the angle

of reflection, this means sin i = sin r .

• The incident ray, the reflected ray and the normal all lie in the same plane.

Normal

• Rules for image position- when an object is placed in front of a plane mirror:

• The image formed is the same size as the object.

• The image is as far behind the mirror as the object is in front.

• The line joining the image to the object is at right angles to the mirror.

• The image is virtual and laterally inverted.

A

C

O

BP

2

M1

M2

If the glancing angle is g, then the angle of deviation of a ray by a plane mirror is 2g.

Angle COB = 2gg

X

The mirror is rotated through an angle to a position M2 from M1. The ray is now reflected in an OP direction.

The glancing angle with M2 is g + . Hence the new angle of deviation COP = 2 (g +)

Therefore the deviation of a reflected ray through an angle BOP = COP - COB = 2 (g + ) - 2g = 2

Deviation of a reflected ray by a rotated mirror.

In a number of instruments a beam of light is used as a pointer; this has a negligible weight and so is sensitive to deflections of the moving system.A mirror galvanometer is used for measuring very small electric

currents by using a rotating mirror.

The formula to calculate the deflection of the ray of light from a rotating mirror when a small current flows through it, is as below:

tan = x / d where x = deflection of the spot of light on scale and d = distance between the beam of light and the mirror.

Angle of incidence

Angle of refraction

Optically more dense medium, e.g. glass

Optically less dense medium, e.g. air

i

r

Sine of angle of incidence (sin i)

Refractive index n = -----------------------------------------Sine of angle of refraction (sin r)

Speed of light in less dense medium (C1)

Refractive index n = ------------------------------------------------------

Speed of light in more dense medium (C2)

Wavelength of light in less dense medium (1)

Refractive index n = -------------------------------------------------------------

Wavelength of light in more dense medium (2)

Material Refractive index

Diamond 2.42

Ruby 1.76

Glass 1.50

Glycerol 1.47

Water 1.33

Ice 1.31

Air 1.00STP

STD = standard temperature and pressure

CRITIC

AL ANGLE

90

TOTAL INTERNAL

REFLECTION

AIR

GLASS

LESS DENSE MEDIUM

MORE DENSE MEDIUM

1

Sin (C) = -------

nWhere n = refractive index of the interface

C r

Incid

ent r

ayInternally

reflected rayMore dense medium

Less dense medium

The critical angle of the glass/air boundary is about 42°.

Thus, whenever light which is travelling in glass is incident on such a boundary at an angle of more than 42, it undergoes a total internal reflection.

Uses: View finder systems of many cameras, prismatic binoculars and periscopes.

CLADDING n2 less dense

CORE n1 more dense

Refractive index of core n1

Refractive index of fibre optics = ----------------------------------------

Refractive index of cladding n2

The transmission of light through a glass fibre is one of the most important applications of total internal reflection.

There is no energy lost at the reflecting surface, therefore it is possible to send a light wave along a glass fibre and for the light to be reflected millions of times without any loss of brightness.

Glass fibres have a core a few micrometers (.000001m = 1 micron) in diameter and are covered with a layer of glass cladding.

The refractive indices of the core and the cladding are different.

When the light beam travels down the fibre, total internal reflection occurs at the interface between the core and the cladding, as long as the angle of incidence is large enough.

USES OF FIBRE OPTICS

• To see inside the human body, for endoscopy inspection and keyhole surgery.

• For lighting road signs.

• In security fences.

• For telephone and cable TV.

100 Fibre optics 1 cable

Polyurethane sheath

Q1. What are the critical angles for the interfaces between the following materials and the air? a) ruby (n= 1.76)

b) diamond (n= 2.42). (air n = 1.00 at STP)

Q2. Light is travelling along an optical fibre of refractive index

1.52. What must be the R.I. Of the cladding if the critical

angle is to be 85º?

Q3. What is meant by polarisation of a wave? Explain whether or not a) sound waves b) electromagnetic waves can be

polarised.

Q4. Describe an experiment to determine the frequency and the wavelength of a sound wave in air.

Q5. Distinguish between longitudinal waves and transverse waves.Q6. The figure shows a cross-section through a bicycle reflector in

the form of a transparent red plastic prism of refractive index (R.I.) 1.6.

90

45

X

A)

B) Ray X is incident normally on one face of the prism, as shown in the figure. Use the result of your calculation in A to complete the path of the ray X through the plastic prism.

SOLVE OLD QUESTION PAPERS TO GET SUCCESS