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.

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Presentation transcript:

Higher Physics – Unit 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. 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. 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. i r i r

Diffraction Waves bend round corners. long wavelengthshort 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.

s1s1 s2s2 two troughs meet crest lines from s 1 crest lines from s 2 two crests meet crest and trough meet Troughs are not drawn, but they would be between two crests: 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 1 st order maxima 2 nd order maxima 1 st order maxima 2 nd order maxima 3 rd order maxima double 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 Fringe Path Difference central maxima0 1 st maximaλ 2 nd maxima2λ2λ 3 rd maxima3λ3λ n th maximanλnλ

s1s1 s2s2 Waves from s 1 and s 2 are travelling the same distance, so there is no path difference for the central maxima. s1s1 s2s2 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. s1s1 s2s2 path difference

Path Difference for Minima Fringe Path Difference 1 st minima½ λ 2 nd minima1½ λ1½ λ 3 rd minima2½ λ n th minima(n - ½) λ

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. s1s1 s2s2 path difference s1s1 s2s2

Example 1 Find the wavelength of the signals used in the following experiment, where x represents the 3 rd minima. 51 cm 61 cm x Minima

Example 2 Find the wavelength of microwaves used in the experiment shown. 75 cm 84 cm max Maxima n = 3 (since central maxima is n = 0)

Diffraction Grating A diffraction grating is a slide with multiple slits equally spaced. (typically per mm). θ θ path difference s1s1 s2s2 s3s3 d d to maxima 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.

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.

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.

Results θ 1m 0.05m

Effect of Colour on Gratings Monochromatic light beams consist of one single frequency of light. Here are some wavelengths of monochromatic light: ColourWavelength (nm) red700 green550 blue500 violet400 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. RedGreen

BlueViolet 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) 1 st 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: prism white 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.