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A –Level Physics: Waves and Quanta: Wave Phase and Interference

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1 A –Level Physics: Waves and Quanta: Wave Phase and Interference

2 Objectives: Additional skills gained: Practical Planning
Integrating GCSE content

3 Starter: Definitions Define the following: Wavelength Period Frequency
Wavelength: the distance between two successive identical points that have the same pattern of oscillation. Period: the time taken for one complete pattern of oscillation Frequency: (1/period) or the number of oscillations per unit time You have 10mins

4 State what 180° would be in rads and wavelengths
Phase All points in a wave will be at some point through their cycle. One full cycle (360°) is one full wavelength (λ) which is 2π rads Each peak would be ¼ of a wavelength (90 °/ λ 𝟒 / π 𝟐 ) State what 180° would be in rads and wavelengths Delta

5 Practice Question C ( as it is 1.5 wavelengths, aka 3pi)

6 So in this example, the top wave is π/2 ahead of the bottom one!
Phase Difference π/2 Phase Difference is the difference in phase(measured in radians) between one point of a wave and the same point on another wave of identical frequency (coherent) that started at the same time So in this example, the top wave is π/2 ahead of the bottom one! Delta

7 Young’s Double Slit Experiment
Robert Young was the first person to measure the wavelength of light as he believed that light travelled in the form of a wave. His double-slit experiment utilises constructive and destructive interference of waves that are diffracted through two slits Why would you need to use a monochromatic source of light? What measurements would you need to take to calculate the wavelength? 10m

8 Superpositioning If two waves interfere in phase the waves will make a resultant with a larger amplitude This is known as constructive interference and only works when they have phase differences of an EVEN number (rads) e.g. (0, 2π, 4π) Delta If their phase difference is an ODD number (rads) Then they interact destructively (e.g. π, 3π, 5π)

9 Path Difference Path difference is the difference in distance that two waves have travelled when they reach the same point (P). Measured in wavelengths P Constructive= Whole number wavelengths path difference Delta Destructive= Half wavelengths path difference

10 One full cycle (360°) is one full wavelength (λ) which is 2π rads
Summary Path Difference Phase Difference In degrees In radians CONSTRUCTIVE Whole Number of Wavelengths e.g. 0, λ, 2λ, 3λ 0, 360, 720 0, 2π, 4π DESTRUCTIVE Half Wavelengths e.g. ½λ, 1 ½λ, 2 ½ λ 180, 540, 900 π, 3π, 5π REMEMBER One full cycle (360°) is one full wavelength (λ) which is 2π rads

11 Interference Peak Trough
Where the peaks cross, we know the waves must be in phase (green dots) Wave 1 Where it is a peak for one wave but the trough for another then the waves at that point are out of phase (red dots) Easy: if a point that has both line cross = constructive and if only on one line=destructive Wave 2

12 Interference Pattern So if a screen is placed away from the interfering waves you’ll get a pattern like this… There is a central maximum and fringes of brightness (maxima) and fringes of darkness (minima) If the source was sound, you’d hear the source as loud in the maxima and quiet in the minima

13 Calculating the Wavelength
λ = 𝑆𝑥 𝐷 Fringe Separation (x) Note: when taking the measurement of fringe separation you need to take it from the centre of the bright fringe to the centre of the next. To reduce uncertainty you should take the measurement of a number of fringes and then divide by the number (e.g. 6 fringes, then divide by 6). Slit separation (S) mms-1 or 9.67x10-4 ms-1 Screen Distance (D)

14 Calculate the wavelength of the laser light
Practice Question λ = 𝑆𝑥 𝐷 A helium-neon laser produces a beam of light that passes through a double slit and creates a fringe pattern on a screen 4m from the slits. The slits are placed 1.0mm apart and when the distance across 20 fringes are measured, they come to 51mm. Calculate the wavelength of the laser light Fringe Separation (x) Slit separation (S) 2. 1x10-3 X 2.55x10-3 divided by 4m = 638nm Screen Distance (D)

15 Independent Study Research how interference is utilized in CDs/DVDs and write at least ½ page of notes based upon this research.


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