Interference of waves
When two or more travel in the same or different directions in a given space, variations in the size of the resulting disturbance occur at points where the meet or overlap.
Interference is the superposition of two waves originating from two coherent sources.
Two waves are in coherent if they are of the same frequency, same amplitude and in phase (same phase)
There are two types of interference :
(a) Constructive interference occurs when the crests or troughs of both waves coincide to produce a wave with maximum amplitude.
(b) Destructive interference occurs when the crests of one wave coincide with the trough of the other waves to produce a wave with zero amplitude.
Interference of the waves is a result or obeys the principle of superposition.
Principle of superposition
Principle of superposition state that “ When two waves move simultaneously and coincide at a point the sum of the displacements at that point is equal to the sum of the displacements of the individual waves by vector method”.
(a) Superposition of two crests
Constructive interference
Constructive interference is a point at which two crest coincide.
(b) Superposition of two troughs
Constructive interference
Constructive interference is a point at which two trough coincide.
(c) Superposition of a crest and a trough
Destructive interference
Destructive interference is a point at which a crest and a trough coincide
Analysing Interference of water waves
To produce the interference pattern of the water waves in a ripple tank we can use:
(a) Two dippers operated from the same motor,
Or
(b) A plane water waves passing through two slits.
The figure shows the interference pattern produced by two sources of water P and Q.
1. A cross (x) in the figure is the points where the crests coincide crests.
2. Mark with a star () in the figure is the points where the troughs coincide troughs .
The points (x) and () is called as constructive interference.
constructive interference is a point where two crest coincide or two trough coincide.
At that points constructive interference occurs.
3. A circle (○) in the figure is the points where a crest coincide with a trough.
The points (○) is called as Destructive Interference.
Destructive Interference is a point where the total displacement is zero.
At the points Destructive Interference occurs.
4. Connect the points (x) and (_{*}) to produce several lines that we call as node lines
5. Connect the points (○) to produce several lines that we call as antinode lines
6.The value of l can be calculated from where :
λ = wavelength
a = source separation
x = distance between two successive constructive interference or Destructive Interference
D = distance between source and screen.
How to change the pattern of interference of waves?
The pattern of interference depends on the distances between two consecutive nodes or antinodes lines , x.
How to change x ?
Conclusion : As λ increases , x increases
As D increases , x increases
As a increases , x decreases
As λ increases , x increases
The experiment to investigate the relationship between the distance between to coherent sources and the distance between two consecutive node lines
Hypothesis:
The distance between two consecutive node lines
increases as the distance between to coherent sources decreases
Aim of the experiment :
To investigate the relationship between the distance between to coherent sources and the distance between two consecutive node lines
Variables in the experiment:
Manipulated variable: the distance between to coherent sources
Responding variable: the distance between two consecutive node lines
Fixed variable: frequency of vibrator or the wavelength
List of apparatus and materials:
Ripple tank, lamp, motor ,wooden bar , power supply ,white paper , spherical dippers ,metre rule and mechanical stroboscope.
Arrangement of the apparatus:
The procedure of the experiment which include the method of controlling the manipulated variable and the method of measuring the responding variable.
By using a metre rule , the distance between two dippers is measured = a
The power supply is switched on to produce two circular waves from the dippers
The waves are freeze by a mechanical stroboscope.
The waves are sketched on the screen.
By using the metre rule , the distance between two consecutive node lines is measured = x
The experiment is repeated 5 times for with different distances between two dippers
Tabulate the data:
a 






x 






Analysis the data:
Plot the graph x against a
Example 1
In an experiment to investigate the pattern interference of a water waves, the distance between two spherical dippers is 2.5 cm and the distance between two consecutive antinodes lines is 5.0 cm. What is the wavelength of the water waves if when the distance from two dippers to the point of measurement is 10 cm.
Solution
Interference of light waves
When light from the same source passes through two narrow slits which are close
together the effect known as interference can be seen as the bright and dark fringes.
The bright fringes to be formed by constructive Interference and the dark fringes to be
formed by destructive interference.
Interference fringes produce by using Doubleslit
interference Young’s experiment.
For all practical purposes, monochromatic light is used.
Monochromatic light which is light of only one colour or one wavelength.
The formula for interference of light waves is,
Where,
λ = wavelength of light waves
a = slit separation
x = distance between two successive bright or dark fringes
D = distance between double slit and screen.
Example 2
In a doubleslit interference experiment with blue light the distance between the screen and double slit is 1.2 m and slit separation is 2 x 10^{4} m . Six successive bright fringes at a distance 1.2 x 10 ^{2 }m is formed on the screen. Calculate the wavelength of the blue light.
Solution
The experiment to investigate the relationship between the wavelength of the light waves distance and the distance between two consecutive bright fringes
Hypothesis:
The distance between two consecutive bright fringes
increases as the wavelength of light waves increases.
Aim of the experiment :
To investigate the relationship between the wavelength of the light waves distance and the distance between two consecutive bright fringes
Variables in the experiment:
Manipulated variable: The wavelength of the light waves( colour of light)
Responding variable: the distance between two consecutive bright fringes
Fixed variable: slit separation and the distance between double slit and screen.
List of apparatus and materials:
Source of light,colour filter, screen, single slit, double slit and metre rule.
Arrangement of the apparatus:
The procedure of the experiment which include the method of controlling the manipulated variable and the method of measuring the responding variable.
A green filter is placed between the light source and the slits.
The source of light is switched on.
The interference pattern formed on the screen is observed and drawn.
By using a metre rule the distance across 6 consecutive bright fringes is measured.
The distance between two consecutive bright fringes is calculated ,
The experiment is repeated 5 times for with different colour filters
Tabulate the data:
λ 






x 






Analysis the data:
Plot the graph x against λ
The arrangement of colour of light in order of wavelength
Interference of sound waves
Like other types of waves ,sound waves can also give interference effects. Interference of sound waves produce regions of louder sound by constructive interference and regions quiet by destructive interference.
When two similar loudspeakers are connected to the same audiofrequency generator they will produce interference effects.
The formula for interference of sound waves is,
Where,
λ = wavelength of sound waves
a = distance between two loudspeakers
x = distance between two successive loud regions or quiet region.
D = distance between the listener from the loudspeaker.
Example 3
In an experiment on the interference of sound wave a listener at distance 5.0 from the loudspeaker. The distance between two loudspeakers is 2.0 m. The loudspeakers are connected to an audiofrequency generator to produce sound waves at a frequency of 0.8 kHz.
Calculate
(a) the wavelength of sound waves when the speed of sound is 320 ms^{1}
(b) the distance between two successive loud regions
Solution:
The experiment to investigate the relationship between distance of the listener from the loudspeaker and the distance between two successive loud regions
Hypothesis:
the distance between two successive loud regions increases as distance between the listener from the loudspeaker increases .
Aim of the experiment :
To investigate the relationship between distance between the listener from the loudspeaker and the distance between two successive loud regions
Variables in the experiment:
Manipulated variable: Responding variable: distance between the listener from the loudspeaker
Responding variable: the distance between two successive loud regions
Fixed variable: the wavelength of sound waves , distance between two loudspeakers
List of apparatus and materials:
Two loudspeakers, audiofrequency generator ,connection wires and metre rule.
Arrangement of the apparatus:
The procedure of the experiment which include the method of controlling the manipulated variable and the method of measuring the responding variable.
By using a metre rule the distance between the listener from the loudspeaker is measured= D
The audiofrequency generator is switched on.
The listener is requested to walk in a straight path and the distance between two successive loud regions is measured by a metre rule = x
The experiment is repeated 5 times for with different distances between the listener from the loudspeaker
Tabulate the data:
D 






x 






Analysis the data:
Plot the graph x against D
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