## WAVE MOTION-2

DERIVATION OF SNELL’S LAW ON THE BASIS OF HUYGEN’S PRINCIPLE
-Consider a plane wave front AA’ which is incident on a transparent medium MM’
The position of a new wave front after time t may be found by applying Huygen’s principle with points on AA’ as centers.
Those wavelets originating near the upper end of AA’ travel with a speed medium and those wavelets originating near the lower end of AA’ travel with a speed medium.
Hence, after time = t:
è Distance travelled in the medium =
è Distance travelled in the medium =
RELATIONSHIP BETWEEN i and r
Where i = angle of incidence r = angle of reflection
From âˆ†AOQ
Sin i = ——————– (1)
From âˆ†AOR
Sin r = ————— (2)
Dividing
For a given pair of media, this ratio is constant which expresses Snell‘s law. From V =
– Equation (3) becomes
1η2
v
– Where Wavelength in the medium
= wavelength in the medium
1η2 = refractive index of the second medium with respect to the first.
Definition
-The refractive index (η) of material is the ratio of the sine of the angle of incidence to the sine of the angle of refraction.
Problem 36
Monochromatic light of a wavelength 800nm enters a glass plate of refractive index 1.5
Calculate;
1. The speed of light in glass
2. The frequency of the light
3. The wavelength of light in glass
Given that C = 3 and 1nm = 1m
INTERFERENCE OF LIGHT WAVES
Interference is a situation in which two or more waves overlap in space. In a region where two or more light waves cross, superposition occurs giving reinforcement (addition) of the waves at some points and cancellation (subtraction) at others.
The resulting effect is called an interference pattern or system of fringes
Definition
Interference pattern/system of fringes is a set of light bands (fringes) and dark bands (fringes) formed on a screen when interference of light waves occurs.
INTERFERENCE PATTERN
————- B
————- D
————- B
————- D
————- B
————- D
————- B
• Where B = Bright band (fringe)
D = Dark band (fringe)
In order to bring about the interference of light waves the first task is to produce two coherence sources of light.
Coherent source of light are sources producing light of the same wavelength, frequency and amplitude.
• The sources have constant phase difference
• Two independent sources of light cannot be coherent.
• It is because the emission of light from any source is from a very larger number of atoms and the emission from each atom is random.
• Therefore, there is no stable phase relationship between radiations from two independent sources.
HOW TO PRODUCE COHERENT SOURCES OF LIGHT
• -Coherent sources of light are obtained by splitting up the light from a single source (primary source) into two parts (secondary source).
METHODS OF PRODUCING INTERFERENCE PATTERN
• Interference pattern can be produced by either of the following methods
1. Young‘s double slit
2. Lloyd‘s mirror
3. Fresnel‘s biprism
4. Newton‘s rings
5. Wedge fringes
In this experiment monochromatic light is passed through a slit ―s‖ and the light emerging from this slit is used to illuminate two adjacent slits
By allowing light from these two slits to fall on the screen, a series of alternate bright and dark bands / fringes are formed on the screen.
The bright bands/ fringes represent constructive interference while the dark bands / fringes represent destructive interference.
CONSTRUCTIVE INTERFERENCE
Constructive interference occurs when the interfering wave are in phase.
Waves are said to be in phase if their maximum and minimum values occur at the same instant. Example
separated waves resultant displacement
-When constructive interference occur the two interfering waves combine together to give a wave of larger amplitude and hence bright fringe is formed.
DESTRUCTIVE INTERFERENCE
Destructive interference occurs when the two interfering waves are out of phase.
Waves are said to be out of phase if the maximum of one wave and minimum the other wave are formed at the same instant.
Example
When destructive interference occurs the two interfering waves cancel each other to produce nothing and hence a dark fringe is formed.
THEORY OF YOUNG’S DOUBLE SLIT EXPERIMENT
Consider the following below
Let A and B be two close slits separated by a distance, a
Let O be central bright fringe
Suppose the bright fringe to be formed at point P a distance from O.
CONCEPT
If the two coherent monochromatic light sources are A and B then a bright fringe can be seen at P only if the path difference (BP – AP) is a whole number of wavelength.
=
Where n = 0, 1, 2, 3, ………………..
(2) If the difference is an odd number of half wavelength then darkness is obtained at the point considered.
I.e. BP – AP =
here n = 0, 1, 2, 3, ………………………
In the figure above, for a bright fringe to be seen at P, we have:
Path difference = BP – AP =
• But, AP = NP è Path difference = BP – NP = v BN = ———— (1)
• Let be angular displacement of the bright fright from central bright fringe.
• Consider âˆ†ABN and âˆ†PMO è m(BN) = m(PMO) =
• From âˆ†BAN
Sin
• Where and AB = a
Sin =
Since is very small angle sin (in radian)
v ————- (2) – From âˆ†PMO
è
• Where D = Distance from the slits to the screen
• Since is a very small angle tan (in radian)
——————- (3) è equation (2) = equation (3)
——————— (4)
• Where n = 0, 1, 2, 3, ……………….
• Equation (4) above gives the displacement of bright fringe from central bright fringe at O.
NOTE.
• Central bright fringe, n= 0
• First bright fringe, n = 1
• Second bright fringe, n = 2
e.t.c
• Similarly, the displacement from O for a dark fringe is given by:

——————– (5)
• First dark fringe, n = 0
• Second dark fringe, n = 1
• Third dark fringe, n = 2
e.t.c
DISTANCE BETWEEN TWO CONSECUTIVE BRIGHT / DARK FRINGES (FRINGE
WIDTH)
Let be displacement of bright fringe from O.
• For the next bright fringe
—————- (2)
• The distance between two consecutive bright fringes () is given by:
• The same formula applies for the separation between two consecutive dark fringes.
All fringes are of the same width.
SOME POINTS ABOUT YOUNG’S INTERFERENCE FRINGES
1. If the source slit ―S‖ is put very close to the double slits the fringe separation does not change but the fringe intensity will increases.
2. If the double slits and are very close so that their separation distance approaches zero, then the fringe separation increases
3. If one of the double slits is covered up, then no fringes can be seen on the screen.
4. If the source slits S is made wider, then the fringes on the screen disappears.
5. If white light is used instead of monochromatic light source then the central fringe at O is white but the other bright fringe on either sides of the central fringe are colored with violet near ―O‖ and red far away from ―O‖.
ANGULAR WIDTH OF A FRINGE
If P is the position of the first bright fringe then OP =
a fringe
Since is very small, therefore tan (in radian)
——————- (1)
From the expression of fringe width
————– (2)
Problem 37
Two slits are at a distance of 0.2mm apart and the screen is at a dist
ance of 1m. The third bright fringe is found to be displaced 7.5mm from the central fringe. Find the wavelength of the light used.
Problem 38
A yellow light from a sodium vapour lamp of wavelength 5893is directed upon two narrow slits of 0.1cm apart. Find the position on the screen 100cm away from the slits.
Problem 39
In Young’s experiment, the distance of the screen from the two slits is 1.0m. When light of wavelength 6000 is allowed to fall on the slits the width of the fringes obtained on a screen is 2.0mm. Determine:
1. The distance between the two slits and
2. The width of the fringes if the wavelength of the incident light is 4800.
Problem 40
In double slit experiment, light has a frequency of 6 . The distance between the centres of adjacent bright fringes is 0.75mm. What is the distance between the slits if the screen is 1.5m away?
Given that the speed of light in free space 3 .
Problem 41
In a Young’s double slit experiment, two narrow slits 0.8mm apart are illuminated by the same source of yellow light (). How far apart are the adjacent bright bands in the interference pattern observed on a screen 2m away?
Problem 42
In young’s double slit experiment the angular width of a fringe formed on a distant screen is 0.1. The wavelength of light used is 6000. What is the spacing between the slits?
Problem 43
A
beam of light consisting of two wavelength 6500 and 5200 is used to obtain interference fringes in a Young’s double slit experiment.
1. Find the distance of the third bright fringe on the screen from the central maximum to the wavelength 6500.
2. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
The distance between the slits is 2mm and the distance between the plane of the slits and the screen is 120cm.
Problem 45
A Young’s double slit arrangement produces interference fringe for sodium light ( ) that are
0.20 apart. What is the angular fringe separation if the entire arrangement is immersed in water? Given that refractive index of water = 4/3
Problem 46
In Young’s double slit experiment the distance between the slits is 1mm and the distance of the screen from the slits is 1m. If light of wavelength 6000 is used, then find the distance between the second dark fringe and the forth bright fringe.
FRINGE SHIFT
-When a thin transparent plate of thickness (t) and refractive index ( is introduced in the path of one of the interfering waves ( say in the path P.) then the effective path in air is increased by ()t due to the introduction of the plate.
• Effective path difference in air
=
=
• But
• Effective path difference in air
=
• For a bright fringe, path difference =
è è v ————— (1) – For the next bright fringe, we have:-
—————– (2)
• The fringe width is given by :
v —————— (3)
• From equation (1)
• In the absence of the plate t = 0
è
• Let âˆ†X be the fringe shift due to introduction of the plate.
• âˆ†X =
• From equation (5)
v ——————————- (6)
• Substitute equation (6) in equation (5)
Thus, with the introduction of the transparent plate in the path of one of the slits, the entire fringe pattern is displaced through a distance given by either equation (5) or equation (7) towards the side on which the plates is placed.
Problem 47
In the figure below are two coherent light sources in a Young’s two slit experiment separated by a distance 0.5mm and O is a point equidistant from at a distance 0.8m from the slits. When a thin parallel sided piece of glass (G) of thickness 3.6 m is placed near as shown, the central fringe system moves from O to point P. Calculate OP. (the wavelength of light used = 6.0 m).
CONDITIONS NECESSARY FOR SUSTAINED INTERFERENCE OF LIGHT WAVES
• Sustained interference is an interference pattern in which the positions of maxima and minima remain fixed.
• In order for light wave to produce an interference pattern which can be observed on a screen the following condition must be satisfied.
(1)The light should be monochromatic.
If this is not so, fringe of different color will overlap.
(2)The two sources producing interference must be coherent.
(3)The two interfering waves must have the same plane of polarization.
(4)To observe interference fringe clearly. It necessary that the fringe width is sufficiently large. This is possible if:
1. The two coherent sources are parallel and close to each other
2. The distance between slits and screen is reasonably large.
A slit S1 illuminated by monochromatic light and placed closed to a plane mirror MN
Interference occurs between direct light from the slit and light reflected from the mirror MN
The reflected ray MA appear to be coming from a virtual image S2 of S1
S1 and S2 act as two coherent sources of, the expression giving the fringe spacing is the same as for the double slit
Except that dark fringe will be obtained at N instead of a bright fringe
Bi prism
This is a glass prism with an obtuse angle that functions as two acute angle prisms placed.
A double image of a single object is always formed by means of this prism
The device was used by Fresnel to produce two coherent beams for interference experiment
Monochromatic light from a narrow slit S falls on a bi prism as shown in the figure above.
Two virtual images S1 and S2 of S are formed by refraction at each half of the bi prism and these acts as coherent source which are close together.
An interference pattern, similar to that given by the double slit but brighter, is obtained in the shaded region where the two refracted beans overlap.
The theory and the expression for the fringes spacing are the same as for young’s method
NEWTON’S RINGS
This is an interference effect discovered by Newton‘s
Interference fringes are formed by placing a slightly Plano-convex lens on a flat glass plate H
A glass plate G reflects the light that comes from the monochromatic light source towards the lens L downwards
Both reflected rays are going to be analyzed by the traveling microscope M and at the point of contact of the lens a dark spot is seen surrounded by a series of alternate bright and dark rings. These are called Newton’s rings.
##### Definition
Newton‘s ring is a circular interference fringes formed between a lens and a glass plate with which the lens is contact. Concept
Consider the air film PA t
Some of the incident light is reflected at P towards the microscope M
Some of the light passes to point A where it is reflected by glass plate H towards the microscope M.
For nth dark ring Path difference = 2t = nλ
When n 0, 1, 2, 3,
Central dark spot, n 0
At constant pressure,
i.e. At constant pressure the mean free path of a gas molecule λ T
At constant pressure the mean free path of a gas molecule is directly proportional to its absolute temperature T.
λ = KT
Where K constant of proportionality
If λ1 and λ2 are mean free path of a gas molecule at temperature T1 and T2 respectively:
And λ1 = KT1 λ2 = KT2
2. Pressure P of the gas
From equation (3)
At constant temperature the mean free path of a gas molecule is inversely proportional to pressure. Pλ = K
Where K constant of proportionality
Generally, we may write
P1λ1= λ2P2

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## 1 Comment

• ### Yusufu Lungwa, June 10, 2024 @ 8:08 pmReply

Nampenda kusoma mungu in bleeding me

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