The center of the fringes is at O, where the perpendicular bisector MO of S1 S2 meets T. Here the path difference S2O – S1O = O, since the paths are equal. On either side of O, the fringes cover only a very small distance as the wavelength of light is very small. So if P is the nth bright fringe from O, with S1 S2very small (0.4 mm, for example) and D very large relatively (1.0 m, for shown. The line MR to P is also parallel to S1A and S2B.
The path difference from S1 and S2 to P is the distance S2 X, where S2X is the perpendicular from S1 to S2B. From the geometry, angle S2 S1 X = θ angle OMP in the large triangle OMP.
From the small triangle S2 S1 X Sin θ = S2X/S2S = nλ/d
From the large triangle OMP Tan θ = Θp/θM = xn/D
Where is the distance of the nth fringe from center O. Now, the angle O is very small (only of the order of 0.5 ). So tan θ = Sin θ from trigonometry.
Then xn = nλ
D d
xn = nλD
d
For the (n-1) the fringe next to the nth fringe, since λ D and d are constant
nn-1 = (n-1) λD
d
So seperation of fringes, x = xn - x n-1 = λD/d
MEASUREMENT OF WAVELENGTH BY YOUNG’S INTERFERENCE FRINGES
A laboratory experiment to measure wavelength by Young’s interference fringes is show in Figure 19.8. Light from ka small filament lamp is focused by a lens
On to a narrow slit S, such as that in the collimator of a spectrometer. Two narrow slits A, B about 0.5 millimetre apart, are placed a short distance in front of S, and the light coming from A, B is viewed in a low-powered microscope or eyepiece M about one metre away. Some coloured interference fringes are then observed by M. A red and then a blue filter, F, placed in front of the slits, produces red and they blue fringes. Observation shows that the separation x of the red fringes is more than that of the blue fringes. Now λ = dx/D, where x is the separation of the fringes or λ__x .
So the wavelength of red light is longer than that of blue light.
An approximate value of the wavelength of red or blue light can be found by placing a Perspex rule R in front of the eyepiece and moving it until the graduations are clearly seen, Figure 19.8. the average distance, x between the fringes is then measured on R. the distance d between the slits can be found by magnifying the distance by a converging lens, or by using a traveling microscope. The distance D from the slits to the Perspex rule, where the fringes are formed, is measured with a metre rule. The wavelength can then be calculated from λ = dx/D; it is of the order 6 x 10-7
m. Further details of the experiment can be obtained from Advanced Level Practical Physics by Nelkon and Ogborn (Heinemann).
Measurements can also be made using a spectrometer, with the collimator and telescope adjusted for parallel light. The narrow collimator slit is illuminated by the sodium light, for example, and the double slits placed on the table. Young’s fringes can be seen through the telescope after alignment. From the theory on page 522, the angular separation of the fringes is λ/d. So by measuring the average angular separation of a number of fringes with the telescope, λ can be calculated from λ= dθ, if d is know or measured.
The wavelengths of the extreme colours of the visible spectrum vary with the observer. This may be 4 x 10 m (400 nm) for violet and 7.5 x 10-7 m (750nm) for red; an average value for visible light is 5.5 x 10-7 m (550 nm) which is a wavelength in the green.
APPEARANCE OF YOUNG’S INTERFERENCE FRINGES
The experiment just outlined can also be used to demonstrate the following points :-
1. If the source slit s is moved nearer the double slits the separation of the fringes is unaffected but their brightness increases. This can be seen from the formula x (separation) = λD/d since D and d are constant.
-
If the distance apart d of the slits is diminished, keeping S fixed, the separation of the fringes increases. This follows from x= λD/d
3. If the source slit S is widened the fringes gradually disappear. The slit S is then equivalent to a large number of narrow slits, each producing its own fringe system at different places. The bright and dark fringes of different systems therefore overlap, giving rise to uniform illumination. It can be show that, to produce interference fringes which are recognizable, the slit width of S must be less than λD/d, where D is the distance of S from the two slits A, B.
- If one of the slits, A or B, is covered up, the fringes disappear.
5. If white light is used, the central fringe is white, and the fringes either side are coloured. Blue is the colour nearer to the central fringe and red is farther away. The path difference to a point O on the perpendicular bisector of the two slits A, B is zero for all colours, and so each colour produces a bright fringe here. As they overlap, a white fringe is formed. Farther away from O, in a direction parallel to the slits, the shortest visible wavelengths, blue,
Produce a bright fringe first.
6. The intensity of a light wave is proportional to A 2 , where A is the amplitude of the wave. In constructive interference, the resultant amplitude of the wave from both slits is 2A if a is the amplitude of each wave. So the intensity (brightness) of the bright fringe is proportional to (2A) or 4A . the intensity of the light from one slit is proportional to A . So light energy is transferred from the dark fringe positions to the bright fringe positions in the Young experiment.
INTRODUCTION
It was a well established fact that light exhibits wave-nature which explains its phenomenon of interference and diffraction.
INTERFERENCE OF LIGHT
When there is a single source of light, the distribution of light energy in the surroundings medium is uniform in all directions. But when we have two coherent sources of light emitting continuous waves of same amplitude, same wavelength and same phase difference (or constant phase difference), the distribution of light energy does not remain same in all direction.
At some points, where the crest of one wave falls on crest of other, resultant amplitude is maximum. At certain other point the crest of one wave falls on the trough of other. Resultant amplitude becomes minimum and zero and hence intensity of light is minimum. This kind of modification in energy distribution is called interference.
Interference of light is the phenomenon of redistribution of light energy on account of superposition of light waves from two coherent sources.
At the points where the resultant intensity of light is maximum, interference is said to be constructive. At the points where the resultant intensity of light is minimum, the interference is said to be destructive.
YOUNG’S DOUBLE SLIT EXPERIMENT
Young in 1802, demonstrated the phenomenon of interference of light by a simple experiment.
S is a narrow narrow slit (of width about 1mm) illuminated by a monochromatic sources of light. At a suitable distance (about 10cm) from S, there are two fine slits A and B about 0.5 mm apart placed symmetrically parallel to S. When a screen is placed at a large distance (about 2m) from the slits A and B, alternate dark and bright bands running parallel to the lengths of slits appear on the screen. These are interference bands or interference fringes, the bands disappear when one of the slits A or B is covered.
The appearance of the fringes is due to the interference of light.
According to Huyens principle, the monochromatic source of light illuminating the slit S sends spherical wave fronts. Let the solid arcs represent the crest and the dotted arcs represents the troughs. These wave fronts reach slit A and B in turn become sources of secondary wavelets.
Thus, the two waves of same wavelength, same frequency and zero phase difference are given out by a A and B. The full line semicircles represents the crest and the dotted line semicircles represents trough. The dots represents positions of constructive interference, where crest of on wave falls on the crest of other and the trough on trough. The resultant amplitude and hence intensity of light is maximum at these positions. The lines joining the dots C,E,G on the screen.
Similarly, the crosses represent the positions of destructive interference, where the crest of one wave falls on trough of other and vice-versa. The resultant amplitude and hence the intensity of light is minimum at these positions. The lines joining the crosses lead to points D, F on the screen, thus, we have bright bands are C,E,G and dark bands at D and F. the bright and the dark bands placed alternately are equally spaced. These bands are also called interference fringes.
If S is a source of white light, interference fringes are coloured and their width are unequal.
CONDITIONS FOR CONSTRUCTIVE AND DESTRUCTIVE INTERFERENCE
Let the waves from two coherent sources of light be presented as :
y1 = a sin wt ……………………….1
y2 = b sin(wt+Q) …………………………2
Where a and b are the respective amplitude of two waves and Q is the constant phase angle by which second wave leads the first wave.
According to superposition principle, the displacement (y) of the resultant wave at time (t) would be as follows
Y = y1 + y2 = a sin wt + b sin(wt + Q)
= a sin wt + b sin wt cos Q + b cos wt sin Q
y = sin wt (a + b cos Q) + b cos wt sin Q …………………..3
Putting a + b cos Q = R Cosθ ………………….4
b sin Q = R sinθ …………………….5
y = R ( sin wt cosθ + cos wt sinθ )
y = R sin (wt + θ )
Thus, the resultant wave is a harmonic wave of amplitude R.
Squaring eq.4 and 5 and adding, we get
R2 ( cos2θ + sin2θ ) = ( a+bcos Q) 2 + (b sin Q) 2
R2 = a2+b2+2abcosQ
R= √ a2+b2+2abcosQ ……………………..6
As resultant intensity I is directly proportional to the square of the amplitude of the
resultant wave.
I α R 2
I α (a 2 +b 2 +2abcosQ)
For constructive interference I would be maximum, for which
CosQ = max = +1
Q =0, 2 π , 4 π ………
Q = 2n π where n = 0,1,2 ……………..
If x is the path difference between the two waves reaching point P, corresponding to phase difference Q, then
X= (λ / 2 π )Q = (λ / 2 π) X (2n π ) = n λ ………………………7
i.e. = n λ
Hence condition for constructive interference is a point that phase difference between the two waves reaching the point should be zero or an even int3egral multiple of . Equivalently, path difference between the two waves reaching the point should be zero or an integral multiple of full wavelength.
For Destructive Interference
It should be maximum
Cos Q = min. = -1
.Q = π, 3 π , 5 π …………
or Q = (2n-1) where n = 1,2………………….
The corresponding path difference between the two waves :
x = λ/2n x (2n-1) π = (2n-1)λ/ 2 ……………………..8
Hence, the condition for constructive interference at a point is that the phase difference between the two waves reaching the point should be an odd integral multiple of or path difference between the two waves reaching the point should be an odd integral multiple of half the wavelength.
-
Interference is constructive, when cos Q=1
From 7
R= √ a2+b2+2ab.1
i.e. R = (a+b) 2 = a+ b = sum of the amplitudes of waves, which would be max.
R(max) = (a+b)
As I α R2 (max)
I (max) α (a+b) 2
Again, interference is destructive, cos Q = -1
From (7), R= √ a2+b2+ 2ab(-1)
i.e. R= = √ (a-b) 2 = (a-b)=difference of the amplitude of two waves, which would be min.
R (min) = a-b
I(min) α R2 (min)
I (min) α (a-b) 2 ………..9
From eq.(8) & (9), we get :
I(max)/I (min) = (a+b) 2 / (a-b) 2
When b = a
R (max) = a+b = a+a = 2a
R (min) = a-b = a-a= 0
Therefore, I (min) = 0 i.e. dark bands will be perfectly dark and the control between the dark and bright will be the best.
From (7) R2 = a2 + b2 + 2ab cos Q
Taking a2 = I’; b2 = I” and R = I (r)
The resultant intensity as a result of superposition of the two waves, may be put as:
I(r) = I’ +I” + 2I’I” cos Q
2. If w’ and w” are the widths of two slits from which intensities of light I’ and I” emanate, then
w’/w” = I’/I” = a2 /b2………………………10
EXPRESSION FOR FRINGE WIDTH
OR THEORY OF INTERFERENCE OF LIGHT
Suppose A and B are the two fine slits distance d apart. Let them be illuminated by a strong source of monochromatic light of wavelength . MN is a screen at a distance D from the slits. The two waves starting from A & B superimpose upon each other, resulting in interference pattern on the screen, placed parallel to slits A & B.
The intensities at a point on the screen will depend upon the path difference between the two waves arranging that point. The point C on the screen is at equal distance from A & B. Therefore, the path difference between the two waves reaching C is zero and the point C is at maximum intensity. It is called central maximum.
Consider a point P at a distance between two waves arriving at P;
= BP – AP ……………………….11
Let O be the midpoint of AB, and
AB=EF=d, AE=BF=D
PF=PC – EC=x – d/2
And PF=PC+CF=x+d/2
In triangle BPF BP= [ BF2 + PF2]1/2
= [ D2 + (x + d/2) 2]1/2
= D [ 1+ (x+d/2) 2/D2] 1/2
Expanding binomially, we get
BP = D [ 1 + (x +d/2) 2 /2D2] ………………….12
In triangle APE AP = [AE2 + PE2] 1/2
= [ D2 + (x-d/2) 2] 1/2
= D[ 1+ (x-d/2) 2/D2] 1/2
Expanding binomially, we get
AP = D[1 + (x-d/2) 2 /2D2] ……………13
Putting in eq(11), we get path difference
BP-AP =xd/D
Now, the intensity at a point P is maximum or minimum according as the path difference (BP – AP) is an integral multiple of wavelength or an odd integral multiple of half wavelength. Thus for bright fringes (maxima)
Path difference = xd/D = nπ ….14
Where n = 0, 1, 2, 3 …………….
Or x = (nλD)/d ……………………15
Hence bright interference fringes are formed as detailed below :
For n = 0, x(0) = 0 i.e. at C…………………Central Maxima
For n= 1, x(1) = λD/d……………1st bright fringe
For n = 2, x(2) = 2λD/d ……………….2nd bright fringe
For n = n, x(n) = nλD/d…………………..nth bright fringe
The separation between the centers of two consecutive bright fringes is the width of a dark fringe.
β’ = x(n) – x(n-1)
β’ = nλ( D/d) –(n-1) ( λD/d) = λD/d …………….16
Similarly for dark fringes (minima)
Path difference,
Xd/D =(2n-1) λ /2
x= (2n-12) λ D/2d………………..17
Hence, dark interference fringes are formed as detailed below :
for n= 1, x(1) = λD/2d ……….1st Dark Fringe
for n= 2, x(2) = 3λD/2d ……….2nd Dark Fringe
for n= n, x(n) = (2n-1)λD/2d ……….nth Dark Fringe
Comparison with the above shows that dark interference are situated in between bright interference fringes. The separation between the centers of two consecutive dark fringes is the width of a bright fringe.
β” = x(n) –x(n-1) = (2n-1) λ D/2d - [2 (n-1) -1] D λ/2d
= D/d ……………..18
from (17) and (18) we get
β’ = β” = λ D/d ……………….19
Hence, all bright and dark fringes are of equal width.
Further, at sites of constructive interference,
I(max) α R² (max) α (a+b)² = constant = 4a², when a = b
Hence, all bright interference bands have the same intensity.
At the sites of a destructive interference
I(min) α R² (min) α (a-b)² = constant = 0, when a = b i.e.
all dark bands have the same intensities.
ACKNOWLEDGMENTS
I would like to extend my gratitude to my physics teachers Mrs. Ritu Sharma and Mrs. Indira Saini who spared their precious time and gave needful advice, without which this project would not have been completed. I am equally grateful to our lab assistant Mr.Deshbandhu for his timely help and guidance.
GAURAV GUPTA
PHYSICS
PROJECT
BY :
GAURAV GUPTA
XII SCIENCE-A
ROLL NO.
CERTIFICATE
This is to certify that my student Gaurav Gupta of Class XII SC A has completed the project titled “To study the interference of light using Helium – Neon Diode Laser”, sincerely and to the best of his efforts.
Mrs.Ritu Sharma
Department of Physics