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# Study the interference of light using Helium - Neon Diode Laser.

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Introduction

## TO STUDY THE INTERFERENCE OF LIGHT BY USING HELIUM -DIODE LASERCOHERENT SOURCES

As we see later, light waves from a sodium lamp, for example, are due to energy changes in the sodium atoms.  The emitted waves occur in bursts lasting about second. The light waves produced by the different atoms are out of phase with each other, as they are omitted randomly and rapidly.  We  call such sources of light waves as these atoms incoherent sources on account of the continual change of phase.

Two sodium lamps X and Y both emit light waves of the same colour or wavelength. But owing to the random emission of light waves from their atoms, their resultant light waves are constantly out of phase.  So X and Y are incoherent sources.  Coherent sources are those which emit light waves of the same wavelength or frequency which are always in phase with each other or have a constant phase difference. As we now show, two coherent sources can together produce the phenomenon of interference.

INTERFERENCE OF LIGHT WAVES, CONSTRUCTIVE INTERFERENCE

Suppose two sources of  light, A, B have exactly the same wavelength and amplitude of vibration, and that their vibration are always in phase with each other, fig.1. The two sources A and B are therefore coherent sources. fig.1

Their combined effect at a point is obtained by adding algebraically the displacements at the point due to the sources individually.  This is known as the principle of superposition.  So, their resultant effect at X, for example, is the algebraic sum of the vibrations at X due to the source A alone and the vibrations at X due to the source B alone.  If X is equidistant from A and B, the vibrations at x are to the two sources are always in phase as (i)

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Middle

λD/d since D and d are constant.
1. 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.

1.      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  .

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Conclusion

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  ...read more.

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