• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17

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

Extracts from this document...




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.


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.



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)

...read more.


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  .

...read more.


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.


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



BY :





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.

This student written piece of work is one of many that can be found in our AS and A Level Waves & Cosmology section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Waves & Cosmology essays

  1. Peer reviewed

    The Electro magnetic spectrum.

    3 star(s)

    The current advice is to keep calls short. People who work on aircraft carrier decks wear special suits which reflect microwaves, to avoid being "cooked" by the powerful radar units in modern military planes. Infrared Infra-red waves are just below visible red light in the electromagnetic spectrum ("Infra" means "below").

  2. Estimating the wavelength of light using a double-slit and a plane diffraction grating

    Using a plane diffraction grating 3. Tabulate the results: For a grating of 3000 lines per cm, Grating separation d=(1/3000) cm=3.33x10 Colour Blue(1st order) Blue(1st order)(2) Blue(2nd order) x/m 0.155 0.15 0.314 tan? 0.155 0.15 0.314 sin? 0.153 0.148 0.300 ?=dsin?/m 5.09x10-7 4.93x10-7 4.995x10-7 Colour Green(1st order) Green(1st order)(2) Green(2nd order) x/m 0.17 0.16 0.354 tan?

  1. The aim of my coursework is to calculate the wavelength of red laser light ...

    This sketch below shows the part where the light builds maxima of fringes on the wall: First of all we need to calculate ? by using trigonometry: Tan ? = Distance between central maximum and 1st order spectrum Distance between grating and wall ?

  2. Refractive index by tracing light rays

    for light between the water-glass interface. The result was recorded in a table. 7. The experiment was repeated with different positions of the strip. Precautions: 1. The glass block should not be moved even for a little when pins have not been all placed for a ray.

  1. Free essay

    OCR Physics B Research Project - The Expanding Universe

    Justification for this assumption was required and it soon arrived as a result of meticulous observation and measurement of the stars by Edwin Hubble. But first, another important idea was announced. In 1927, a Belgian cosmologist named Georges Lema�tre reasoned that if the universe is expanding, it must have previously been smaller.

  2. Investigate any relationship present between the distance between a solar cell and a lamp, ...

    When increasing this, it would output more light, and when decreasing the power, less energy would be output in the light. You can vary this variable using a power pack, and it is also very easy to control with this equipment, therefore I will also measure this variable.

  1. Waves and Cosmology - AQA GCE Physics Revision Notes

    It acts as a force mediator. * The exchange particle can?t be detected in its transfer or it can?t act as the mediator of force. The larger the rest masses of the exchange particle, the lower the time it can be in flight without detection, therefore the lower the range.

  2. I intend to investigate whether any correlation exists between the wavelength of light exerted ...

    Since I am only changing the photons energy, it is imperative that the number of photons emitted from the LED are constant between different wavelength LED?s. The LEDs will therefore have to be calibrated,. Manufacturer tolerances of LED?s range from +- 10% on this factor so to attempt to maintain

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work