Interference of Light Waves and Young’s Double Slit Experiment

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Interference of Light Waves :-

When two light waves from different sources meet together, then the distribution of energy due to one wave is distributed by the other. This modification of light energy due to superposition of two light wave is called interference of light waves.

Coherent sources :-

Those sources of light which emit light waves continuously of same wavelength, time period, frequency, amplitude and have zero phase difference or constant phase difference are coherent sources.

Interference of light waves may be of

Two types –

  1. Constructive Interference.
  2. Destructive Interference.
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Fig. 1 :- Interference of light waves.

Constructive Interference :-

When two light waves are superpose with each other in such a way that the crest of one wave falls on the second wave and through of one wave falls on the through of the second wave, then the resultant wave has larger amplitude and it is called constructive interference.

Destructive Interference :-

When two light waves superpose with each other in such a way that the crest of one wave coincides through of the second wave, then the amplitude, intensity of resultant wave becomes zero and it is called destructive interference.

Essential conditions for Constructive Interference :-

  1. For Constructive Interference phase difference of the waves at the point should be even multiple of ╧А

i. e ╧Ж = 0, +2╧А, -2╧А, +4╧А, -4╧А, +8╧А, -8╧А ……

Or

╧Ж = +2n╧А, -2n╧А

where n = 0, 1, 2, …..

2. For Constructive Interference path difference of the waves should be even multiple of ╬╗/2

If light sources SтВБ and SтВВ , distance from the point P is yтВБ and yтВВ respectively, then the path deference

тИЖP = SтВБP – SтВВP = yтВВ – yтВБ

Since the value of phase difference between separate two points from ╬╗ distance 2╧А

Hence the phase difference for тИЖP path difference

╧Ж = (2╧А/╬╗) тИЖP

We know that ╧Ж = 2╧Аn

Hence (2╧А/╬╗) тИЖP =2n╧А

тИЖP = n╬╗

Or

тИЖP = 2n╬╗/2

3. For Constructive Interference the resultant Intensity should be maximum.

For maximum Intensity

cos╧Ж = +1

If the displacement produced by The source SтВБ and SтВВ at the point P is given by

yтВБ = a cos╧Йt

yтВВ = a cos╧Йt

Then the resultant displacement at Point P would be given by

y = yтВБ + yтВВ = 2 a cos╧Йt

Since the Intensity is proportional to the square of the amplitude, the resultant Intensity will be given by

I = 4IтВА

Where IтВА is the Intensity of each source and IтВА proportional a┬▓

Essential conditions for Destructive Interference :-

  1. For Destructive Interference phase difference of the waves at a point should be odd multiple of ╧А

i. e ╧Ж = +╧А, -╧А, +3╧А, -3╧А, +5╧А, -5╧А, +7╧А, -7╧А …..

Or

╬ж = (2n +1)╧А

where n =0, 1, 2, 3, ….

2. For Destructive Interference path difference of the waves at a point should be odd multiple of ╬╗/2

╧Ж = (2n +1)╧А

The phase difference тИЖP path difference

╧Ж = (2╧А/╬╗)тИЖP

= (2n +1) ╧А

Or

тИЖP = (n + 1/2) ╬╗

Or

тИЖP = (2n +1) ╬╗/2

3. For Destructive Interference the resultant Intensity should be minimum.

For minimum Intensity

cos╧Ж = -1

For Destructive Interference the resultant Intensity will be zero.

Young’s Double Slit Experiment :-

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Fig. 2 :- Interference of light waves and Young’s double slit experiments.

Young’s performed an experiment to prove the wave nature of Light by explaining the phenomenon of interference. He used to coherent sources to perform in this experiment.

He used a light bulb, two small slits SтВБ and SтВВ , Source S .

Here in the above figure, we and say that the slits are placed very close to each other and are separated by the distance d.

There is a screen placed in front of the setup. He observed that alternate dark and light bands were formed on the screen, such bands are called fringes.

  • The source S illuminate the Source SтВБ and SтВВ therefore the light from SтВБ and SтВВ becomes coherent because both SтВБ and SтВВ receive their light from same source S . So if there is any change in the phase, the change will reflect in both SтВБ and SтВБ when both the slits are open, fringes are formed.
  • We can say that SтВБ and SтВВ will always remain in phase. In the double slit experiment consecutive bright as well as dark fringes are seen on the screen as a consequence of the type of interference of light waves.
  • The Interference fringe minimum occur for Path difference = (2n +1)╬╗
  • The Interference fringe maximum occur for path difference = n╬╗

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