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Interference of Light, an integral portion in Wave Optics, implies the phenomenon of redistribution of light energy in a medium when two or more light waves having similar frequencies and same coherent sources superimpose with each other. The scientist named Thomas Young by performing the original version of the modern double-slit experiment (Young’s double-slit interferometer) made it clearer how the interference of light can be understood.
Key Terms: Diffraction, Single Slit Diffraction, Young’s Single Slit Experiment, Fringe Width, Huygens' guideline, interference of light
Definition of Interference of light
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Due to the superposition of two light waves of the same frequency travelling in the same direction, the resultant intensity of light is different from the sum of the intensities of two waves at any point. Due to the superposition of two light waves, the redistribution of light is called interference of light.
When the resultant intensity is maximum, the interference is said to be constructive and when the resultant intensity is minimum or zero, interference is said to be destructive.
Conditions of Interference
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Let S1 and S2 be two similar slits very close to each other and very close to the monochromatic source of light. The two sources are in the same phase.
Let a1 and a2 be the amplitudes of light emitting from sources S1 and S2 respectively. The waves arrive at P, destination point, via paths S1P and S2P respectively which makes them superposed at P with Ø phase difference and x is the path difference. The displacement at P by the simple harmonic waves travelling from S1 and S2 can be represented as-
y1 = a1sin(wt – kx)
y2 = a2sin(wt – kx + Φ)
The resultant displacement at point P, by the principle of superposition, is-
y = y1 + y2
= a1sin(wt – kx) + a2sin(wt – kx + Φ)
= a1sin(wt – kx) + a2sin(wt – kx + Φ)sinΦ + a2sin(wt – kx)cosΦ
= sin(wt – kx)(a1 + a2cosΦ) + cos(wt – kx)(a2sinΦ)
from which we obtain the following- I = I1 + I2 + 2 \(\sqrt{I_1I_2}\)
cos ΦThus we conclude that the resultant intensity, at any given point, will depend on the phase difference Ø between the two waves.
Phase difference for the two waves is as follows- Φ = 2mr
| Topics Related to Wave Optics | ||
|---|---|---|
| Carnot engine | Huygens Principle | Polarisation |
| Diffraction | Coherent and Incoherent Addition of Waves | Polarization of Light |
Constructive Interference
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Those points where two waves meet in the same phase, the resultant intensity is maximum, hence interference at those points are said to be constructive. The resultant is maximum at points where two waves meet in the phase difference of 0, 2π, 4π and so on. The path difference is as follows.
Path Difference = mλ
Destructive Interference
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The point at which the resultant intensity is minimum is said to have a destructive interference. The phase difference for two waves to meet in dissimilar phases is π, 3π, 5π and so on. The path difference is as follows-
Path Difference = (2m – 1)\(\frac{\lambda}{2}\)
Young’s double slit Experiment
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Light's nature of interference
Light's nature of interference can be explained by the young’s double slit experiment.
As shown in the figure, he used a source of light S and two slits, S1 and S2, at a distance d from each other and a screen is set up in front of the set up.
The monochromatic light from source S falls on the slits S1 and S2 which are at a distance d from each other, and act as coherent sources of light, due to which the waves coming from them superimpose and a pattern of interference is obtained. A patch of alternate bright and dark bands are obtained on the screen in front of them, where these bands are called fringes.
young’s double slit experiment
Characteristics of Young’s double-slit experiment
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- The central fringe is always at 00, hence it is always bright.
- Due to a slit, the fringe pattern obtained is brighter.
- The minima will not be completely dark if the slits are of unequal lengths. For very large widths, there is a uniform illumination.
- There is no interference pattern observed if one slit is illuminated with red light and the other is illuminated with blue light.
- If the two sources are object and its reflected image then the central fringe is dark rather than a bright one.
- Path difference between the two waves at a point is given by
Δ = \(\frac{xd}{D}\) = d sinθ
Where x is the position of point P from central maxima.
Following the diagram as below-
Young’s double-slit experiment Diagram
Maxima at P are n, where n=0, 1, 2….
Minima at P is Δ = \(\frac{(2n-1)\lambda}{2}\)
Where n is 1, 2….
Condition for observing sustained interference
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- Interference will not be sustained if the phase difference between the interfering waves is not constant.
- If the frequency and wavelengths of two waves is not equal, the phase difference will not be constant, hence interference is not sustained.
- The source must be monochromatic in nature as it eliminates overlapping of patterns as each wavelength is associated with one interference pattern.
- The eyes cannot resolve fringe resulting in uniform illumination if the fringe length is small. Hence the source must always be close to each other.
Fringes of YDSE
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As all fringes are of same width, width of each fringe is- β = \(\frac{\lambda D}{d}\)and its angular width is- θ = \(\frac{\lambda}{d} = \frac{\beta}{D}\)
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The fringes of the YDSE changes with medium according to the following – \(\beta = \frac{\lambda D}{d}\) β ∝ λ
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With increase in the separation of the slits, the fringe width decreases.
- Position of nth bright fringe from the central maxima- \(y = \frac{n \lambda D}{2(\mu -1) \alpha Z_1}\)
- Position of the nth dark fringe from the central maxima is- \((2n - 1) = \frac{\lambda D}{2D}\)
Things to Remember
- Interference of Light, an integral portion in Wave Optics, implies the phenomenon of redistribution of light energy in a medium when two or more light waves having similar frequencies and same coherent sources superimpose with each other.
- Due to the superposition of two light waves, the redistribution of light is called interference of light.
- Those points where two waves meet in the same phase, the resultant intensity is maximum, hence interference at those points are said to be constructive.
- The point at which the resultant intensity is minimum is said to have a destructive interference.
- Light's nature of interference can be explained by the young’s double slit experiment.
- Interference will not be sustained if the phase difference between the interfering waves is not constant.
- If the frequency and wavelengths of two waves is not equal, the phase difference will not be constant, hence interference is not sustained.
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Previous Year Questions
- When hydrogen atom is in its first excited level, its radius, is….
- When an electron does transition from n=4 to n=2 , then emitted line spectrum will be…...
- α -particle consists of….
- An electron of a stationary hydrogen atom passes from the fifth energy level….
- Complete the equation for the following fission process…. [NEET 1998]
- using non-relativistic approach, the speed of electron in this orbit will be…. [NEET 2015]
- When the glancing angle of incidence of light on a material is...[COMEDK UGET 2004]
- Two waves having intensity ratio 25 : 4 produce interference. The ratio of maximum to minima intensity is...[COMEDK UGET 2004]
- In Young's double slit experiment,1st dark fringe occurs directly...[COMEDK UGET 2009]
- In the diffraction pattern due to a single slit linear width of the central max...[COMEDK UGET 2007]
- In Newton ring experiment, monochromatic light is replaced by white light...[COMEDK UGET 2008]
- In diffraction through a single slit experiment, slit width is halved...[COMEDK UGET 2004]
- Which of the following is false for interference of light?...[JKCET 2012]
- Which of the following is true for the minimum angular separation of two stars...[JKCET 2012]
- Values for Brewster's angle can be...[JKCET 2015]
- Unpolarized light falls on two polarizing sheets placed one on top of other….[JKCET 2013]
- In Young's double slit experiment using monochromatic light of wavelength \(\lambda\) ….[AMUEEE 2018]
- In a single-slit diffraction experiment, the width of the slit is reduced by...[JKCET 2012]
- the wavelength of light illuminating the slits is….[JKCET 2013]
- Colours in thin films are due to….[JKCET 2008]
Sample Questions
Ques. What are ‘coherent sources’ which are required to produce interference pattern in Young’s double slit experiment? (Delhi 2014c)
Ans. Coherent sources are those two monochromatic sources which produce light waves having the same frequency and zero or a constant phase difference. If there are two sources, S1 and S2 will be coherent sources if their frequencies are the same and the phase difference between them is zero or constant.
Ques. State the effect on the interference fringes in Young’s double slit experiment due to each of the following operations? Justify your answers.
(a) the screen is moved away from the plane of the slits
(b) the separation of slits is increased.
(c) the source slit is moved closer to the plane of the double slit. (CBSE 2020)
Ans. (a) Fringe width is \(\beta = \frac{\lambda D}{d}\) ,
Therefore, D is increased as the fringe width increases.
(b) Here λ decreases, ∴ β decreases thus when d is increased the fringe width decreases.
(c) Let us assume, s is the size of the source and S denotes the distant of the source from the plane of the two slits, then for interference fringes, the condition is
\(\frac{s}{S} < \frac{\lambda}{D}\)
The fringe disappears because
\(\frac{s}{S} < \frac{\lambda}{D}\)
Ques. (i) Write two points to differentiate between interference and diffraction fringes.
(ii) In Young’s double-slit experiment, fringes are obtained on a screen placed a few distance away from the slits. If the screen is moved towards the slits by 5 cm, the fringe width changes by 30 mm. Given that the slits are 1 mm apart, calculate the wavelength of the light used. (CBSE 2018)
Ans. (i) The two points are as follows:
- The interference pattern possesses a number of equally speed dark and bright bands. On the other hand, the diffraction pattern has a central maxima and is twice wide as other maxima. Away from the centre intensity of maxima falls.
- Diffraction is obtained by superposing two waves that originate from each point of a slit. While, the interference pattern is obtained by superposing two waves that originate from two narrow slits.
(ii) Δβ = 30μm = 30 x 10−6m, d = 1mm = 10−3m, ΔD = 5cm = 5×10−2 m
\(\beta = \frac{\lambda D}{d}\)
or \(\Delta \beta = \frac{\lambda \Delta}{Dd}\) or λ = \(\frac{\Delta \beta d}{\Delta D}\)
λ = \(\frac{30 \times 10^{−6} \times 10^{−3 }}{5 \times 10^{−2}}\)
λ = 6 × 10−7 m
λ = 6000Å
Ques. Draw the intensity pattern for single slit diffraction and double slit interference. And write two differences between interference and diffraction patterns. (CBSE 2017)
Ans.
Differences between interference and diffraction patterns are,
(i) Interference fringes have the same intensity while diffraction fringes have different intensities.
(ii) Fringe width is of the same size in case of interference of light, while it is not the same in diffraction.
Ques. (a) Two monochromatic waves emanating from two coherent sources possess the displacements represented by y1 = a coswt and y2 = a cos(wt+Φ)
Where, Φ is the phase difference between the two waves. Show that the resultant intensity at the point due to their superposition is represented by given by I = 4Iocos2 Φ/2, where Io = a2
(b) Hence Obtain the conditions for constructive and destructive interference. (All India 2014c)
Ans. Given,
Ques. (a) Describe any two characteristic features that differentiate between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double slit experiment.
(b) Due to a single slit experiment in the diffraction, the aperture of the slit is 3 mm. If monochromatic light having the wavelength of 620 nm is incident normally on the slit. Calculate the separation between the first order minimum and third order maximum on one side of the screen. The distance between the slit and screen is 1.5 m. (CBSE 2019)
Ans. (a) The difference between the interference and diffraction phenomena
| Interference | Diffraction |
|---|---|
| i) The phenomenon of redistribution of light energy in a medium due to superposition of light waves from two similar sources is called interference of light. | i) The phenomenon of bending of lights around corners of obstacles in a path of light for which light penetrates into the geometrical shadow of an obstacle is called diffraction of light. |
| ii) All bright fringes possess equal intensity. | ii) Intensity of successive bright fringes decreases. |
| iii) The intensity of all dark fringes is zero. | iii) Intensity of dark fringes is not zero. |
| iv) Fringes that are obtained from interference of monochromatic light have equal width. | iv) The fringes that are obtained from diffraction of monochromatic light do not have equal width. |
The intensity at a point of the interference pattern in Young’s double slit experiment- considering the following figure where S is a monochromatic source of light. Again, A and B are two narrow slits which generate two coherent sources of light P which is any point on the screen.
Here, a and b are amplitudes of two waves and Φ is the phase difference. Based on the principle of superposition, the resultant displacement is equal to the vector sum of the displacement of individual waves.
Therefore, y = y1 + y2 = a sin ωt + b sin (ωt + Φ)
By using trigonometry identity: sin(a+b) = sina cosb + cosa sinb
we get, y = a sin ωt + B sin ωt Cos Φ + b sin Φ Cos ωt
Substituting,
a + b Cos Φ = R Cos θ
and b Sin Φ R Sin θ
therefore,
y = Sin ωt R Cos θ + Cos ωt R Sin θ
= R( Sin ωt Cos θ + Cos ωt Sin θ)
= R Sin (ωt + θ)
(b) Aperture of slit = 3 mm,
Wavelength of incident light, λ = 620 nm,
Distance between screen and slit, D = 1.5m,
For minimum diffraction, a sinθ. n λ
Small θ = n λ
For first order minima, n=1,
For third-order maxima, n=3,
Substituting the values of λ, D and a
Therefore, the separation between first order minima and third order maxima is 0.775 mm.
Ques. (i) In Young’s double slit experiment, deduce the condition (a) constructive, and (b) destructive interference at a point on screen. Draw a graph that shows variation of intensity in the interference pattern against the position ‘x’ on screen.
(ii) By pointing out three distinguishing features, compare the interference pattern observed in Young’s double slit experiment with a single slit diffraction pattern. (CBSE 2016)
Ans. (i) Let S1 and S2 be the two slits which are separated by a distance d.
GG’ is the screen which is at a distance D from S1 and S2.
The point C is equidistant from both the slits.
The light intensity will be maximum at this point as the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays appearing from the slits is given by,
S1 = S2P - S1P
Now, S1 S2= d, S2 F = D
For Constructive interference, the path difference is an integral multiple of wavelengths, i.e., path difference = nλ
(ii) Three distinguishing features are as follows:-
- In the interference pattern, each bright fringes have the same intensity. While, in the diffraction pattern, the bright fringes are not of same intensity.
- In interference pattern, the dark fringes are of zero or small intensity for which bright and dark fringes can be easily distinguishable. Again, in the diffraction, all the dark fringes are not of zero intensity.
- The width of almost all fringes are the same in interference pattern but in diffraction they are of different widths.
Ques. (a) Explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally by using Huygen’s construction of secondary wavelets.
(b) Show that angular width of the first diffraction fringe is half that of the central fringe.
(c) Explain why the maxima at becomes weaker with increasing n. (CBSE 2015)
Ans. (a) A single narrow slit is elucidated by a monochromatic source of light. The diffraction pattern is formed on the screen placed in front of the slits. There is a central bright region which is called central maximum where all the waves reaching are in phase hence, the intensity is maximum. There are alternate dark and bright regions on both sides of the central maximum and the intensity becomes weaker away from the centre. The intensity at any point P on the screen depends on the path difference between the waves generated from different parts of the wave-front at the slit.
According to the figure, the path difference (BP - AP) between the two edges of the slit can be calculated
Angle θ is zero at the point C (central point) on the screen for which all path differences are zero and hence all the parts of slit are in phase.
For this, the intensity at point C is maximum. If this is path difference, then P will be the point of minimum intensity as the whole wave-front can be considered to be divided into two equal halves, CA and CB. If the path difference between the secondary waves from A and B is λ, then the path difference between the secondary waves from A and C reaching P will be λ/2. Also for all the points in upper half AC, there is a corresponding point in the lower half CB due to which the path difference between the secondary waves reaching P is λ/2. Therefore destructive interference takes place at point P and thus P is a point of first secondary minimum.
(b) 
(c) Part of the slit contributing towards maximum decreases as n increases. This is the reason why maxima gets weaker with increasing n.
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